OP Malhotra Class 12 Maths Solutions Chapter 6 Matrices Exercise 6 (H)

S Chand Class 12 ICSE Maths Solutions Chapter 6 Matrices Ex 6(h)

Question 1. Show that A is a singular matrix if
(i) \( A = \left[\begin{array}{ll} 3 & 6 \\ 2 & 4 \end{array}\right]\)
(ii) \( A = \left[\begin{array}{rrr} 1 & 1 & 3 \\ 2 & 2 & 6 \\ 2 & -3 & 1 \end{array}\right]\)
Answer:
(i) Given \( A = \left[\begin{array}{ll} 3 & 6 \\ 2 & 4 \end{array}\right]\)
First, find the determinant of A:
\( |A| = \left|\begin{array}{ll} 3 & 6 \\ 2 & 4 \end{array}\right] = (3 \times 4) - (6 \times 2) \)
\( \implies |A| = 12 - 12 = 0 \)
Since the determinant \( |A| = 0 \), matrix A is a singular matrix. A singular matrix does not have an inverse.
(ii) Given \( A = \left[\begin{array}{rrr} 1 & 1 & 3 \\ 2 & 2 & 6 \\ 2 & -3 & 1 \end{array}\right]\)
Now, find the determinant of A by expanding along the first row (R1):
\( |A| = 1 \times \left|\begin{array}{rr} 2 & 6 \\ -3 & 1 \end{array}\right] - 1 \times \left|\begin{array}{rr} 2 & 6 \\ 2 & 1 \end{array}\right] + 3 \times \left|\begin{array}{rr} 2 & 2 \\ 2 & -3 \end{array}\right] \)
\( \implies |A| = 1((2 \times 1) - (6 \times -3)) - 1((2 \times 1) - (6 \times 2)) + 3((2 \times -3) - (2 \times 2)) \)
\( \implies |A| = 1(2 + 18) - 1(2 - 12) + 3(-6 - 4) \)
\( \implies |A| = 1(20) - 1(-10) + 3(-10) \)
\( \implies |A| = 20 + 10 - 30 \)
\( \implies |A| = 0 \)
Since the determinant \( |A| = 0 \), matrix A is a singular matrix. This means it cannot be inverted.
In simple words: To show a matrix is singular, you need to calculate its determinant. If the determinant is zero, the matrix is singular. We calculated the determinant for both matrices, and in both cases, it came out to be zero, proving they are singular.

🎯 Exam Tip: Remember that a matrix is singular if and only if its determinant is zero. For a \(2 \times 2\) matrix, this is straightforward; for a \(3 \times 3\) matrix, choose any row or column to expand the determinant calculation for easier computation.

Question 2. Find x if \( \left[\begin{array}{rrr} 8 & -6 & 2 \\ -6 & 7 & -4 \\ 2 & -4 & x \end{array}\right] \) is a singular matrix.
Answer:
Let \( A = \left[\begin{array}{rrr} 8 & -6 & 2 \\ -6 & 7 & -4 \\ 2 & -4 & x \end{array}\right]\)
Since matrix A is a singular matrix, its determinant must be equal to 0.
\( |A| = 0 \)
Now, calculate the determinant of A by expanding along the first row (R1) and set it to zero:
\( 8 \times \left|\begin{array}{rr} 7 & -4 \\ -4 & x \end{array}\right] - (-6) \times \left|\begin{array}{rr} -6 & -4 \\ 2 & x \end{array}\right] + 2 \times \left|\begin{array}{rr} -6 & 7 \\ 2 & -4 \end{array}\right] = 0 \)
\( \implies 8((7 \times x) - (-4 \times -4)) + 6((-6 \times x) - (-4 \times 2)) + 2((-6 \times -4) - (7 \times 2)) = 0 \)
\( \implies 8(7x - 16) + 6(-6x + 8) + 2(24 - 14) = 0 \)
\( \implies 56x - 128 - 36x + 48 + 48 - 28 = 0 \)
Now, group the x terms and the constant terms together:
\( \implies (56x - 36x) + (-128 + 48 + 48 - 28) = 0 \)
\( \implies 20x - 60 = 0 \)
Add 60 to both sides:
\( \implies 20x = 60 \)
Divide by 20 to find x:
\( \implies x = \frac{60}{20} \)
\( \implies x = 3 \)
In simple words: Because a singular matrix always has a determinant of zero, we set the determinant of the given matrix to zero. Then, we solved the resulting equation to find the value of x, which came out to be 3.

🎯 Exam Tip: When finding an unknown variable in a singular matrix, the key is to correctly compute the determinant expression and then solve the linear equation. Double-check all multiplications and subtractions to avoid calculation errors.

Question 3. For each of the following matrices, determine wheter the inverse exists. If it exists, find it.
(i) \( \left[\begin{array}{ll} 4 & 5 \\ 2 & 3 \end{array}\right]\)
(ii) \( \left[\begin{array}{rr} 2 & -6 \\ -1 & 3 \end{array}\right]\)
(iii) \( \left[\begin{array}{ll} 2 & 6 \\ 1 & 1 \end{array}\right]\)
(iv) \( \left[\begin{array}{rr} \cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{array}\right]\)
Answer:
(i) Let \( A = \left[\begin{array}{ll} 4 & 5 \\ 2 & 3 \end{array}\right]\)
First, find the determinant of A:
\( |A| = (4 \times 3) - (5 \times 2) \)
\( \implies |A| = 12 - 10 = 2 \)
Since \( |A| = 2 \neq 0 \), the inverse of A exists. A non-zero determinant means the matrix is invertible.
To find the inverse, we need the adjoint of A. First, find the cofactors:
Cofactor of \( a_{11} \) (4) is 3.
Cofactor of \( a_{12} \) (5) is \(-2\).
Cofactor of \( a_{21} \) (2) is \(-5\).
Cofactor of \( a_{22} \) (3) is 4.
The matrix of cofactors is \( C = \left[\begin{array}{rr} 3 & -2 \\ -5 & 4 \end{array}\right]\)
The adjoint of A (adj A) is the transpose of the cofactor matrix:
\( \text{adj } A = C^T = \left[\begin{array}{rr} 3 & -5 \\ -2 & 4 \end{array}\right]\)
Now, find the inverse using the formula \( A^{-1} = \frac{1}{|A|} \text{adj } A \):
\( A^{-1} = \frac{1}{2}\left[\begin{array}{rr} 3 & -5 \\ -2 & 4 \end{array}\right]\)
(ii) Let \( A = \left[\begin{array}{rr} 2 & -6 \\ -1 & 3 \end{array}\right]\)
First, find the determinant of A:
\( |A| = (2 \times 3) - (-6 \times -1) \)
\( \implies |A| = 6 - 6 = 0 \)
Since \( |A| = 0 \), the inverse of A does not exist. This matrix is singular.
(iii) Let \( A = \left[\begin{array}{ll} 2 & 6 \\ 1 & 1 \end{array}\right]\)
First, find the determinant of A:
\( |A| = (2 \times 1) - (6 \times 1) \)
\( \implies |A| = 2 - 6 = -4 \)
Since \( |A| = -4 \neq 0 \), the inverse of A exists.
To find the inverse, we need the adjoint of A. First, find the cofactors:
Cofactor of \( a_{11} \) (2) is 1.
Cofactor of \( a_{12} \) (6) is \(-1\).
Cofactor of \( a_{21} \) (1) is \(-6\).
Cofactor of \( a_{22} \) (1) is 2.
The matrix of cofactors is \( C = \left[\begin{array}{rr} 1 & -1 \\ -6 & 2 \end{array}\right]\)
The adjoint of A (adj A) is the transpose of the cofactor matrix:
\( \text{adj } A = C^T = \left[\begin{array}{rr} 1 & -6 \\ -1 & 2 \end{array}\right]\)
Now, find the inverse using the formula \( A^{-1} = \frac{1}{|A|} \text{adj } A \):
\( A^{-1} = \frac{1}{-4}\left[\begin{array}{rr} 1 & -6 \\ -1 & 2 \end{array}\right] \)
We can also write this as:
\( A^{-1} = -\frac{1}{4}\left[\begin{array}{rr} 1 & -6 \\ -1 & 2 \end{array}\right] = \frac{1}{4}\left[\begin{array}{rr} -1 & 6 \\ 1 & -2 \end{array}\right]\)
(iv) Let \( A = \left[\begin{array}{rr} \cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{array}\right]\)
First, find the determinant of A:
\( |A| = (\cos \theta \times \cos \theta) - (-\sin \theta \times \sin \theta) \)
\( \implies |A| = \cos^2 \theta - (-\sin^2 \theta) \)
\( \implies |A| = \cos^2 \theta + \sin^2 \theta \)
Using the trigonometric identity \( \cos^2 \theta + \sin^2 \theta = 1 \):
\( \implies |A| = 1 \)
Since \( |A| = 1 \neq 0 \), the inverse of A exists.
To find the inverse, we need the adjoint of A. First, find the cofactors:
Cofactor of \( a_{11} \) \( (\cos \theta) \) is \( \cos \theta \).
Cofactor of \( a_{12} \) \( (-\sin \theta) \) is \( -(\sin \theta) = -\sin \theta \).
Cofactor of \( a_{21} \) \( (\sin \theta) \) is \( -(-\sin \theta) = \sin \theta \).
Cofactor of \( a_{22} \) \( (\cos \theta) \) is \( \cos \theta \).
The matrix of cofactors is \( C = \left[\begin{array}{rr} \cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{array}\right]\)
The adjoint of A (adj A) is the transpose of the cofactor matrix:
\( \text{adj } A = C^T = \left[\begin{array}{rr} \cos \theta & \sin \theta \\ -\sin \theta & \cos \theta \end{array}\right]\)
Now, find the inverse using the formula \( A^{-1} = \frac{1}{|A|} \text{adj } A \):
\( A^{-1} = \frac{1}{1}\left[\begin{array}{rr} \cos \theta & \sin \theta \\ -\sin \theta & \cos \theta \end{array}\right]\)
\( \implies A^{-1} = \left[\begin{array}{rr} \cos \theta & \sin \theta \\ -\sin \theta & \cos \theta \end{array}\right]\)
In simple words: An inverse of a matrix can only exist if its determinant is not zero. We calculated the determinant for each matrix. If it was zero, no inverse exists. If it was not zero, we then found the cofactors and transposed them to get the adjoint matrix. Finally, we divided the adjoint by the determinant to get the inverse.

🎯 Exam Tip: Always calculate the determinant first. If it is zero, you can immediately state that the inverse does not exist, saving time. For trigonometric matrices, remember your basic identities like \( \cos^2 \theta + \sin^2 \theta = 1 \).

Question 4.
(a) If \( A = \left[\begin{array}{ll} 2 & x \\ 4 & 2 \end{array}\right], x \neq 1 \), calculate
(i) \( A^2 \)
(ii) \( (A^2)^{-1} \)
(b) Find the sum of the matrix \( \left[\begin{array}{ll} 2 & 3 \\ 5 & 7 \end{array}\right] \) and its multiplicative inverse.
Answer:
(a) Given \( A = \left[\begin{array}{ll} 2 & x \\ 4 & 2 \end{array}\right]\)
(i) To calculate \( A^2 \), multiply A by A:
\( A^2 = A \cdot A = \left[\begin{array}{ll} 2 & x \\ 4 & 2 \end{array}\right]\left[\begin{array}{ll} 2 & x \\ 4 & 2 \end{array}\right]\)
\( \implies A^2 = \left[\begin{array}{cc} (2 \times 2) + (x \times 4) & (2 \times x) + (x \times 2) \\ (4 \times 2) + (2 \times 4) & (4 \times x) + (2 \times 2) \end{array}\right]\)
\( \implies A^2 = \left[\begin{array}{cc} 4+4x & 2x+2x \\ 8+8 & 4x+4 \end{array}\right]\)
\( \implies A^2 = \left[\begin{array}{cc} 4+4x & 4x \\ 16 & 4x+4 \end{array}\right]\)
This is the matrix \( A^2 \).
(ii) To calculate \( (A^2)^{-1} \), first let \( B = A^2 \). So, \( B = \left[\begin{array}{cc} 4+4x & 4x \\ 16 & 4x+4 \end{array}\right]\)
Find the determinant of B:
\( |B| = ((4+4x)(4x+4)) - (4x \times 16) \)
\( \implies |B| = (4+4x)^2 - 64x \)
\( \implies |B| = (16+32x+16x^2) - 64x \)
\( \implies |B| = 16 - 32x + 16x^2 \)
\( \implies |B| = 16(1 - 2x + x^2) \)
\( \implies |B| = 16(1 - x)^2 \)
Since it is given that \( x \neq 1 \), then \( (1 - x)^2 \neq 0 \). So, \( |B| = 16(1 - x)^2 \neq 0 \).
Thus, the inverse of \( A^2 \) (which is B) exists.
Now, find the cofactors of B:
Cofactor of \( b_{11} \) \( (4+4x) \) is \( 4x+4 \).
Cofactor of \( b_{12} \) \( (4x) \) is \( -16 \).
Cofactor of \( b_{21} \) \( (16) \) is \( -4x \).
Cofactor of \( b_{22} \) \( (4x+4) \) is \( 4+4x \).
The matrix of cofactors is \( C_B = \left[\begin{array}{cc} 4x+4 & -16 \\ -4x & 4x+4 \end{array}\right]\)
The adjoint of B (adj B) is the transpose of \( C_B \):
\( \text{adj } B = C_B^T = \left[\begin{array}{cc} 4x+4 & -4x \\ -16 & 4x+4 \end{array}\right]\)
Finally, \( (A^2)^{-1} = B^{-1} = \frac{1}{|B|} \text{adj } B \):
\( (A^2)^{-1} = \frac{1}{16(x-1)^2}\left[\begin{array}{cc} 4x+4 & -4x \\ -16 & 4x+4 \end{array}\right]\)
(b) Let \( A = \left[\begin{array}{ll} 2 & 3 \\ 5 & 7 \end{array}\right]\)
First, find the determinant of A:
\( |A| = (2 \times 7) - (3 \times 5) \)
\( \implies |A| = 14 - 15 = -1 \)
Since \( |A| = -1 \neq 0 \), the inverse of A exists.
Now, find the cofactors of A:
Cofactor of \( a_{11} \) (2) is 7.
Cofactor of \( a_{12} \) (3) is \(-5\).
Cofactor of \( a_{21} \) (5) is \(-3\).
Cofactor of \( a_{22} \) (7) is 2.
The matrix of cofactors is \( C = \left[\begin{array}{rr} 7 & -5 \\ -3 & 2 \end{array}\right]\)
The adjoint of A (adj A) is the transpose of C:
\( \text{adj } A = C^T = \left[\begin{array}{rr} 7 & -3 \\ -5 & 2 \end{array}\right]\)
Now, calculate \( A^{-1} = \frac{1}{|A|} \text{adj } A \):
\( A^{-1} = \frac{1}{-1}\left[\begin{array}{rr} 7 & -3 \\ -5 & 2 \end{array}\right] \)
\( \implies A^{-1} = \left[\begin{array}{rr} -7 & 3 \\ 5 & -2 \end{array}\right]\)
Finally, find the sum of A and \( A^{-1} \):
\( A + A^{-1} = \left[\begin{array}{ll} 2 & 3 \\ 5 & 7 \end{array}\right] + \left[\begin{array}{rr} -7 & 3 \\ 5 & -2 \end{array}\right]\)
\( \implies A + A^{-1} = \left[\begin{array}{cc} 2+(-7) & 3+3 \\ 5+5 & 7+(-2) \end{array}\right]\)
\( \implies A + A^{-1} = \left[\begin{array}{cc} 2-7 & 6 \\ 10 & 7-2 \end{array}\right]\)
\( \implies A + A^{-1} = \left[\begin{array}{cc} -5 & 6 \\ 10 & 5 \end{array}\right]\)
In simple words: For part (a), we first multiplied matrix A by itself to get \( A^2 \). Then, to find the inverse of \( A^2 \), we calculated its determinant, found its cofactors, and then its adjoint, and finally divided the adjoint by the determinant. For part (b), we first found the inverse of matrix A. After getting the inverse, we added it to the original matrix A to find their sum.

🎯 Exam Tip: When multiplying matrices, ensure you are correctly multiplying rows by columns. For inverses, carefully compute each cofactor and remember to transpose the cofactor matrix to get the adjoint. Small arithmetic errors are common in these steps.

Question 5.
(i) Find a matrix X for which \( X\left[\begin{array}{rr} 3 & 2 \\ 1 & -1 \end{array}\right]=\left[\begin{array}{ll} 4 & 1 \\ 2 & 3 \end{array}\right] \). Also find the inverse of \( \left[\begin{array}{ll} 4 & 1 \\ 2 & 3 \end{array}\right] \).
(ii) Find 2 x 2 matrix M such that \( \left[\begin{array}{rr} -5 & 2 \\ 15 & -7 \end{array}\right] M=\left[\begin{array}{ll} 5 & 0 \\ 0 & 5 \end{array}\right] \).
Answer:
(i) Let the given equation be \( X A = B \), where \( A = \left[\begin{array}{rr} 3 & 2 \\ 1 & -1 \end{array}\right] \) and \( B = \left[\begin{array}{ll} 4 & 1 \\ 2 & 3 \end{array}\right] \).
To find X, we can multiply both sides of \( XA = B \) by \( A^{-1} \) from the right: \( X = B A^{-1} \).
First, find the inverse of A. Calculate the determinant of A:
\( |A| = (3 \times -1) - (2 \times 1) \)
\( \implies |A| = -3 - 2 = -5 \)
Since \( |A| = -5 \neq 0 \), \( A^{-1} \) exists.
Find the cofactors of A:
Cofactor of \( a_{11} \) (3) is \(-1\).
Cofactor of \( a_{12} \) (2) is \(-1\).
Cofactor of \( a_{21} \) (1) is \(-2\).
Cofactor of \( a_{22} \) (-1) is 3.
The matrix of cofactors is \( C_A = \left[\begin{array}{rr} -1 & -1 \\ -2 & 3 \end{array}\right]\)
The adjoint of A (adj A) is the transpose of \( C_A \):
\( \text{adj } A = C_A^T = \left[\begin{array}{rr} -1 & -2 \\ -1 & 3 \end{array}\right]\)
Now, calculate \( A^{-1} = \frac{1}{|A|} \text{adj } A \):
\( A^{-1} = \frac{1}{-5}\left[\begin{array}{rr} -1 & -2 \\ -1 & 3 \end{array}\right] = \frac{1}{5}\left[\begin{array}{rr} 1 & 2 \\ 1 & -3 \end{array}\right]\)
Now, find X using \( X = B A^{-1} \):
\( X = \left[\begin{array}{ll} 4 & 1 \\ 2 & 3 \end{array}\right] \frac{1}{5}\left[\begin{array}{rr} 1 & 2 \\ 1 & -3 \end{array}\right]\)
\( \implies X = \frac{1}{5}\left[\begin{array}{ll} (4 \times 1) + (1 \times 1) & (4 \times 2) + (1 \times -3) \\ (2 \times 1) + (3 \times 1) & (2 \times 2) + (3 \times -3) \end{array}\right]\)
\( \implies X = \frac{1}{5}\left[\begin{array}{cc} 4+1 & 8-3 \\ 2+3 & 4-9 \end{array}\right]\)
\( \implies X = \frac{1}{5}\left[\begin{array}{rr} 5 & 5 \\ 5 & -5 \end{array}\right]\)
\( \implies X = \left[\begin{array}{rr} 1 & 1 \\ 1 & -1 \end{array}\right]\)

Next, find the inverse of the matrix \( \left[\begin{array}{ll} 4 & 1 \\ 2 & 3 \end{array}\right] \). Let this matrix be D.
\( D = \left[\begin{array}{ll} 4 & 1 \\ 2 & 3 \end{array}\right]\)
Calculate the determinant of D:
\( |D| = (4 \times 3) - (1 \times 2) \)
\( \implies |D| = 12 - 2 = 10 \)
Since \( |D| = 10 \neq 0 \), \( D^{-1} \) exists.
Find the cofactors of D:
Cofactor of \( d_{11} \) (4) is 3.
Cofactor of \( d_{12} \) (1) is \(-2\).
Cofactor of \( d_{21} \) (2) is \(-1\).
Cofactor of \( d_{22} \) (3) is 4.
The matrix of cofactors is \( C_D = \left[\begin{array}{rr} 3 & -2 \\ -1 & 4 \end{array}\right]\)
The adjoint of D (adj D) is the transpose of \( C_D \):
\( \text{adj } D = C_D^T = \left[\begin{array}{rr} 3 & -1 \\ -2 & 4 \end{array}\right]\)
Now, calculate \( D^{-1} = \frac{1}{|D|} \text{adj } D \):
\( D^{-1} = \frac{1}{10}\left[\begin{array}{rr} 3 & -1 \\ -2 & 4 \end{array}\right]\)

(ii) Let the given equation be \( A M = B \), where \( A = \left[\begin{array}{rr} -5 & 2 \\ 15 & -7 \end{array}\right] \) and \( B = \left[\begin{array}{ll} 5 & 0 \\ 0 & 5 \end{array}\right] \).
To find M, we can multiply both sides of \( AM = B \) by \( A^{-1} \) from the left: \( M = A^{-1} B \).
First, find the inverse of A. Calculate the determinant of A:
\( |A| = (-5 \times -7) - (2 \times 15) \)
\( \implies |A| = 35 - 30 = 5 \)
Since \( |A| = 5 \neq 0 \), \( A^{-1} \) exists.
Find the cofactors of A:
Cofactor of \( a_{11} \) (-5) is \(-7\).
Cofactor of \( a_{12} \) (2) is \(-15\).
Cofactor of \( a_{21} \) (15) is \(-2\).
Cofactor of \( a_{22} \) (-7) is \(-5\).
The matrix of cofactors is \( C_A = \left[\begin{array}{rr} -7 & -15 \\ -2 & -5 \end{array}\right]\)
The adjoint of A (adj A) is the transpose of \( C_A \):
\( \text{adj } A = C_A^T = \left[\begin{array}{rr} -7 & -2 \\ -15 & -5 \end{array}\right]\)
Now, calculate \( A^{-1} = \frac{1}{|A|} \text{adj } A \):
\( A^{-1} = \frac{1}{5}\left[\begin{array}{rr} -7 & -2 \\ -15 & -5 \end{array}\right]\)
Now, find M using \( M = A^{-1} B \):
\( M = \frac{1}{5}\left[\begin{array}{rr} -7 & -2 \\ -15 & -5 \end{array}\right]\left[\begin{array}{ll} 5 & 0 \\ 0 & 5 \end{array}\right]\)
\( \implies M = \frac{1}{5}\left[\begin{array}{ll} (-7 \times 5) + (-2 \times 0) & (-7 \times 0) + (-2 \times 5) \\ (-15 \times 5) + (-5 \times 0) & (-15 \times 0) + (-5 \times 5) \end{array}\right]\)
\( \implies M = \frac{1}{5}\left[\begin{array}{rr} -35+0 & 0-10 \\ -75+0 & 0-25 \end{array}\right]\)
\( \implies M = \frac{1}{5}\left[\begin{array}{rr} -35 & -10 \\ -75 & -25 \end{array}\right]\)
Divide each element by 5:
\( \implies M = \left[\begin{array}{rr} -7 & -2 \\ -15 & -5 \end{array}\right]\)
In simple words: For both parts, the main idea is to use matrix inverses to solve for an unknown matrix. We found the inverse of the matrix next to the unknown one, then multiplied it with the other side of the equation. Remember to multiply the inverse on the correct side (left or right) of the equation.

🎯 Exam Tip: Pay close attention to the order of matrix multiplication, as matrix multiplication is not commutative (AB is generally not BA). If you have \( XA=B \), then \( X=BA^{-1} \); if you have \( AX=B \), then \( X=A^{-1}B \).

Question 6. Find the inverse of the following matrices and verify your result.
(i) \( \left[\begin{array}{rr} 2 & 5 \\ -3 & 1 \end{array}\right]\)
(ii) \( \left[\begin{array}{rr} -4 & 3 \\ 5 & -5 \end{array}\right]\)
Answer:
(i) Let \( A = \left[\begin{array}{rr} 2 & 5 \\ -3 & 1 \end{array}\right]\)
First, find the determinant of A:
\( |A| = (2 \times 1) - (5 \times -3) \)
\( \implies |A| = 2 - (-15) \)
\( \implies |A| = 2 + 15 = 17 \)
Since \( |A| = 17 \neq 0 \), the inverse exists.
Find the cofactors of A:
Cofactor of \( a_{11} \) (2) is 1.
Cofactor of \( a_{12} \) (5) is \(-(-3)) = 3\).
Cofactor of \( a_{21} \) (-3) is \(-5\).
Cofactor of \( a_{22} \) (1) is 2.
The matrix of cofactors is \( C = \left[\begin{array}{rr} 1 & 3 \\ -5 & 2 \end{array}\right]\)
The adjoint of A (adj A) is the transpose of C:
\( \text{adj } A = C^T = \left[\begin{array}{rr} 1 & -5 \\ 3 & 2 \end{array}\right]\)
Now, calculate \( A^{-1} = \frac{1}{|A|} \text{adj } A \):
\( A^{-1} = \frac{1}{17}\left[\begin{array}{rr} 1 & -5 \\ 3 & 2 \end{array}\right]\)

Verification:
To verify, we need to show that \( A^{-1}A = I \) and \( AA^{-1} = I \).
\( A^{-1}A = \frac{1}{17}\left[\begin{array}{rr} 1 & -5 \\ 3 & 2 \end{array}\right]\left[\begin{array}{rr} 2 & 5 \\ -3 & 1 \end{array}\right]\)
\( \implies A^{-1}A = \frac{1}{17}\left[\begin{array}{cc} (1 \times 2) + (-5 \times -3) & (1 \times 5) + (-5 \times 1) \\ (3 \times 2) + (2 \times -3) & (3 \times 5) + (2 \times 1) \end{array}\right]\)
\( \implies A^{-1}A = \frac{1}{17}\left[\begin{array}{cc} 2+15 & 5-5 \\ 6-6 & 15+2 \end{array}\right]\)
\( \implies A^{-1}A = \frac{1}{17}\left[\begin{array}{rr} 17 & 0 \\ 0 & 17 \end{array}\right]\)
\( \implies A^{-1}A = \left[\begin{array}{rr} 1 & 0 \\ 0 & 1 \end{array}\right] = I \)

\( AA^{-1} = \left[\begin{array}{rr} 2 & 5 \\ -3 & 1 \end{array}\right]\frac{1}{17}\left[\begin{array}{rr} 1 & -5 \\ 3 & 2 \end{array}\right]\)
\( \implies AA^{-1} = \frac{1}{17}\left[\begin{array}{cc} (2 \times 1) + (5 \times 3) & (2 \times -5) + (5 \times 2) \\ (-3 \times 1) + (1 \times 3) & (-3 \times -5) + (1 \times 2) \end{array}\right]\)
\( \implies AA^{-1} = \frac{1}{17}\left[\begin{array}{cc} 2+15 & -10+10 \\ -3+3 & 15+2 \end{array}\right]\)
\( \implies AA^{-1} = \frac{1}{17}\left[\begin{array}{rr} 17 & 0 \\ 0 & 17 \end{array}\right]\)
\( \implies AA^{-1} = \left[\begin{array}{rr} 1 & 0 \\ 0 & 1 \end{array}\right] = I \)
Since \( A^{-1}A = AA^{-1} = I \), the inverse is verified.

(ii) Let \( A = \left[\begin{array}{rr} -4 & 3 \\ 5 & -5 \end{array}\right]\)
First, find the determinant of A:
\( |A| = (-4 \times -5) - (3 \times 5) \)
\( \implies |A| = 20 - 15 = 5 \)
Since \( |A| = 5 \neq 0 \), the inverse exists.
Find the cofactors of A:
Cofactor of \( a_{11} \) (-4) is \(-5\).
Cofactor of \( a_{12} \) (3) is \(-(5) = -5\).
Cofactor of \( a_{21} \) (5) is \(-(3) = -3\).
Cofactor of \( a_{22} \) (-5) is \(-4\).
The matrix of cofactors is \( C = \left[\begin{array}{rr} -5 & -5 \\ -3 & -4 \end{array}\right]\)
The adjoint of A (adj A) is the transpose of C:
\( \text{adj } A = C^T = \left[\begin{array}{rr} -5 & -3 \\ -5 & -4 \end{array}\right]\)
Now, calculate \( A^{-1} = \frac{1}{|A|} \text{adj } A \):
\( A^{-1} = \frac{1}{5}\left[\begin{array}{rr} -5 & -3 \\ -5 & -4 \end{array}\right]\)

Verification:
To verify, we need to show that \( A^{-1}A = I \) and \( AA^{-1} = I \).
\( A^{-1}A = \frac{1}{5}\left[\begin{array}{rr} -5 & -3 \\ -5 & -4 \end{array}\right]\left[\begin{array}{rr} -4 & 3 \\ 5 & -5 \end{array}\right]\)
\( \implies A^{-1}A = \frac{1}{5}\left[\begin{array}{cc} (-5 \times -4) + (-3 \times 5) & (-5 \times 3) + (-3 \times -5) \\ (-5 \times -4) + (-4 \times 5) & (-5 \times 3) + (-4 \times -5) \end{array}\right]\)
\( \implies A^{-1}A = \frac{1}{5}\left[\begin{array}{cc} 20-15 & -15+15 \\ 20-20 & -15+20 \end{array}\right]\)
\( \implies A^{-1}A = \frac{1}{5}\left[\begin{array}{rr} 5 & 0 \\ 0 & 5 \end{array}\right]\)
\( \implies A^{-1}A = \left[\begin{array}{rr} 1 & 0 \\ 0 & 1 \end{array}\right] = I \)

\( AA^{-1} = \left[\begin{array}{rr} -4 & 3 \\ 5 & -5 \end{array}\right]\frac{1}{5}\left[\begin{array}{rr} -5 & -3 \\ -5 & -4 \end{array}\right]\)
\( \implies AA^{-1} = \frac{1}{5}\left[\begin{array}{cc} (-4 \times -5) + (3 \times -5) & (-4 \times -3) + (3 \times -4) \\ (5 \times -5) + (-5 \times -5) & (5 \times -3) + (-5 \times -4) \end{array}\right]\)
\( \implies AA^{-1} = \frac{1}{5}\left[\begin{array}{cc} 20-15 & 12-12 \\ -25+25 & -15+20 \end{array}\right]\)
\( \implies AA^{-1} = \frac{1}{5}\left[\begin{array}{rr} 5 & 0 \\ 0 & 5 \end{array}\right]\)
\( \implies AA^{-1} = \left[\begin{array}{rr} 1 & 0 \\ 0 & 1 \end{array}\right] = I \)
Since \( A^{-1}A = AA^{-1} = I \), the inverse is verified.
In simple words: For each matrix, we first calculated its determinant to check if the inverse exists. If it did, we found the cofactors and then the adjoint. The inverse was then found by dividing the adjoint by the determinant. To verify, we multiplied the original matrix by its inverse in both orders, making sure the result was the identity matrix (I).

🎯 Exam Tip: Verification is crucial for inverse matrix problems. It confirms your calculations are correct and ensures you get full marks. The product of a matrix and its inverse must always yield the identity matrix, in both multiplication orders.

Question 7.
(a) Find the inverse of each of the following matrices and verify your result.
(i) \( \left[\begin{array}{rrr} 1 & 4 & 3 \\ 2 & 5 & 4 \\ 1 & -3 & -2 \end{array}\right]\)
(ii) \( \left[\begin{array}{ccr} 1 & 0 & 0 \\ 0 & \cos \alpha & \sin \alpha \\ 0 & \sin \alpha & -\cos \alpha \end{array}\right]\)
(b) Verify that \( AA^{-1} = A^{-1}A = I \), if \( A = \left[\begin{array}{rrr} 1 & 2 & -2 \\ -1 & 3 & 0 \\ 0 & -2 & 1 \end{array}\right] \).
(c) If \( A = \left[\begin{array}{rrr} 3 & 0 & 2 \\ 1 & 5 & 9 \\ -6 & 4 & 7 \end{array}\right] \) and \( AB = BA = I \), find B.
Answer:
(a) (i) Let \( A = \left[\begin{array}{rrr} 1 & 4 & 3 \\ 2 & 5 & 4 \\ 1 & -3 & -2 \end{array}\right]\)
First, find the determinant of A by expanding along R1:
\( |A| = 1 \times \left|\begin{array}{rr} 5 & 4 \\ -3 & -2 \end{array}\right] - 4 \times \left|\begin{array}{rr} 2 & 4 \\ 1 & -2 \end{array}\right] + 3 \times \left|\begin{array}{rr} 2 & 5 \\ 1 & -3 \end{array}\right] \)
\( \implies |A| = 1((5 \times -2) - (4 \times -3)) - 4((2 \times -2) - (4 \times 1)) + 3((2 \times -3) - (5 \times 1)) \)
\( \implies |A| = 1(-10 + 12) - 4(-4 - 4) + 3(-6 - 5) \)
\( \implies |A| = 1(2) - 4(-8) + 3(-11) \)
\( \implies |A| = 2 + 32 - 33 \)
\( \implies |A| = 1 \)
Since \( |A| = 1 \neq 0 \), the inverse of A exists.
Now, find the cofactors of A:
Cofactor of \( a_{11} \) is \( (5 \times -2) - (4 \times -3) = -10 + 12 = 2 \).
Cofactor of \( a_{12} \) is \( -((2 \times -2) - (4 \times 1)) = -(-4 - 4) = -(-8) = 8 \).
Cofactor of \( a_{13} \) is \( (2 \times -3) - (5 \times 1) = -6 - 5 = -11 \).
Cofactor of \( a_{21} \) is \( -((4 \times -2) - (3 \times -3)) = -(-8 + 9) = -(1) = -1 \).
Cofactor of \( a_{22} \) is \( (1 \times -2) - (3 \times 1) = -2 - 3 = -5 \).
Cofactor of \( a_{23} \) is \( -((1 \times -3) - (4 \times 1)) = -(-3 - 4) = -(-7) = 7 \).
Cofactor of \( a_{31} \) is \( (4 \times 4) - (3 \times 5) = 16 - 15 = 1 \).
Cofactor of \( a_{32} \) is \( -((1 \times 4) - (3 \times 2)) = -(4 - 6) = -(-2) = 2 \).
Cofactor of \( a_{33} \) is \( (1 \times 5) - (4 \times 2) = 5 - 8 = -3 \).
The matrix of cofactors is \( C = \left[\begin{array}{rrr} 2 & 8 & -11 \\ -1 & -5 & 7 \\ 1 & 2 & -3 \end{array}\right]\)
The adjoint of A (adj A) is the transpose of C:
\( \text{adj } A = C^T = \left[\begin{array}{rrr} 2 & -1 & 1 \\ 8 & -5 & 2 \\ -11 & 7 & -3 \end{array}\right]\)
Now, calculate \( A^{-1} = \frac{1}{|A|} \text{adj } A \):
\( A^{-1} = \frac{1}{1}\left[\begin{array}{rrr} 2 & -1 & 1 \\ 8 & -5 & 2 \\ -11 & 7 & -3 \end{array}\right] = \left[\begin{array}{rrr} 2 & -1 & 1 \\ 8 & -5 & 2 \\ -11 & 7 & -3 \end{array}\right]\)

Verification:
\( A^{-1}A = \left[\begin{array}{rrr} 2 & -1 & 1 \\ 8 & -5 & 2 \\ -11 & 7 & -3 \end{array}\right]\left[\begin{array}{rrr} 1 & 4 & 3 \\ 2 & 5 & 4 \\ 1 & -3 & -2 \end{array}\right]\)
\( \implies A^{-1}A = \left[\begin{array}{ccc} (2-2+1) & (8-5-3) & (6-4-2) \\ (8-10+2) & (32-25-6) & (24-20-4) \\ (-11+14-3) & (-44+35+9) & (-33+28+6) \end{array}\right]\)
\( \implies A^{-1}A = \left[\begin{array}{rrr} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right] = I \)

\( AA^{-1} = \left[\begin{array}{rrr} 1 & 4 & 3 \\ 2 & 5 & 4 \\ 1 & -3 & -2 \end{array}\right]\left[\begin{array}{rrr} 2 & -1 & 1 \\ 8 & -5 & 2 \\ -11 & 7 & -3 \end{array}\right]\)
\( \implies AA^{-1} = \left[\begin{array}{ccc} (2+32-33) & (-1-20+21) & (1+8-9) \\ (4+40-44) & (-2-25+28) & (2+10-12) \\ (2-24+22) & (-1-15+14) & (1-6+6) \end{array}\right]\)
\( \implies AA^{-1} = \left[\begin{array}{rrr} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right] = I \)
The inverse is verified.

(ii) Let \( A = \left[\begin{array}{ccr} 1 & 0 & 0 \\ 0 & \cos \alpha & \sin \alpha \\ 0 & \sin \alpha & -\cos \alpha \end{array}\right]\)
First, find the determinant of A by expanding along R1:
\( |A| = 1 \times \left|\begin{array}{rr} \cos \alpha & \sin \alpha \\ \sin \alpha & -\cos \alpha \end{array}\right] - 0 + 0 \)
\( \implies |A| = (\cos \alpha \times -\cos \alpha) - (\sin \alpha \times \sin \alpha) \)
\( \implies |A| = -\cos^2 \alpha - \sin^2 \alpha \)
\( \implies |A| = -(\cos^2 \alpha + \sin^2 \alpha) \)
Using the identity \( \cos^2 \alpha + \sin^2 \alpha = 1 \):
\( \implies |A| = -1 \)
Since \( |A| = -1 \neq 0 \), the inverse of A exists.
Now, find the cofactors of A:
Cofactor of \( a_{11} \) is \( -\cos^2 \alpha - \sin^2 \alpha = -1 \).
Cofactor of \( a_{12} \) is \( -(0 \times -\cos \alpha - 0 \times \sin \alpha) = 0 \).
Cofactor of \( a_{13} \) is \( (0 \times \sin \alpha - 0 \times \cos \alpha) = 0 \).
Cofactor of \( a_{21} \) is \( -(0 \times -\cos \alpha - 0 \times \sin \alpha) = 0 \).
Cofactor of \( a_{22} \) is \( (1 \times -\cos \alpha - 0 \times 0) = -\cos \alpha \).
Cofactor of \( a_{23} \) is \( -(1 \times \sin \alpha - 0 \times 0) = -\sin \alpha \).
Cofactor of \( a_{31} \) is \( (0 \times \sin \alpha - 0 \times \cos \alpha) = 0 \).
Cofactor of \( a_{32} \) is \( -(1 \times \sin \alpha - 0 \times 0) = -\sin \alpha \).
Cofactor of \( a_{33} \) is \( (1 \times \cos \alpha - 0 \times 0) = \cos \alpha \).
The matrix of cofactors is \( C = \left[\begin{array}{ccc} -1 & 0 & 0 \\ 0 & -\cos \alpha & -\sin \alpha \\ 0 & -\sin \alpha & \cos \alpha \end{array}\right]\)
The adjoint of A (adj A) is the transpose of C:
\( \text{adj } A = C^T = \left[\begin{array}{ccc} -1 & 0 & 0 \\ 0 & -\cos \alpha & -\sin \alpha \\ 0 & -\sin \alpha & \cos \alpha \end{array}\right]\)
Now, calculate \( A^{-1} = \frac{1}{|A|} \text{adj } A \):
\( A^{-1} = \frac{1}{-1}\left[\begin{array}{ccc} -1 & 0 & 0 \\ 0 & -\cos \alpha & -\sin \alpha \\ 0 & -\sin \alpha & \cos \alpha \end{array}\right]\)
\( \implies A^{-1} = \left[\begin{array}{ccc} 1 & 0 & 0 \\ 0 & \cos \alpha & \sin \alpha \\ 0 & \sin \alpha & -\cos \alpha \end{array}\right]\)

Verification:
\( A^{-1}A = \left[\begin{array}{ccc} 1 & 0 & 0 \\ 0 & \cos \alpha & \sin \alpha \\ 0 & \sin \alpha & -\cos \alpha \end{array}\right]\left[\begin{array}{ccr} 1 & 0 & 0 \\ 0 & \cos \alpha & \sin \alpha \\ 0 & \sin \alpha & -\cos \alpha \end{array}\right]\)
\( \implies A^{-1}A = \left[\begin{array}{ccc} 1 & 0 & 0 \\ 0 & \cos^2 \alpha + \sin^2 \alpha & \cos \alpha \sin \alpha - \sin \alpha \cos \alpha \\ 0 & \sin \alpha \cos \alpha - \cos \alpha \sin \alpha & \sin^2 \alpha + \cos^2 \alpha \end{array}\right]\)
\( \implies A^{-1}A = \left[\begin{array}{rrr} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right] = I \)

\( AA^{-1} = \left[\begin{array}{ccr} 1 & 0 & 0 \\ 0 & \cos \alpha & \sin \alpha \\ 0 & \sin \alpha & -\cos \alpha \end{array}\right]\left[\begin{array}{ccc} 1 & 0 & 0 \\ 0 & \cos \alpha & \sin \alpha \\ 0 & \sin \alpha & -\cos \alpha \end{array}\right]\)
\( \implies AA^{-1} = \left[\begin{array}{ccc} 1 & 0 & 0 \\ 0 & \cos^2 \alpha + \sin^2 \alpha & \cos \alpha \sin \alpha - \sin \alpha \cos \alpha \\ 0 & \sin \alpha \cos \alpha - \cos \alpha \sin \alpha & \sin^2 \alpha + \cos^2 \alpha \end{array}\right]\)
\( \implies AA^{-1} = \left[\begin{array}{rrr} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right] = I \)
The inverse is verified.

(b) Let \( A = \left[\begin{array}{rrr} 1 & 2 & -2 \\ -1 & 3 & 0 \\ 0 & -2 & 1 \end{array}\right]\)
First, find the determinant of A by expanding along C1:
\( |A| = 1 \times \left|\begin{array}{rr} 3 & 0 \\ -2 & 1 \end{array}\right] - (-1) \times \left|\begin{array}{rr} 2 & -2 \\ -2 & 1 \end{array}\right] + 0 \)
\( \implies |A| = 1((3 \times 1) - (0 \times -2)) + 1((2 \times 1) - (-2 \times -2)) \)
\( \implies |A| = 1(3 - 0) + 1(2 - 4) \)
\( \implies |A| = 3 + (-2) \)
\( \implies |A| = 1 \)
Since \( |A| = 1 \neq 0 \), the inverse of A exists.
Now, find the cofactors of A:
Cofactor of \( a_{11} \) is \( (3 \times 1) - (0 \times -2) = 3 \).
Cofactor of \( a_{12} \) is \( -((-1 \times 1) - (0 \times 0)) = -(-1) = 1 \).
Cofactor of \( a_{13} \) is \( (-1 \times -2) - (3 \times 0) = 2 - 0 = 2 \).
Cofactor of \( a_{21} \) is \( -((2 \times 1) - (-2 \times -2)) = -(2 - 4) = -(-2) = 2 \).
Cofactor of \( a_{22} \) is \( (1 \times 1) - (-2 \times 0) = 1 - 0 = 1 \).
Cofactor of \( a_{23} \) is \( -((1 \times -2) - (2 \times 0)) = -(-2 - 0) = -(-2) = 2 \).
Cofactor of \( a_{31} \) is \( (2 \times 0) - (-2 \times 3) = 0 - (-6) = 6 \).
Cofactor of \( a_{32} \) is \( -((1 \times 0) - (-2 \times -1)) = -(0 - 2) = -(-2) = 2 \).
Cofactor of \( a_{33} \) is \( (1 \times 3) - (2 \times -1) = 3 - (-2) = 5 \).
The matrix of cofactors is \( C = \left[\begin{array}{rrr} 3 & 1 & 2 \\ 2 & 1 & 2 \\ 6 & 2 & 5 \end{array}\right]\)
The adjoint of A (adj A) is the transpose of C:
\( \text{adj } A = C^T = \left[\begin{array}{rrr} 3 & 2 & 6 \\ 1 & 1 & 2 \\ 2 & 2 & 5 \end{array}\right]\)
Now, calculate \( A^{-1} = \frac{1}{|A|} \text{adj } A \):
\( A^{-1} = \frac{1}{1}\left[\begin{array}{rrr} 3 & 2 & 6 \\ 1 & 1 & 2 \\ 2 & 2 & 5 \end{array}\right] = \left[\begin{array}{rrr} 3 & 2 & 6 \\ 1 & 1 & 2 \\ 2 & 2 & 5 \end{array}\right]\)

Verification:
\( A^{-1}A = \left[\begin{array}{rrr} 3 & 2 & 6 \\ 1 & 1 & 2 \\ 2 & 2 & 5 \end{array}\right]\left[\begin{array}{rrr} 1 & 2 & -2 \\ -1 & 3 & 0 \\ 0 & -2 & 1 \end{array}\right]\)
\( \implies A^{-1}A = \left[\begin{array}{ccc} (3-2+0) & (6+6-12) & (-6+0+6) \\ (1-1+0) & (2+3-4) & (-2+0+2) \\ (2-2+0) & (4+6-10) & (-4+0+5) \end{array}\right]\)
\( \implies A^{-1}A = \left[\begin{array}{rrr} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right] = I \)

\( AA^{-1} = \left[\begin{array}{rrr} 1 & 2 & -2 \\ -1 & 3 & 0 \\ 0 & -2 & 1 \end{array}\right]\left[\begin{array}{rrr} 3 & 2 & 6 \\ 1 & 1 & 2 \\ 2 & 2 & 5 \end{array}\right]\)
\( \implies AA^{-1} = \left[\begin{array}{ccc} (3+2-4) & (2+2-4) & (6+4-10) \\ (-3+3+0) & (-2+3+0) & (-6+6+0) \\ (0-2+2) & (0-2+2) & (0-4+5) \end{array}\right]\)
\( \implies AA^{-1} = \left[\begin{array}{rrr} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right] = I \)
The verification is complete.

(c) If \( A = \left[\begin{array}{rrr} 3 & 0 & 2 \\ 1 & 5 & 9 \\ -6 & 4 & 7 \end{array}\right] \) and \( AB = BA = I \), find B.
Given \( AB = BA = I \), this means that B is the inverse of A (i.e., \( B = A^{-1} \)).
First, find the determinant of A by expanding along R1:
\( |A| = 3 \times \left|\begin{array}{rr} 5 & 9 \\ 4 & 7 \end{array}\right] - 0 \times \left|\begin{array}{rr} 1 & 9 \\ -6 & 7 \end{array}\right] + 2 \times \left|\begin{array}{rr} 1 & 5 \\ -6 & 4 \end{array}\right] \)
\( \implies |A| = 3((5 \times 7) - (9 \times 4)) - 0 + 2((1 \times 4) - (5 \times -6)) \)
\( \implies |A| = 3(35 - 36) + 2(4 - (-30)) \)
\( \implies |A| = 3(-1) + 2(4 + 30) \)
\( \implies |A| = -3 + 2(34) \)
\( \implies |A| = -3 + 68 = 65 \)
Since \( |A| = 65 \neq 0 \), \( A^{-1} \) (and thus B) exists.
Now, find the cofactors of A:
Cofactor of \( a_{11} \) is \( (5 \times 7) - (9 \times 4) = 35 - 36 = -1 \).
Cofactor of \( a_{12} \) is \( -((1 \times 7) - (9 \times -6)) = -(7 + 54) = -61 \).
Cofactor of \( a_{13} \) is \( (1 \times 4) - (5 \times -6) = 4 - (-30) = 34 \).
Cofactor of \( a_{21} \) is \( -((0 \times 7) - (2 \times 4)) = -(0 - 8) = 8 \).
Cofactor of \( a_{22} \) is \( (3 \times 7) - (2 \times -6) = 21 - (-12) = 33 \).
Cofactor of \( a_{23} \) is \( -((3 \times 4) - (0 \times -6)) = -(12 - 0) = -12 \).
Cofactor of \( a_{31} \) is \( (0 \times 9) - (2 \times 5) = 0 - 10 = -10 \).
Cofactor of \( a_{32} \) is \( -((3 \times 9) - (2 \times 1)) = -(27 - 2) = -25 \).
Cofactor of \( a_{33} \) is \( (3 \times 5) - (0 \times 1) = 15 - 0 = 15 \).
The matrix of cofactors is \( C = \left[\begin{array}{rrr} -1 & -61 & 34 \\ 8 & 33 & -12 \\ -10 & -25 & 15 \end{array}\right]\)
The adjoint of A (adj A) is the transpose of C:
\( \text{adj } A = C^T = \left[\begin{array}{rrr} -1 & 8 & -10 \\ -61 & 33 & -25 \\ 34 & -12 & 15 \end{array}\right]\)
Now, calculate \( B = A^{-1} = \frac{1}{|A|} \text{adj } A \):
\( B = \frac{1}{65}\left[\begin{array}{rrr} -1 & 8 & -10 \\ -61 & 33 & -25 \\ 34 & -12 & 15 \end{array}\right]\)
In simple words: For (a), we found the inverse of both 3x3 matrices by calculating the determinant, then all cofactors, forming the adjoint, and finally dividing by the determinant. Verification ensures the product of the matrix and its inverse is the identity matrix. For (b), we followed the same steps to find the inverse of A and confirmed that multiplying A by its inverse gives the identity matrix. For (c), since \( AB = BA = I \), B must be the inverse of A, so we calculated \( A^{-1} \) using the standard method.

🎯 Exam Tip: For \(3 \times 3\) matrices, ensure careful calculation of all nine cofactors, remembering the alternating sign pattern. A single sign error can invalidate the entire inverse. For verification, always perform both \(A^{-1}A\) and \(AA^{-1}\) multiplications.

Question 7. (a) Find the inverse of each of the following matrices and verify your result.
(i) \( \begin{pmatrix} 1 & 4 & 3 \\ 2 & 5 & 4 \\ 1 & -3 & -2 \end{pmatrix} \)
(ii) \( \begin{pmatrix} 1 & 0 & 0 \\ 0 & \cos \alpha & \sin \alpha \\ 0 & \sin \alpha & -\cos \alpha \end{pmatrix} \)

Answer:
(ii) Let \( A = \begin{pmatrix} 1 & 0 & 0 \\ 0 & \cos \alpha & \sin \alpha \\ 0 & \sin \alpha & -\cos \alpha \end{pmatrix} \)
Now, let's find the determinant of A, denoted as \( |A| \). We will expand along the first row \( R_1 \):
\( |A| = 1 \begin{vmatrix} \cos \alpha & \sin \alpha \\ \sin \alpha & -\cos \alpha \end{vmatrix} - 0 \begin{vmatrix} 0 & \sin \alpha \\ 0 & -\cos \alpha \end{vmatrix} + 0 \begin{vmatrix} 0 & \cos \alpha \\ 0 & \sin \alpha \end{vmatrix} \)
\( |A| = 1 ( (\cos \alpha)(-\cos \alpha) - (\sin \alpha)(\sin \alpha) ) \)
\( |A| = -\cos^2 \alpha - \sin^2 \alpha \)
\( |A| = -(\cos^2 \alpha + \sin^2 \alpha) \)
Since \( \cos^2 \alpha + \sin^2 \alpha = 1 \), we have:
\( |A| = -1 \)
Because \( |A| = -1 \neq 0 \), the inverse of matrix A exists.
Next, we find the cofactors of each element in matrix A.
Cofactors of the first row \( R_1 \):
Cofactor of \( a_{11} = \begin{vmatrix} \cos \alpha & \sin \alpha \\ \sin \alpha & -\cos \alpha \end{vmatrix} = -\cos^2 \alpha - \sin^2 \alpha = -1 \)
Cofactor of \( a_{12} = - \begin{vmatrix} 0 & \sin \alpha \\ 0 & -\cos \alpha \end{vmatrix} = -(0 - 0) = 0 \)
Cofactor of \( a_{13} = \begin{vmatrix} 0 & \cos \alpha \\ 0 & \sin \alpha \end{vmatrix} = (0 - 0) = 0 \)
So, the cofactors of \( R_1 \) are \( -1, 0, 0 \).
Cofactors of the second row \( R_2 \):
Cofactor of \( a_{21} = - \begin{vmatrix} 0 & 0 \\ \sin \alpha & -\cos \alpha \end{vmatrix} = -(0 - 0) = 0 \)
Cofactor of \( a_{22} = \begin{vmatrix} 1 & 0 \\ 0 & -\cos \alpha \end{vmatrix} = (-\cos \alpha - 0) = -\cos \alpha \)
Cofactor of \( a_{23} = - \begin{vmatrix} 1 & 0 \\ 0 & \sin \alpha \end{vmatrix} = -(\sin \alpha - 0) = -\sin \alpha \)
So, the cofactors of \( R_2 \) are \( 0, -\cos \alpha, -\sin \alpha \).
Cofactors of the third row \( R_3 \):
Cofactor of \( a_{31} = \begin{vmatrix} 0 & 0 \\ \cos \alpha & \sin \alpha \end{vmatrix} = (0 - 0) = 0 \)
Cofactor of \( a_{32} = - \begin{vmatrix} 1 & 0 \\ 0 & \sin \alpha \end{vmatrix} = -(\sin \alpha - 0) = -\sin \alpha \)
Cofactor of \( a_{33} = \begin{vmatrix} 1 & 0 \\ 0 & \cos \alpha \end{vmatrix} = (\cos \alpha - 0) = \cos \alpha \)
So, the cofactors of \( R_3 \) are \( 0, -\sin \alpha, \cos \alpha \).
The adjoint of A (adj A) is the transpose of the cofactor matrix:
\[ \text{adj A} = \begin{pmatrix} -1 & 0 & 0 \\ 0 & -\cos \alpha & -\sin \alpha \\ 0 & -\sin \alpha & \cos \alpha \end{pmatrix}^T = \begin{pmatrix} -1 & 0 & 0 \\ 0 & -\cos \alpha & -\sin \alpha \\ 0 & -\sin \alpha & \cos \alpha \end{pmatrix} \] Now, we can find \( A^{-1} \) using the formula \( A^{-1} = \frac{1}{|A|} \text{adj A} \).
Since \( |A| = -1 \):
\[ A^{-1} = \frac{1}{-1} \begin{pmatrix} -1 & 0 & 0 \\ 0 & -\cos \alpha & -\sin \alpha \\ 0 & -\sin \alpha & \cos \alpha \end{pmatrix} = \begin{pmatrix} 1 & 0 & 0 \\ 0 & \cos \alpha & \sin \alpha \\ 0 & \sin \alpha & -\cos \alpha \end{pmatrix} \] Therefore, \( A^{-1} = \begin{pmatrix} 1 & 0 & 0 \\ 0 & \cos \alpha & \sin \alpha \\ 0 & \sin \alpha & -\cos \alpha \end{pmatrix} \). Notice that in this special case, the inverse matrix is the same as the original matrix A. **Verification:** To verify the result, we need to show that \( A^{-1}A = I \) and \( AA^{-1} = I \), where \( I \) is the identity matrix.
First, let's calculate \( A^{-1}A \):
\[ A^{-1}A = \begin{pmatrix} 1 & 0 & 0 \\ 0 & \cos \alpha & \sin \alpha \\ 0 & \sin \alpha & -\cos \alpha \end{pmatrix} \begin{pmatrix} 1 & 0 & 0 \\ 0 & \cos \alpha & \sin \alpha \\ 0 & \sin \alpha & -\cos \alpha \end{pmatrix} \] \[ = \begin{pmatrix} 1 \cdot 1 + 0 \cdot 0 + 0 \cdot 0 & 1 \cdot 0 + 0 \cdot \cos \alpha + 0 \cdot \sin \alpha & 1 \cdot 0 + 0 \cdot \sin \alpha + 0 \cdot (-\cos \alpha) \\ 0 \cdot 1 + \cos \alpha \cdot 0 + \sin \alpha \cdot 0 & 0 \cdot 0 + \cos \alpha \cdot \cos \alpha + \sin \alpha \cdot \sin \alpha & 0 \cdot 0 + \cos \alpha \cdot \sin \alpha + \sin \alpha \cdot (-\cos \alpha) \\ 0 \cdot 1 + \sin \alpha \cdot 0 + (-\cos \alpha) \cdot 0 & 0 \cdot 0 + \sin \alpha \cdot \cos \alpha + (-\cos \alpha) \cdot \sin \alpha & 0 \cdot 0 + \sin \alpha \cdot \sin \alpha + (-\cos \alpha) \cdot (-\cos \alpha) \end{pmatrix} \] \[ = \begin{pmatrix} 1 & 0 & 0 \\ 0 & \cos^2 \alpha + \sin^2 \alpha & \cos \alpha \sin \alpha - \sin \alpha \cos \alpha \\ 0 & \sin \alpha \cos \alpha - \cos \alpha \sin \alpha & \sin^2 \alpha + \cos^2 \alpha \end{pmatrix} \] \[ = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix} = I \] Now, let's calculate \( AA^{-1} \):
\[ AA^{-1} = \begin{pmatrix} 1 & 0 & 0 \\ 0 & \cos \alpha & \sin \alpha \\ 0 & \sin \alpha & -\cos \alpha \end{pmatrix} \begin{pmatrix} 1 & 0 & 0 \\ 0 & \cos \alpha & \sin \alpha \\ 0 & \sin \alpha & -\cos \alpha \end{pmatrix} \] Since \( A^{-1} \) is equal to \( A \) in this case, the multiplication \( AA^{-1} \) will be the same as \( A^2 \), which is also \( I \).
\[ AA^{-1} = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix} = I \] Both calculations result in the identity matrix \( I \), so the inverse is verified.
In simple words: First, we found the determinant of the matrix. Since it was not zero, we knew an inverse existed. Then, we found the cofactors of all numbers in the matrix and arranged them to get the adjoint matrix. Finally, we divided the adjoint by the determinant to get the inverse. We checked our answer by multiplying the original matrix with its inverse, and it correctly gave the identity matrix.

🎯 Exam Tip: Remember that for an inverse to exist, the determinant of the matrix must be non-zero. Also, for trigonometric matrices, always use identities like \( \sin^2 \alpha + \cos^2 \alpha = 1 \) to simplify results.

Question 7. (b) Verify that \( AA^{-1} = A^{-1}A = I \), if \( A = \left[\begin{array}{rrr} 1 & 2 & -2 \\ -1 & 3 & 0 \\ 0 & -2 & 1 \end{array}\right] \)
Answer:
Let \( A = \begin{pmatrix} 1 & 2 & -2 \\ -1 & 3 & 0 \\ 0 & -2 & 1 \end{pmatrix} \)
First, calculate the determinant of A, \( |A| \), by expanding along the first column \( C_1 \):
\( |A| = 1 \begin{vmatrix} 3 & 0 \\ -2 & 1 \end{vmatrix} - (-1) \begin{vmatrix} 2 & -2 \\ -2 & 1 \end{vmatrix} + 0 \begin{vmatrix} 2 & -2 \\ 3 & 0 \end{vmatrix} \)
\( |A| = 1((3)(1) - (0)(-2)) + 1((2)(1) - (-2)(-2)) + 0 \)
\( |A| = 1(3 - 0) + 1(2 - 4) \)
\( |A| = 3 + (-2) = 1 \)
Since \( |A| = 1 \neq 0 \), the inverse \( A^{-1} \) exists.
Next, we find the cofactors for each element of matrix A.
Cofactors of the first row \( R_1 \):
Cofactor of \( a_{11} = \begin{vmatrix} 3 & 0 \\ -2 & 1 \end{vmatrix} = 3 \)
Cofactor of \( a_{12} = - \begin{vmatrix} -1 & 0 \\ 0 & 1 \end{vmatrix} = -(-1) = 1 \)
Cofactor of \( a_{13} = \begin{vmatrix} -1 & 3 \\ 0 & -2 \end{vmatrix} = 2 \)
Cofactors of the second row \( R_2 \):
Cofactor of \( a_{21} = - \begin{vmatrix} 2 & -2 \\ -2 & 1 \end{vmatrix} = -(2 - 4) = 2 \)
Cofactor of \( a_{22} = \begin{vmatrix} 1 & -2 \\ 0 & 1 \end{vmatrix} = 1 \)
Cofactor of \( a_{23} = - \begin{vmatrix} 1 & 2 \\ 0 & -2 \end{vmatrix} = -(-2) = 2 \)
Cofactors of the third row \( R_3 \):
Cofactor of \( a_{31} = \begin{vmatrix} 2 & -2 \\ 3 & 0 \end{vmatrix} = 6 \)
Cofactor of \( a_{32} = - \begin{vmatrix} 1 & -2 \\ -1 & 0 \end{vmatrix} = -(0 - 2) = 2 \)
Cofactor of \( a_{33} = \begin{vmatrix} 1 & 2 \\ -1 & 3 \end{vmatrix} = 5 \)
Now, we form the adjoint of A, which is the transpose of the cofactor matrix:
\[ \text{adj A} = \begin{pmatrix} 3 & 1 & 2 \\ 2 & 1 & 2 \\ 6 & 2 & 5 \end{pmatrix}^T = \begin{pmatrix} 3 & 2 & 6 \\ 1 & 1 & 2 \\ 2 & 2 & 5 \end{pmatrix} \] Now, calculate \( A^{-1} = \frac{1}{|A|} \text{adj A} \). Since \( |A| = 1 \):
\[ A^{-1} = \frac{1}{1} \begin{pmatrix} 3 & 2 & 6 \\ 1 & 1 & 2 \\ 2 & 2 & 5 \end{pmatrix} = \begin{pmatrix} 3 & 2 & 6 \\ 1 & 1 & 2 \\ 2 & 2 & 5 \end{pmatrix} \] **Verification:** We need to verify that \( A^{-1}A = I \) and \( AA^{-1} = I \).
First, calculate \( A^{-1}A \):
\[ A^{-1}A = \begin{pmatrix} 3 & 2 & 6 \\ 1 & 1 & 2 \\ 2 & 2 & 5 \end{pmatrix} \begin{pmatrix} 1 & 2 & -2 \\ -1 & 3 & 0 \\ 0 & -2 & 1 \end{pmatrix} \] \[ = \begin{pmatrix} 3 \cdot 1 + 2 \cdot (-1) + 6 \cdot 0 & 3 \cdot 2 + 2 \cdot 3 + 6 \cdot (-2) & 3 \cdot (-2) + 2 \cdot 0 + 6 \cdot 1 \\ 1 \cdot 1 + 1 \cdot (-1) + 2 \cdot 0 & 1 \cdot 2 + 1 \cdot 3 + 2 \cdot (-2) & 1 \cdot (-2) + 1 \cdot 0 + 2 \cdot 1 \\ 2 \cdot 1 + 2 \cdot (-1) + 5 \cdot 0 & 2 \cdot 2 + 2 \cdot 3 + 5 \cdot (-2) & 2 \cdot (-2) + 2 \cdot 0 + 5 \cdot 1 \end{pmatrix} \] \[ = \begin{pmatrix} 3 - 2 + 0 & 6 + 6 - 12 & -6 + 0 + 6 \\ 1 - 1 + 0 & 2 + 3 - 4 & -2 + 0 + 2 \\ 2 - 2 + 0 & 4 + 6 - 10 & -4 + 0 + 5 \end{pmatrix} = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix} = I \] Next, calculate \( AA^{-1} \):
\[ AA^{-1} = \begin{pmatrix} 1 & 2 & -2 \\ -1 & 3 & 0 \\ 0 & -2 & 1 \end{pmatrix} \begin{pmatrix} 3 & 2 & 6 \\ 1 & 1 & 2 \\ 2 & 2 & 5 \end{pmatrix} \] \[ = \begin{pmatrix} 1 \cdot 3 + 2 \cdot 1 + (-2) \cdot 2 & 1 \cdot 2 + 2 \cdot 1 + (-2) \cdot 2 & 1 \cdot 6 + 2 \cdot 2 + (-2) \cdot 5 \\ (-1) \cdot 3 + 3 \cdot 1 + 0 \cdot 2 & (-1) \cdot 2 + 3 \cdot 1 + 0 \cdot 2 & (-1) \cdot 6 + 3 \cdot 2 + 0 \cdot 5 \\ 0 \cdot 3 + (-2) \cdot 1 + 1 \cdot 2 & 0 \cdot 2 + (-2) \cdot 1 + 1 \cdot 2 & 0 \cdot 6 + (-2) \cdot 2 + 1 \cdot 5 \end{pmatrix} \] \[ = \begin{pmatrix} 3 + 2 - 4 & 2 + 2 - 4 & 6 + 4 - 10 \\ -3 + 3 + 0 & -2 + 3 + 0 & -6 + 6 + 0 \\ 0 - 2 + 2 & 0 - 2 + 2 & 0 - 4 + 5 \end{pmatrix} = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix} = I \] Since both \( A^{-1}A = I \) and \( AA^{-1} = I \), the verification is complete.
In simple words: We first found the determinant of matrix A to confirm that its inverse exists. Then, we calculated the cofactor matrix and its transpose to get the adjoint matrix. Dividing the adjoint by the determinant gave us the inverse matrix. Finally, we multiplied the original matrix by its inverse (in both orders) and saw that both products resulted in the identity matrix, which proved our inverse was correct.

🎯 Exam Tip: When verifying an inverse, remember to check both \( A^{-1}A \) and \( AA^{-1} \) as a complete verification requires both multiplications to yield the identity matrix \( I \).

Question 7. (c) If \( A = \left[\begin{array}{rrr} 3 & 0 & 2 \\ 1 & 5 & 9 \\ -6 & 4 & 7 \end{array}\right] \) and \( AB = BA = I \), find B.
Answer:
Given that \( AB = BA = I \), this means that B is the inverse of A, i.e., \( B = A^{-1} \).
So, our task is to find the inverse of matrix A.
Let \( A = \begin{pmatrix} 3 & 0 & 2 \\ 1 & 5 & 9 \\ -6 & 4 & 7 \end{pmatrix} \)
First, calculate the determinant of A, \( |A| \), by expanding along the first row \( R_1 \):
\( |A| = 3 \begin{vmatrix} 5 & 9 \\ 4 & 7 \end{vmatrix} - 0 \begin{vmatrix} 1 & 9 \\ -6 & 7 \end{vmatrix} + 2 \begin{vmatrix} 1 & 5 \\ -6 & 4 \end{vmatrix} \)
\( |A| = 3((5)(7) - (9)(4)) - 0 + 2((1)(4) - (5)(-6)) \)
\( |A| = 3(35 - 36) + 2(4 - (-30)) \)
\( |A| = 3(-1) + 2(4 + 30) \)
\( |A| = -3 + 2(34) = -3 + 68 = 65 \)
Since \( |A| = 65 \neq 0 \), the inverse \( A^{-1} \) exists.
Next, find the cofactors for each element of matrix A.
Cofactors of the first row \( R_1 \):
Cofactor of \( a_{11} = \begin{vmatrix} 5 & 9 \\ 4 & 7 \end{vmatrix} = 35 - 36 = -1 \)
Cofactor of \( a_{12} = - \begin{vmatrix} 1 & 9 \\ -6 & 7 \end{vmatrix} = -(7 - (-54)) = -(7 + 54) = -61 \)
Cofactor of \( a_{13} = \begin{vmatrix} 1 & 5 \\ -6 & 4 \end{vmatrix} = 4 - (-30) = 4 + 30 = 34 \)
So, the cofactors of \( R_1 \) are \( -1, -61, 34 \).
Cofactors of the second row \( R_2 \):
Cofactor of \( a_{21} = - \begin{vmatrix} 0 & 2 \\ 4 & 7 \end{vmatrix} = -(0 - 8) = 8 \)
Cofactor of \( a_{22} = \begin{vmatrix} 3 & 2 \\ -6 & 7 \end{vmatrix} = 21 - (-12) = 21 + 12 = 33 \)
Cofactor of \( a_{23} = - \begin{vmatrix} 3 & 0 \\ -6 & 4 \end{vmatrix} = -(12 - 0) = -12 \)
So, the cofactors of \( R_2 \) are \( 8, 33, -12 \).
Cofactors of the third row \( R_3 \):
Cofactor of \( a_{31} = \begin{vmatrix} 0 & 2 \\ 5 & 9 \end{vmatrix} = 0 - 10 = -10 \)
Cofactor of \( a_{32} = - \begin{vmatrix} 3 & 2 \\ 1 & 9 \end{vmatrix} = -(27 - 2) = -25 \)
Cofactor of \( a_{33} = \begin{vmatrix} 3 & 0 \\ 1 & 5 \end{vmatrix} = 15 - 0 = 15 \)
So, the cofactors of \( R_3 \) are \( -10, -25, 15 \).
Now, we form the adjoint of A, which is the transpose of the cofactor matrix:
\[ \text{adj A} = \begin{pmatrix} -1 & -61 & 34 \\ 8 & 33 & -12 \\ -10 & -25 & 15 \end{pmatrix}^T = \begin{pmatrix} -1 & 8 & -10 \\ -61 & 33 & -25 \\ 34 & -12 & 15 \end{pmatrix} \] Finally, we calculate \( B = A^{-1} = \frac{1}{|A|} \text{adj A} \).
\[ B = \frac{1}{65} \begin{pmatrix} -1 & 8 & -10 \\ -61 & 33 & -25 \\ 34 & -12 & 15 \end{pmatrix} \]In simple words: When \( AB = BA = I \), it means matrix B is the inverse of matrix A. So, we needed to find the inverse of A. We started by calculating the determinant of A. Since it was not zero, we knew the inverse existed. Then, we found all the cofactors and arranged them into the adjoint matrix. Finally, we divided the adjoint matrix by the determinant to get B, which is the inverse of A.

🎯 Exam Tip: Recognise that \( AB = BA = I \) immediately implies B is the inverse of A. This saves time by directly computing \( A^{-1} \).

Question 8. Verify that \( (AB)^{-1} = B^{-1}A^{-1} \) for the matrices A and B where
(i) \( A = \left[\begin{array}{rr} 2 & 1 \\ 5 & 3 \end{array}\right] \) and \( B = \left[\begin{array}{rr} 4 & 5 \\ 3 & 4 \end{array}\right] \)
(ii) \( A = \left[\begin{array}{rr} 3 & 2 \\ 7 & 5 \end{array}\right] \) and \( B = \left[\begin{array}{rr} 4 & 6 \\ 3 & 2 \end{array}\right] \)

Answer:
(i) Given matrices are \( A = \begin{pmatrix} 2 & 1 \\ 5 & 3 \end{pmatrix} \) and \( B = \begin{pmatrix} 4 & 5 \\ 3 & 4 \end{pmatrix} \).
First, let's find \( A^{-1} \).
\( |A| = (2)(3) - (1)(5) = 6 - 5 = 1 \). Since \( |A| \neq 0 \), \( A^{-1} \) exists.
Cofactors of \( A \): \( \begin{pmatrix} 3 & -5 \\ -1 & 2 \end{pmatrix} \).
\( \text{adj A} = \begin{pmatrix} 3 & -5 \\ -1 & 2 \end{pmatrix}^T = \begin{pmatrix} 3 & -1 \\ -5 & 2 \end{pmatrix} \).
\( A^{-1} = \frac{1}{|A|} \text{adj A} = \frac{1}{1} \begin{pmatrix} 3 & -1 \\ -5 & 2 \end{pmatrix} = \begin{pmatrix} 3 & -1 \\ -5 & 2 \end{pmatrix} \).
Next, let's find \( B^{-1} \).
\( |B| = (4)(4) - (5)(3) = 16 - 15 = 1 \). Since \( |B| \neq 0 \), \( B^{-1} \) exists.
Cofactors of \( B \): \( \begin{pmatrix} 4 & -3 \\ -5 & 4 \end{pmatrix} \).
\( \text{adj B} = \begin{pmatrix} 4 & -3 \\ -5 & 4 \end{pmatrix}^T = \begin{pmatrix} 4 & -5 \\ -3 & 4 \end{pmatrix} \).
\( B^{-1} = \frac{1}{|B|} \text{adj B} = \frac{1}{1} \begin{pmatrix} 4 & -5 \\ -3 & 4 \end{pmatrix} = \begin{pmatrix} 4 & -5 \\ -3 & 4 \end{pmatrix} \).
Now, calculate \( B^{-1}A^{-1} \):
\[ B^{-1}A^{-1} = \begin{pmatrix} 4 & -5 \\ -3 & 4 \end{pmatrix} \begin{pmatrix} 3 & -1 \\ -5 & 2 \end{pmatrix} \] \[ = \begin{pmatrix} (4)(3) + (-5)(-5) & (4)(-1) + (-5)(2) \\ (-3)(3) + (4)(-5) & (-3)(-1) + (4)(2) \end{pmatrix} \] \[ = \begin{pmatrix} 12 + 25 & -4 - 10 \\ -9 - 20 & 3 + 8 \end{pmatrix} = \begin{pmatrix} 37 & -14 \\ -29 & 11 \end{pmatrix} \] Next, calculate \( AB \):
\[ AB = \begin{pmatrix} 2 & 1 \\ 5 & 3 \end{pmatrix} \begin{pmatrix} 4 & 5 \\ 3 & 4 \end{pmatrix} \] \[ = \begin{pmatrix} (2)(4) + (1)(3) & (2)(5) + (1)(4) \\ (5)(4) + (3)(3) & (5)(5) + (3)(4) \end{pmatrix} \] \[ = \begin{pmatrix} 8 + 3 & 10 + 4 \\ 20 + 9 & 25 + 12 \end{pmatrix} = \begin{pmatrix} 11 & 14 \\ 29 & 37 \end{pmatrix} \] Now, let's find \( (AB)^{-1} \).
\( |AB| = (11)(37) - (14)(29) = 407 - 406 = 1 \). Since \( |AB| \neq 0 \), \( (AB)^{-1} \) exists.
Cofactors of \( AB \): \( \begin{pmatrix} 37 & -29 \\ -14 & 11 \end{pmatrix} \).
\( \text{adj (AB)} = \begin{pmatrix} 37 & -29 \\ -14 & 11 \end{pmatrix}^T = \begin{pmatrix} 37 & -14 \\ -29 & 11 \end{pmatrix} \).
\( (AB)^{-1} = \frac{1}{|AB|} \text{adj (AB)} = \frac{1}{1} \begin{pmatrix} 37 & -14 \\ -29 & 11 \end{pmatrix} = \begin{pmatrix} 37 & -14 \\ -29 & 11 \end{pmatrix} \).
Comparing the results, \( B^{-1}A^{-1} = \begin{pmatrix} 37 & -14 \\ -29 & 11 \end{pmatrix} \) and \( (AB)^{-1} = \begin{pmatrix} 37 & -14 \\ -29 & 11 \end{pmatrix} \).
Thus, \( (AB)^{-1} = B^{-1}A^{-1} \) is verified for part (i).

(ii) Given matrices are \( A = \begin{pmatrix} 3 & 2 \\ 7 & 5 \end{pmatrix} \) and \( B = \begin{pmatrix} 4 & 6 \\ 3 & 2 \end{pmatrix} \).
First, let's find \( A^{-1} \).
\( |A| = (3)(5) - (2)(7) = 15 - 14 = 1 \). Since \( |A| \neq 0 \), \( A^{-1} \) exists.
Cofactors of \( A \): \( \begin{pmatrix} 5 & -7 \\ -2 & 3 \end{pmatrix} \).
\( \text{adj A} = \begin{pmatrix} 5 & -7 \\ -2 & 3 \end{pmatrix}^T = \begin{pmatrix} 5 & -2 \\ -7 & 3 \end{pmatrix} \).
\( A^{-1} = \frac{1}{|A|} \text{adj A} = \frac{1}{1} \begin{pmatrix} 5 & -2 \\ -7 & 3 \end{pmatrix} = \begin{pmatrix} 5 & -2 \\ -7 & 3 \end{pmatrix} \).
Next, let's find \( B^{-1} \).
\( |B| = (4)(2) - (6)(3) = 8 - 18 = -10 \). Since \( |B| \neq 0 \), \( B^{-1} \) exists.
Cofactors of \( B \): \( \begin{pmatrix} 2 & -3 \\ -6 & 4 \end{pmatrix} \).
\( \text{adj B} = \begin{pmatrix} 2 & -3 \\ -6 & 4 \end{pmatrix}^T = \begin{pmatrix} 2 & -6 \\ -3 & 4 \end{pmatrix} \).
\( B^{-1} = \frac{1}{|B|} \text{adj B} = \frac{1}{-10} \begin{pmatrix} 2 & -6 \\ -3 & 4 \end{pmatrix} = -\frac{1}{10} \begin{pmatrix} 2 & -6 \\ -3 & 4 \end{pmatrix} \).
Now, calculate \( B^{-1}A^{-1} \):
\[ B^{-1}A^{-1} = -\frac{1}{10} \begin{pmatrix} 2 & -6 \\ -3 & 4 \end{pmatrix} \begin{pmatrix} 5 & -2 \\ -7 & 3 \end{pmatrix} \] \[ = -\frac{1}{10} \begin{pmatrix} (2)(5) + (-6)(-7) & (2)(-2) + (-6)(3) \\ (-3)(5) + (4)(-7) & (-3)(-2) + (4)(3) \end{pmatrix} \] \[ = -\frac{1}{10} \begin{pmatrix} 10 + 42 & -4 - 18 \\ -15 - 28 & 6 + 12 \end{pmatrix} \] \[ = -\frac{1}{10} \begin{pmatrix} 52 & -22 \\ -43 & 18 \end{pmatrix} = \frac{1}{10} \begin{pmatrix} -52 & 22 \\ 43 & -18 \end{pmatrix} \] Next, calculate \( AB \):
\[ AB = \begin{pmatrix} 3 & 2 \\ 7 & 5 \end{pmatrix} \begin{pmatrix} 4 & 6 \\ 3 & 2 \end{pmatrix} \] \[ = \begin{pmatrix} (3)(4) + (2)(3) & (3)(6) + (2)(2) \\ (7)(4) + (5)(3) & (7)(6) + (5)(2) \end{pmatrix} \] \[ = \begin{pmatrix} 12 + 6 & 18 + 4 \\ 28 + 15 & 42 + 10 \end{pmatrix} = \begin{pmatrix} 18 & 22 \\ 43 & 52 \end{pmatrix} \] Now, let's find \( (AB)^{-1} \).
\( |AB| = (18)(52) - (22)(43) = 936 - 946 = -10 \). Since \( |AB| \neq 0 \), \( (AB)^{-1} \) exists.
Cofactors of \( AB \): \( \begin{pmatrix} 52 & -43 \\ -22 & 18 \end{pmatrix} \).
\( \text{adj (AB)} = \begin{pmatrix} 52 & -43 \\ -22 & 18 \end{pmatrix}^T = \begin{pmatrix} 52 & -22 \\ -43 & 18 \end{pmatrix} \).
\( (AB)^{-1} = \frac{1}{|AB|} \text{adj (AB)} = \frac{1}{-10} \begin{pmatrix} 52 & -22 \\ -43 & 18 \end{pmatrix} = \frac{1}{10} \begin{pmatrix} -52 & 22 \\ 43 & -18 \end{pmatrix} \).
Comparing the results, \( B^{-1}A^{-1} = \frac{1}{10} \begin{pmatrix} -52 & 22 \\ 43 & -18 \end{pmatrix} \) and \( (AB)^{-1} = \frac{1}{10} \begin{pmatrix} -52 & 22 \\ 43 & -18 \end{pmatrix} \).
Thus, \( (AB)^{-1} = B^{-1}A^{-1} \) is verified for part (ii).
In simple words: For both parts of the question, we first calculated the inverse of matrix A and matrix B separately. Then, we multiplied these inverses in reverse order to get \( B^{-1}A^{-1} \). After that, we first multiplied A and B to get the product matrix AB, and then found its inverse \( (AB)^{-1} \). In both cases, the results for \( (AB)^{-1} \) and \( B^{-1}A^{-1} \) matched, proving the property.

🎯 Exam Tip: This property, \( (AB)^{-1} = B^{-1}A^{-1} \), is crucial in matrix algebra. Always remember that the order of the inverses is reversed when taking the inverse of a product.

Question 9. Given \( A = \left[\begin{array}{rr} 2 & -3 \\ -4 & 7 \end{array}\right] \), compute \( A^{-1} \) and show that \( 2A^{-1} = 9I - A \).
Answer:
Given matrix \( A = \begin{pmatrix} 2 & -3 \\ -4 & 7 \end{pmatrix} \).
First, let's compute the determinant of A, \( |A| \):
\( |A| = (2)(7) - (-3)(-4) = 14 - 12 = 2 \).
Since \( |A| = 2 \neq 0 \), the inverse \( A^{-1} \) exists.
Next, find the cofactors of A:
Cofactor of \( a_{11} = 7 \)
Cofactor of \( a_{12} = -(-4) = 4 \)
Cofactor of \( a_{21} = -(-3) = 3 \)
Cofactor of \( a_{22} = 2 \)
The cofactor matrix is \( \begin{pmatrix} 7 & 4 \\ 3 & 2 \end{pmatrix} \).
The adjoint of A is the transpose of the cofactor matrix:
\[ \text{adj A} = \begin{pmatrix} 7 & 4 \\ 3 & 2 \end{pmatrix}^T = \begin{pmatrix} 7 & 3 \\ 4 & 2 \end{pmatrix} \] Now, calculate \( A^{-1} = \frac{1}{|A|} \text{adj A} \):
\[ A^{-1} = \frac{1}{2} \begin{pmatrix} 7 & 3 \\ 4 & 2 \end{pmatrix} = \begin{pmatrix} 7/2 & 3/2 \\ 4/2 & 2/2 \end{pmatrix} = \begin{pmatrix} 3.5 & 1.5 \\ 2 & 1 \end{pmatrix} \] Next, we need to show that \( 2A^{-1} = 9I - A \).
First, calculate \( 2A^{-1} \):
\[ 2A^{-1} = 2 \cdot \frac{1}{2} \begin{pmatrix} 7 & 3 \\ 4 & 2 \end{pmatrix} = \begin{pmatrix} 7 & 3 \\ 4 & 2 \end{pmatrix} \] Now, calculate \( 9I - A \), where \( I = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} \) is the identity matrix.
\[ 9I - A = 9 \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} - \begin{pmatrix} 2 & -3 \\ -4 & 7 \end{pmatrix} \] \[ = \begin{pmatrix} 9 & 0 \\ 0 & 9 \end{pmatrix} - \begin{pmatrix} 2 & -3 \\ -4 & 7 \end{pmatrix} \] \[ = \begin{pmatrix} 9 - 2 & 0 - (-3) \\ 0 - (-4) & 9 - 7 \end{pmatrix} = \begin{pmatrix} 7 & 3 \\ 4 & 2 \end{pmatrix} \] Comparing the results, \( 2A^{-1} = \begin{pmatrix} 7 & 3 \\ 4 & 2 \end{pmatrix} \) and \( 9I - A = \begin{pmatrix} 7 & 3 \\ 4 & 2 \end{pmatrix} \).
Since both expressions yield the same matrix, \( 2A^{-1} = 9I - A \) is verified.
In simple words: We first found the inverse of matrix A by calculating its determinant and adjoint. Then, we multiplied this inverse by 2. Separately, we calculated 9 times the identity matrix minus matrix A. When we compared both results, they were exactly the same, which proved the given relationship. This identity helps link a matrix, its inverse, and the identity matrix in a useful way.

🎯 Exam Tip: When asked to verify an identity involving matrices, compute both sides of the equation separately and show that they are equal. Be careful with scalar multiplication and matrix subtraction.

Question 10.
(i) Given \( A = \left[\begin{array}{rrr} 2 & 4 & -1 \\ -1 & 0 & 2 \end{array}\right] \), \( B = \left[\begin{array}{rr} 3 & 4 \\ -1 & 2 \\ 2 & 1 \end{array}\right] \), find \( (AB)^{-1} \).
(ii) If \( A = \left[\begin{array}{III} 5 & 0 & 4 \\ 2 & 3 & 2 \\ 1 & 2 & 1 \end{array}\right] \), \( B^{-1} = \left[\begin{array}{III} 1 & 3 & 3 \\ 1 & 4 & 3 \\ 1 & 3 & 4 \end{array}\right] \), find \( (AB)^{-1} \).
(iii) If \( A^{-1} = \frac{1}{10}\left[\begin{array}{rrr} 10 & -10 & 2 \\ 0 & 5 & -4 \\ 0 & 0 & 2 \end{array}\right] \), \( B = \left[\begin{array}{rrr} 1 & -1 & 2 \\ 0 & 2 & -3 \\ 3 & -2 & 4 \end{array}\right] \), find \( (BA)^{-1} \).
Answer:
(i) Given matrices are \( A = \begin{pmatrix} 2 & 4 & -1 \\ -1 & 0 & 2 \end{pmatrix} \) and \( B = \begin{pmatrix} 3 & 4 \\ -1 & 2 \\ 2 & 1 \end{pmatrix} \).
First, calculate the product \( AB \):
\[ AB = \begin{pmatrix} 2 & 4 & -1 \\ -1 & 0 & 2 \end{pmatrix} \begin{pmatrix} 3 & 4 \\ -1 & 2 \\ 2 & 1 \end{pmatrix} \] \[ = \begin{pmatrix} (2)(3) + (4)(-1) + (-1)(2) & (2)(4) + (4)(2) + (-1)(1) \\ (-1)(3) + (0)(-1) + (2)(2) & (-1)(4) + (0)(2) + (2)(1) \end{pmatrix} \] \[ = \begin{pmatrix} 6 - 4 - 2 & 8 + 8 - 1 \\ -3 + 0 + 4 & -4 + 0 + 2 \end{pmatrix} = \begin{pmatrix} 0 & 15 \\ 1 & -2 \end{pmatrix} \] Let \( C = AB = \begin{pmatrix} 0 & 15 \\ 1 & -2 \end{pmatrix} \). Now we need to find \( C^{-1} \).
Calculate the determinant of C, \( |C| \):
\( |C| = (0)(-2) - (15)(1) = 0 - 15 = -15 \).
Since \( |C| = -15 \neq 0 \), the inverse \( (AB)^{-1} \) exists.
Next, find the cofactors of C:
Cofactor of \( c_{11} = -2 \)
Cofactor of \( c_{12} = -1 \)
Cofactor of \( c_{21} = -15 \)
Cofactor of \( c_{22} = 0 \)
The adjoint of C is the transpose of the cofactor matrix:
\[ \text{adj C} = \begin{pmatrix} -2 & -1 \\ -15 & 0 \end{pmatrix}^T = \begin{pmatrix} -2 & -15 \\ -1 & 0 \end{pmatrix} \] Finally, calculate \( (AB)^{-1} = C^{-1} = \frac{1}{|C|} \text{adj C} \):
\[ (AB)^{-1} = \frac{1}{-15} \begin{pmatrix} -2 & -15 \\ -1 & 0 \end{pmatrix} = \frac{1}{15} \begin{pmatrix} 2 & 15 \\ 1 & 0 \end{pmatrix} \]
(ii) Given \( A = \begin{pmatrix} 5 & 0 & 4 \\ 2 & 3 & 2 \\ 1 & 2 & 1 \end{pmatrix} \) and \( B^{-1} = \begin{pmatrix} 1 & 3 & 3 \\ 1 & 4 & 3 \\ 1 & 3 & 4 \end{pmatrix} \). We need to find \( (AB)^{-1} \).
We know the property \( (AB)^{-1} = B^{-1}A^{-1} \). So we first need to find \( A^{-1} \).
Calculate the determinant of A, \( |A| \), by expanding along the first row \( R_1 \):
\( |A| = 5 \begin{vmatrix} 3 & 2 \\ 2 & 1 \end{vmatrix} - 0 \begin{vmatrix} 2 & 2 \\ 1 & 1 \end{vmatrix} + 4 \begin{vmatrix} 2 & 3 \\ 1 & 2 \end{vmatrix} \)
\( |A| = 5((3)(1) - (2)(2)) - 0 + 4((2)(2) - (3)(1)) \)
\( |A| = 5(3 - 4) + 4(4 - 3) \)
\( |A| = 5(-1) + 4(1) = -5 + 4 = -1 \).
Since \( |A| = -1 \neq 0 \), the inverse \( A^{-1} \) exists.
Next, find the cofactors for each element of matrix A.
Cofactors of the first row \( R_1 \):
Cofactor of \( a_{11} = \begin{vmatrix} 3 & 2 \\ 2 & 1 \end{vmatrix} = 3 - 4 = -1 \)
Cofactor of \( a_{12} = - \begin{vmatrix} 2 & 2 \\ 1 & 1 \end{vmatrix} = -(2 - 2) = 0 \)
Cofactor of \( a_{13} = \begin{vmatrix} 2 & 3 \\ 1 & 2 \end{vmatrix} = 4 - 3 = 1 \)
Cofactors of the second row \( R_2 \):
Cofactor of \( a_{21} = - \begin{vmatrix} 0 & 4 \\ 2 & 1 \end{vmatrix} = -(0 - 8) = 8 \)
Cofactor of \( a_{22} = \begin{vmatrix} 5 & 4 \\ 1 & 1 \end{vmatrix} = 5 - 4 = 1 \)
Cofactor of \( a_{23} = - \begin{vmatrix} 5 & 0 \\ 1 & 2 \end{vmatrix} = -(10 - 0) = -10 \)
Cofactors of the third row \( R_3 \):
Cofactor of \( a_{31} = \begin{vmatrix} 0 & 4 \\ 3 & 2 \end{vmatrix} = 0 - 12 = -12 \)
Cofactor of \( a_{32} = - \begin{vmatrix} 5 & 4 \\ 2 & 2 \end{vmatrix} = -(10 - 8) = -2 \)
Cofactor of \( a_{33} = \begin{vmatrix} 5 & 0 \\ 2 & 3 \end{vmatrix} = 15 - 0 = 15 \)
Now, we form the adjoint of A, which is the transpose of the cofactor matrix:
\[ \text{adj A} = \begin{pmatrix} -1 & 0 & 1 \\ 8 & 1 & -10 \\ -12 & -2 & 15 \end{pmatrix}^T = \begin{pmatrix} -1 & 8 & -12 \\ 0 & 1 & -2 \\ 1 & -10 & 15 \end{pmatrix} \] Now, calculate \( A^{-1} = \frac{1}{|A|} \text{adj A} \):
\[ A^{-1} = \frac{1}{-1} \begin{pmatrix} -1 & 8 & -12 \\ 0 & 1 & -2 \\ 1 & -10 & 15 \end{pmatrix} = \begin{pmatrix} 1 & -8 & 12 \\ 0 & -1 & 2 \\ -1 & 10 & -15 \end{pmatrix} \] Finally, calculate \( (AB)^{-1} = B^{-1}A^{-1} \):
\[ (AB)^{-1} = \begin{pmatrix} 1 & 3 & 3 \\ 1 & 4 & 3 \\ 1 & 3 & 4 \end{pmatrix} \begin{pmatrix} 1 & -8 & 12 \\ 0 & -1 & 2 \\ -1 & 10 & -15 \end{pmatrix} \] \[ = \begin{pmatrix} 1 \cdot 1 + 3 \cdot 0 + 3 \cdot (-1) & 1 \cdot (-8) + 3 \cdot (-1) + 3 \cdot 10 & 1 \cdot 12 + 3 \cdot 2 + 3 \cdot (-15) \\ 1 \cdot 1 + 4 \cdot 0 + 3 \cdot (-1) & 1 \cdot (-8) + 4 \cdot (-1) + 3 \cdot 10 & 1 \cdot 12 + 4 \cdot 2 + 3 \cdot (-15) \\ 1 \cdot 1 + 3 \cdot 0 + 4 \cdot (-1) & 1 \cdot (-8) + 3 \cdot (-1) + 4 \cdot 10 & 1 \cdot 12 + 3 \cdot 2 + 4 \cdot (-15) \end{pmatrix} \] \[ = \begin{pmatrix} 1 + 0 - 3 & -8 - 3 + 30 & 12 + 6 - 45 \\ 1 + 0 - 3 & -8 - 4 + 30 & 12 + 8 - 45 \\ 1 + 0 - 4 & -8 - 3 + 40 & 12 + 6 - 60 \end{pmatrix} \] \[ = \begin{pmatrix} -2 & 19 & -27 \\ -2 & 18 & -25 \\ -3 & 29 & -42 \end{pmatrix} \]
(iii) Given \( A^{-1} = \frac{1}{10}\begin{pmatrix} 10 & -10 & 2 \\ 0 & 5 & -4 \\ 0 & 0 & 2 \end{pmatrix} \) and \( B = \begin{pmatrix} 1 & -1 & 2 \\ 0 & 2 & -3 \\ 3 & -2 & 4 \end{pmatrix} \). We need to find \( (BA)^{-1} \).
We know that \( (BA)^{-1} = A^{-1}B^{-1} \). So we first need to find \( B^{-1} \).
Calculate the determinant of B, \( |B| \), by expanding along the first column \( C_1 \):
\( |B| = 1 \begin{vmatrix} 2 & -3 \\ -2 & 4 \end{vmatrix} - 0 \begin{vmatrix} -1 & 2 \\ -2 & 4 \end{vmatrix} + 3 \begin{vmatrix} -1 & 2 \\ 2 & -3 \end{vmatrix} \)
\( |B| = 1((2)(4) - (-3)(-2)) - 0 + 3((-1)(-3) - (2)(2)) \)
\( |B| = 1(8 - 6) + 3(3 - 4) \)
\( |B| = 1(2) + 3(-1) = 2 - 3 = -1 \).
Since \( |B| = -1 \neq 0 \), the inverse \( B^{-1} \) exists.
Next, find the cofactors for each element of matrix B.
Cofactors of the first row \( R_1 \):
Cofactor of \( b_{11} = \begin{vmatrix} 2 & -3 \\ -2 & 4 \end{vmatrix} = 8 - 6 = 2 \)
Cofactor of \( b_{12} = - \begin{vmatrix} 0 & -3 \\ 3 & 4 \end{vmatrix} = -(0 - (-9)) = -9 \)
Cofactor of \( b_{13} = \begin{vmatrix} 0 & 2 \\ 3 & -2 \end{vmatrix} = 0 - 6 = -6 \)
Cofactors of the second row \( R_2 \):
Cofactor of \( b_{21} = - \begin{vmatrix} -1 & 2 \\ -2 & 4 \end{vmatrix} = -(-4 - (-4)) = 0 \)
Cofactor of \( b_{22} = \begin{vmatrix} 1 & 2 \\ 3 & 4 \end{vmatrix} = 4 - 6 = -2 \)
Cofactor of \( b_{23} = - \begin{vmatrix} 1 & -1 \\ 3 & -2 \end{vmatrix} = -(-2 - (-3)) = -(1) = -1 \)
Cofactors of the third row \( R_3 \):
Cofactor of \( b_{31} = \begin{vmatrix} -1 & 2 \\ 2 & -3 \end{vmatrix} = 3 - 4 = -1 \)
Cofactor of \( b_{32} = - \begin{vmatrix} 1 & 2 \\ 0 & -3 \end{vmatrix} = -(-3 - 0) = 3 \)
Cofactor of \( b_{33} = \begin{vmatrix} 1 & -1 \\ 0 & 2 \end{vmatrix} = 2 - 0 = 2 \)
Now, we form the adjoint of B, which is the transpose of the cofactor matrix:
\[ \text{adj B} = \begin{pmatrix} 2 & -9 & -6 \\ 0 & -2 & -1 \\ -1 & 3 & 2 \end{pmatrix}^T = \begin{pmatrix} 2 & 0 & -1 \\ -9 & -2 & 3 \\ -6 & -1 & 2 \end{pmatrix} \] Now, calculate \( B^{-1} = \frac{1}{|B|} \text{adj B} \):
\[ B^{-1} = \frac{1}{-1} \begin{pmatrix} 2 & 0 & -1 \\ -9 & -2 & 3 \\ -6 & -1 & 2 \end{pmatrix} = \begin{pmatrix} -2 & 0 & 1 \\ 9 & 2 & -3 \\ 6 & 1 & -2 \end{pmatrix} \] Finally, calculate \( (BA)^{-1} = A^{-1}B^{-1} \):
\[ (BA)^{-1} = \frac{1}{10}\begin{pmatrix} 10 & -10 & 2 \\ 0 & 5 & -4 \\ 0 & 0 & 2 \end{pmatrix} \begin{pmatrix} -2 & 0 & 1 \\ 9 & 2 & -3 \\ 6 & 1 & -2 \end{pmatrix} \] \[ = \frac{1}{10}\begin{pmatrix} 10(-2) + (-10)(9) + 2(6) & 10(0) + (-10)(2) + 2(1) & 10(1) + (-10)(-3) + 2(-2) \\ 0(-2) + 5(9) + (-4)(6) & 0(0) + 5(2) + (-4)(1) & 0(1) + 5(-3) + (-4)(-2) \\ 0(-2) + 0(9) + 2(6) & 0(0) + 0(2) + 2(1) & 0(1) + 0(-3) + 2(-2) \end{pmatrix} \] \[ = \frac{1}{10}\begin{pmatrix} -20 - 90 + 12 & 0 - 20 + 2 & 10 + 30 - 4 \\ 0 + 45 - 24 & 0 + 10 - 4 & 0 - 15 + 8 \\ 0 + 0 + 12 & 0 + 0 + 2 & 0 + 0 - 4 \end{pmatrix} \] \[ = \frac{1}{10}\begin{pmatrix} -98 & -18 & 36 \\ 21 & 6 & -7 \\ 12 & 2 & -4 \end{pmatrix} \]In simple words: For each part, we used the property \( (AB)^{-1} = B^{-1}A^{-1} \) or \( (BA)^{-1} = A^{-1}B^{-1} \). This meant finding the inverse of individual matrices (A or B), if not already given. We computed each inverse by finding the determinant and then the adjoint matrix. Once we had all necessary inverses, we multiplied them in the correct reverse order to find the inverse of the product matrix. This method is often faster than first multiplying the original matrices and then finding the inverse of the large product.

🎯 Exam Tip: Remember the critical property \( (AB)^{-1} = B^{-1}A^{-1} \) and its variation for \( (BA)^{-1} \). This can greatly simplify calculations, especially with larger matrices. Always calculate individual inverses before multiplying them for the product inverse.

Question 10. (i) Given \( A = \left[\begin{array}{rrr} 2 & 4 & -1 \\ -1 & 0 & 2 \end{array}\right], B = \left[\begin{array}{rr} 3 & 4 \\ -1 & 2 \\ 2 & 1 \end{array}\right], \) find \( (AB)^{-1} \).
Answer:To find \( (AB)^{-1} \), we first need to calculate the product matrix \( AB \). \( AB = \left[\begin{array}{rrr} 2 & 4 & -1 \\ -1 & 0 & 2 \end{array}\right] \left[\begin{array}{rr} 3 & 4 \\ -1 & 2 \\ 2 & 1 \end{array}\right] \) \( AB = \left[\begin{array}{cc} (2)(3)+(4)(-1)+(-1)(2) & (2)(4)+(4)(2)+(-1)(1) \\ (-1)(3)+(0)(-1)+(2)(2) & (-1)(4)+(0)(2)+(2)(1) \end{array}\right] \) \( AB = \left[\begin{array}{cc} 6-4-2 & 8+8-1 \\ -3+0+4 & -4+0+2 \end{array}\right] \) \( AB = \left[\begin{array}{cc} 0 & 15 \\ 1 & -2 \end{array}\right] \) Next, calculate the determinant of \( AB \): \( |AB| = (0)(-2) - (15)(1) = 0 - 15 = -15 \) Since \( |AB| = -15 \neq 0 \), the inverse of \( AB \) exists. Now, find the adjoint of \( AB \). The cofactors of \( AB \) are: Cofactor of \( a_{11} = -2 \) Cofactor of \( a_{12} = -1 \) Cofactor of \( a_{21} = -15 \) Cofactor of \( a_{22} = 0 \) The cofactor matrix is \( \left[\begin{array}{rr} -2 & -1 \\ -15 & 0 \end{array}\right] \). The adjoint of \( AB \) is the transpose of the cofactor matrix: \( \text{adj } AB = \left[\begin{array}{rr} -2 & -15 \\ -1 & 0 \end{array}\right] \) Finally, calculate \( (AB)^{-1} = \frac{1}{|AB|} \text{adj } AB \): \( (AB)^{-1} = \frac{1}{-15} \left[\begin{array}{rr} -2 & -15 \\ -1 & 0 \end{array}\right] \) \( (AB)^{-1} = \frac{1}{15} \left[\begin{array}{rr} 2 & 15 \\ 1 & 0 \end{array}\right] \)In simple words: First, we multiplied the two given matrices A and B to get a new matrix AB. Then, we found a special number called the determinant of AB, which was -15. Because it's not zero, we know an inverse exists. After finding the cofactor matrix and then its transpose (the adjoint), we used these to calculate the final inverse matrix by dividing the adjoint by the determinant. This operation helps us 'undo' the matrix multiplication.

🎯 Exam Tip: Remember that \( (AB)^{-1} = B^{-1}A^{-1} \) is a key property, but in this case, direct calculation of AB and then its inverse is more straightforward since A and B are not square matrices of the same order individually to have direct inverses that are easy to use. Also, always double-check your determinant calculation as a zero determinant means no inverse exists.

Question 10. (ii) If \( A = \left[\begin{array}{III} 5 & 0 & 4 \\ 2 & 3 & 2 \\ 1 & 2 & 1 \end{array}\right], B^{-1} = \left[\begin{array}{III} 1 & 3 & 3 \\ 1 & 4 & 3 \\ 1 & 3 & 4 \end{array}\right], \) find \( (AB)^{-1} \).
Answer:To find \( (AB)^{-1} \), we use the property that \( (AB)^{-1} = B^{-1}A^{-1} \). We are given \( A \) and \( B^{-1} \), so we need to find \( A^{-1} \). First, find the determinant of A: \( |A| = \left|\begin{array}{ccc} 5 & 0 & 4 \\ 2 & 3 & 2 \\ 1 & 2 & 1 \end{array}\right| \) Expanding along R₁: \( |A| = 5 \left|\begin{array}{cc} 3 & 2 \\ 2 & 1 \end{array}\right| - 0 \left|\begin{array}{cc} 2 & 2 \\ 1 & 1 \end{array}\right| + 4 \left|\begin{array}{cc} 2 & 3 \\ 1 & 2 \end{array}\right| \) \( |A| = 5(3 \times 1 - 2 \times 2) - 0 + 4(2 \times 2 - 3 \times 1) \) \( |A| = 5(3 - 4) + 4(4 - 3) \) \( |A| = 5(-1) + 4(1) \) \( |A| = -5 + 4 = -1 \) Since \( |A| = -1 \neq 0 \), \( A^{-1} \) exists. Next, find the cofactors of A: Cofactor of \( a_{11} = \left|\begin{array}{cc} 3 & 2 \\ 2 & 1 \end{array}\right| = 3-4 = -1 \) Cofactor of \( a_{12} = -\left|\begin{array}{cc} 2 & 2 \\ 1 & 1 \end{array}\right| = -(2-2) = 0 \) Cofactor of \( a_{13} = \left|\begin{array}{cc} 2 & 3 \\ 1 & 2 \end{array}\right| = 4-3 = 1 \) Cofactor of \( a_{21} = -\left|\begin{array}{cc} 0 & 4 \\ 2 & 1 \end{array}\right| = -(0-8) = 8 \) Cofactor of \( a_{22} = \left|\begin{array}{cc} 5 & 4 \\ 1 & 1 \end{array}\right| = 5-4 = 1 \) Cofactor of \( a_{23} = -\left|\begin{array}{cc} 5 & 0 \\ 1 & 2 \end{array}\right| = -(10-0) = -10 \) Cofactor of \( a_{31} = \left|\begin{array}{cc} 0 & 4 \\ 3 & 2 \end{array}\right| = 0-12 = -12 \) Cofactor of \( a_{32} = -\left|\begin{array}{cc} 5 & 4 \\ 2 & 2 \end{array}\right| = -(10-8) = -2 \) Cofactor of \( a_{33} = \left|\begin{array}{cc} 5 & 0 \\ 2 & 3 \end{array}\right| = 15-0 = 15 \) The cofactor matrix is: \( C = \left[\begin{array}{rrr} -1 & 0 & 1 \\ 8 & 1 & -10 \\ -12 & -2 & 15 \end{array}\right] \) The adjoint of A is the transpose of the cofactor matrix: \( \text{adj } A = C^T = \left[\begin{array}{rrr} -1 & 8 & -12 \\ 0 & 1 & -2 \\ 1 & -10 & 15 \end{array}\right] \) Now, calculate \( A^{-1} = \frac{1}{|A|} \text{adj } A \): \( A^{-1} = \frac{1}{-1} \left[\begin{array}{rrr} -1 & 8 & -12 \\ 0 & 1 & -2 \\ 1 & -10 & 15 \end{array}\right] \) \( A^{-1} = \left[\begin{array}{rrr} 1 & -8 & 12 \\ 0 & -1 & 2 \\ -1 & 10 & -15 \end{array}\right] \) Finally, calculate \( (AB)^{-1} = B^{-1}A^{-1} \): \( (AB)^{-1} = \left[\begin{array}{rrr} 1 & 3 & 3 \\ 1 & 4 & 3 \\ 1 & 3 & 4 \end{array}\right] \left[\begin{array}{rrr} 1 & -8 & 12 \\ 0 & -1 & 2 \\ -1 & 10 & -15 \end{array}\right] \) \( (AB)^{-1} = \left[\begin{array}{ccc} (1)(1)+(3)(0)+(3)(-1) & (1)(-8)+(3)(-1)+(3)(10) & (1)(12)+(3)(2)+(3)(-15) \\ (1)(1)+(4)(0)+(3)(-1) & (1)(-8)+(4)(-1)+(3)(10) & (1)(12)+(4)(2)+(3)(-15) \\ (1)(1)+(3)(0)+(4)(-1) & (1)(-8)+(3)(-1)+(4)(10) & (1)(12)+(3)(2)+(4)(-15) \end{array}\right] \) \( (AB)^{-1} = \left[\begin{array}{ccc} 1+0-3 & -8-3+30 & 12+6-45 \\ 1+0-3 & -8-4+30 & 12+8-45 \\ 1+0-4 & -8-3+40 & 12+6-60 \end{array}\right] \) \( (AB)^{-1} = \left[\begin{array}{rrr} -2 & 19 & -27 \\ -2 & 18 & -25 \\ -3 & 29 & -42 \end{array}\right] \)In simple words: We used a rule that says to find the inverse of AB, you can find the inverse of B and the inverse of A, then multiply them in reverse order (\( B^{-1}A^{-1} \)). We already had \( B^{-1} \), so we first calculated \( A^{-1} \) by finding its determinant and adjoint. After getting \( A^{-1} \), we multiplied it with \( B^{-1} \) to get our final answer. It's like solving a puzzle in steps using specific matrix rules.

🎯 Exam Tip: For problems involving \( (AB)^{-1} \), always check if \( B^{-1}A^{-1} \) can be used. This often simplifies the calculation significantly compared to finding AB first and then its inverse, especially if A and B are not conformable for direct multiplication (e.g., if A is not square). Remember to calculate the determinant first to confirm if the inverse exists.

Question 10. (iii) If \( A^{-1} = \frac{1}{10}\left[\begin{array}{rrr} 10 & -10 & 2 \\ 0 & 5 & -4 \\ 0 & 0 & 2 \end{array}\right], B = \left[\begin{array}{rrr} 1 & -1 & 2 \\ 0 & 2 & -3 \\ 3 & -2 & 4 \end{array}\right]. \)
Answer:The implied question is to find \( (BA)^{-1} \). To find \( (BA)^{-1} \), we use the property that \( (BA)^{-1} = A^{-1}B^{-1} \). We are given \( A^{-1} \) directly, so we need to find \( B^{-1} \). First, find the determinant of B: \( |B| = \left|\begin{array}{rrr} 1 & -1 & 2 \\ 0 & 2 & -3 \\ 3 & -2 & 4 \end{array}\right| \) Expanding along C₁: \( |B| = 1 \left|\begin{array}{cc} 2 & -3 \\ -2 & 4 \end{array}\right| - 0 \left|\begin{array}{cc} -1 & 2 \\ -2 & 4 \end{array}\right| + 3 \left|\begin{array}{cc} -1 & 2 \\ 2 & -3 \end{array}\right| \) \( |B| = 1(2 \times 4 - (-3) \times (-2)) - 0 + 3((-1) \times (-3) - (2) \times (2)) \) \( |B| = 1(8 - 6) + 3(3 - 4) \) \( |B| = 1(2) + 3(-1) \) \( |B| = 2 - 3 = -1 \) Since \( |B| = -1 \neq 0 \), \( B^{-1} \) exists. Next, find the cofactors of B: Cofactor of \( b_{11} = \left|\begin{array}{cc} 2 & -3 \\ -2 & 4 \end{array}\right| = 8-6 = 2 \) Cofactor of \( b_{12} = -\left|\begin{array}{cc} 0 & -3 \\ 3 & 4 \end{array}\right| = -(0-(-9)) = -9 \) Cofactor of \( b_{13} = \left|\begin{array}{cc} 0 & 2 \\ 3 & -2 \end{array}\right| = 0-6 = -6 \) Cofactor of \( b_{21} = -\left|\begin{array}{cc} -1 & 2 \\ -2 & 4 \end{array}\right| = -(-4-(-4)) = 0 \) Cofactor of \( b_{22} = \left|\begin{array}{cc} 1 & 2 \\ 3 & 4 \end{array}\right| = 4-6 = -2 \) Cofactor of \( b_{23} = -\left|\begin{array}{cc} 1 & -1 \\ 3 & -2 \end{array}\right| = -(-2-(-3)) = -(1) = -1 \) Cofactor of \( b_{31} = \left|\begin{array}{cc} -1 & 2 \\ 2 & -3 \end{array}\right| = 3-4 = -1 \) Cofactor of \( b_{32} = -\left|\begin{array}{cc} 1 & 2 \\ 0 & -3 \end{array}\right| = -(-3-0) = 3 \) Cofactor of \( b_{33} = \left|\begin{array}{cc} 1 & -1 \\ 0 & 2 \end{array}\right| = 2-0 = 2 \) The cofactor matrix is: \( C = \left[\begin{array}{rrr} 2 & -9 & -6 \\ 0 & -2 & -1 \\ -1 & 3 & 2 \end{array}\right] \) The adjoint of B is the transpose of the cofactor matrix: \( \text{adj } B = C^T = \left[\begin{array}{rrr} 2 & 0 & -1 \\ -9 & -2 & 3 \\ -6 & -1 & 2 \end{array}\right] \) Now, calculate \( B^{-1} = \frac{1}{|B|} \text{adj } B \): \( B^{-1} = \frac{1}{-1} \left[\begin{array}{rrr} 2 & 0 & -1 \\ -9 & -2 & 3 \\ -6 & -1 & 2 \end{array}\right] \) \( B^{-1} = \left[\begin{array}{rrr} -2 & 0 & 1 \\ 9 & 2 & -3 \\ 6 & 1 & -2 \end{array}\right] \) Finally, calculate \( (BA)^{-1} = A^{-1}B^{-1} \): \( (BA)^{-1} = \frac{1}{10}\left[\begin{array}{rrr} 10 & -10 & 2 \\ 0 & 5 & -4 \\ 0 & 0 & 2 \end{array}\right] \left[\begin{array}{rrr} -2 & 0 & 1 \\ 9 & 2 & -3 \\ 6 & 1 & -2 \end{array}\right] \) \( (BA)^{-1} = \frac{1}{10}\left[\begin{array}{ccc} (10)(-2)+(-10)(9)+(2)(6) & (10)(0)+(-10)(2)+(2)(1) & (10)(1)+(-10)(-3)+(2)(-2) \\ (0)(-2)+(5)(9)+(-4)(6) & (0)(0)+(5)(2)+(-4)(1) & (0)(1)+(5)(-3)+(-4)(-2) \\ (0)(-2)+(0)(9)+(2)(6) & (0)(0)+(0)(2)+(2)(1) & (0)(1)+(0)(-3)+(2)(-2) \end{array}\right] \) \( (BA)^{-1} = \frac{1}{10}\left[\begin{array}{ccc} -20-90+12 & 0-20+2 & 10+30-4 \\ 0+45-24 & 0+10-4 & 0-15+8 \\ 0+0+12 & 0+0+2 & 0+0-4 \end{array}\right] \) \( (BA)^{-1} = \frac{1}{10}\left[\begin{array}{rrr} -98 & -18 & 36 \\ 21 & 6 & -7 \\ 12 & 2 & -4 \end{array}\right] \)In simple words: We used the property that the inverse of a product of matrices, (BA), is the product of their inverses in reverse order, ( \( A^{-1}B^{-1} \) ). Since \( A^{-1} \) was given, our first task was to find \( B^{-1} \). This involved calculating the determinant of B and then its adjoint matrix. Once we had both \( A^{-1} \) and \( B^{-1} \), we multiplied them together to get the final answer. This process lets us efficiently find the inverse of a complex matrix product.

🎯 Exam Tip: Remember the property \( (AB)^{-1} = B^{-1}A^{-1} \) (and similarly \( (BA)^{-1} = A^{-1}B^{-1} \)) for matrix inverses. This property is vital for solving problems where the inverse of a product is needed, and individual inverses are given or easier to find. Be careful with the order of multiplication!

Question 11. If \( A = \left[\begin{array}{rr} 2 & -3 \\ 4 & 6 \end{array}\right], \) verify that \( (\text{adj. } A)^{-1} = \text{adj. } (A^{-1}) \).
Answer:Given matrix \( A = \left[\begin{array}{rr} 2 & -3 \\ 4 & 6 \end{array}\right] \). First, calculate the determinant of A: \( |A| = (2)(6) - (-3)(4) = 12 - (-12) = 12 + 12 = 24 \) Since \( |A| = 24 \neq 0 \), \( A^{-1} \) exists. Next, find the cofactors of A: Cofactor of \( a_{11} = 6 \) Cofactor of \( a_{12} = -4 \) Cofactor of \( a_{21} = 3 \) Cofactor of \( a_{22} = 2 \) The cofactor matrix is \( \left[\begin{array}{rr} 6 & -4 \\ 3 & 2 \end{array}\right] \). The adjoint of A is the transpose of the cofactor matrix: \( \text{adj } A = \left[\begin{array}{rr} 6 & 3 \\ -4 & 2 \end{array}\right] \) Now, find \( (\text{adj } A)^{-1} \). First, find the determinant of \( \text{adj } A \): \( |\text{adj } A| = (6)(2) - (3)(-4) = 12 - (-12) = 12 + 12 = 24 \) Since \( |\text{adj } A| = 24 \neq 0 \), \( (\text{adj } A)^{-1} \) exists. Next, find the cofactors of \( \text{adj } A \): Cofactor of \( 6 = 2 \) Cofactor of \( 3 = 4 \) Cofactor of \( -4 = -3 \) Cofactor of \( 2 = 6 \) The cofactor matrix of \( \text{adj } A \) is \( \left[\begin{array}{rr} 2 & 4 \\ -3 & 6 \end{array}\right] \). The adjoint of \( \text{adj } A \) is the transpose of its cofactor matrix: \( \text{adj } (\text{adj } A) = \left[\begin{array}{rr} 2 & -3 \\ 4 & 6 \end{array}\right] \) So, \( (\text{adj } A)^{-1} = \frac{1}{|\text{adj } A|} \text{adj } (\text{adj } A) \): \( (\text{adj } A)^{-1} = \frac{1}{24} \left[\begin{array}{rr} 2 & -3 \\ 4 & 6 \end{array}\right] \) Now, we need to find \( A^{-1} \) and then \( \text{adj } (A^{-1}) \). \( A^{-1} = \frac{1}{|A|} \text{adj } A = \frac{1}{24} \left[\begin{array}{rr} 6 & 3 \\ -4 & 2 \end{array}\right] \) \( A^{-1} = \left[\begin{array}{rr} \frac{6}{24} & \frac{3}{24} \\ \frac{-4}{24} & \frac{2}{24} \end{array}\right] = \left[\begin{array}{rr} \frac{1}{4} & \frac{1}{8} \\ \frac{-1}{6} & \frac{1}{12} \end{array}\right] \) Next, find the cofactors of \( A^{-1} \): Cofactor of \( \frac{1}{4} = \frac{1}{12} \) Cofactor of \( \frac{1}{8} = \frac{1}{6} \) Cofactor of \( \frac{-1}{6} = \frac{-1}{8} \) Cofactor of \( \frac{1}{12} = \frac{1}{4} \) The cofactor matrix of \( A^{-1} \) is \( \left[\begin{array}{rr} \frac{1}{12} & \frac{1}{6} \\ \frac{-1}{8} & \frac{1}{4} \end{array}\right] \). The adjoint of \( A^{-1} \) is the transpose of its cofactor matrix: \( \text{adj } (A^{-1}) = \left[\begin{array}{rr} \frac{1}{12} & \frac{-1}{8} \\ \frac{1}{6} & \frac{1}{4} \end{array}\right] \) Let's express \( (\text{adj } A)^{-1} \) as \( \left[\begin{array}{rr} \frac{2}{24} & \frac{-3}{24} \\ \frac{4}{24} & \frac{6}{24} \end{array}\right] = \left[\begin{array}{rr} \frac{1}{12} & \frac{-1}{8} \\ \frac{1}{6} & \frac{1}{4} \end{array}\right] \). Comparing \( (\text{adj } A)^{-1} \) and \( \text{adj } (A^{-1}) \), we see that they are equal. Therefore, \( (\text{adj } A)^{-1} = \text{adj } (A^{-1}) \). This property is demonstrated using a 2x2 matrix.In simple words: This problem asked us to check if two special matrix operations give the same result. We first found the "adjoint" of matrix A, then its inverse. Separately, we found the inverse of A, and then its "adjoint". When we compared the final matrices from both calculations, they were identical, showing that this mathematical rule holds true for matrix A.

🎯 Exam Tip: This problem verifies a useful property: \( (\text{adj } A)^{-1} = \text{adj } (A^{-1}) \). Understanding how to calculate determinants, cofactors, adjoints, and inverses for both a matrix and its adjoint/inverse is crucial. Always perform calculations carefully, especially when dealing with fractions in matrix elements.

Question 12. If \( A = \left[\begin{array}{rr} -1 & -1 \\ 2 & -2 \end{array}\right], \) show that \( A^2 + 3A + 4I = 0 \). Hence find \( A^{-1} \).
Answer:Given matrix \( A = \left[\begin{array}{rr} -1 & -1 \\ 2 & -2 \end{array}\right] \). First, calculate \( A^2 \): \( A^2 = A \times A = \left[\begin{array}{rr} -1 & -1 \\ 2 & -2 \end{array}\right] \left[\begin{array}{rr} -1 & -1 \\ 2 & -2 \end{array}\right] \) \( A^2 = \left[\begin{array}{cc} (-1)(-1)+(-1)(2) & (-1)(-1)+(-1)(-2) \\ (2)(-1)+(-2)(2) & (2)(-1)+(-2)(-2) \end{array}\right] \) \( A^2 = \left[\begin{array}{cc} 1-2 & 1+2 \\ -2-4 & -2+4 \end{array}\right] = \left[\begin{array}{rr} -1 & 3 \\ -6 & 2 \end{array}\right] \) Now, calculate \( 3A \): \( 3A = 3 \left[\begin{array}{rr} -1 & -1 \\ 2 & -2 \end{array}\right] = \left[\begin{array}{rr} -3 & -3 \\ 6 & -6 \end{array}\right] \) And for \( 4I \), where I is the identity matrix of order 2: \( 4I = 4 \left[\begin{array}{rr} 1 & 0 \\ 0 & 1 \end{array}\right] = \left[\begin{array}{rr} 4 & 0 \\ 0 & 4 \end{array}\right] \) Now, substitute these into the equation \( A^2 + 3A + 4I \): \( A^2 + 3A + 4I = \left[\begin{array}{rr} -1 & 3 \\ -6 & 2 \end{array}\right] + \left[\begin{array}{rr} -3 & -3 \\ 6 & -6 \end{array}\right] + \left[\begin{array}{rr} 4 & 0 \\ 0 & 4 \end{array}\right] \) \( A^2 + 3A + 4I = \left[\begin{array}{cc} -1-3+4 & 3-3+0 \\ -6+6+0 & 2-6+4 \end{array}\right] \) \( A^2 + 3A + 4I = \left[\begin{array}{cc} 0 & 0 \\ 0 & 0 \end{array}\right] = O \) So, we have shown that \( A^2 + 3A + 4I = O \). To find \( A^{-1} \), we can use this equation. Multiply the equation \( A^2 + 3A + 4I = O \) by \( A^{-1} \) (pre-multiplying both sides): \( A^{-1}(A^2 + 3A + 4I) = A^{-1}O \) \( A^{-1}A^2 + A^{-1}(3A) + A^{-1}(4I) = O \) \( (A^{-1}A)A + 3(A^{-1}A) + 4(A^{-1}I) = O \) We know that \( A^{-1}A = I \) and \( A^{-1}I = A^{-1} \), and \( IA = A \). \( IA + 3I + 4A^{-1} = O \) \( A + 3I + 4A^{-1} = O \) Now, solve for \( 4A^{-1} \): \( 4A^{-1} = -A - 3I \) \( A^{-1} = \frac{1}{4}(-A - 3I) \) \( A^{-1} = \frac{1}{4} \left( -\left[\begin{array}{rr} -1 & -1 \\ 2 & -2 \end{array}\right] - 3\left[\begin{array}{rr} 1 & 0 \\ 0 & 1 \end{array}\right] \right) \) \( A^{-1} = \frac{1}{4} \left( \left[\begin{array}{rr} 1 & 1 \\ -2 & 2 \end{array}\right] - \left[\begin{array}{rr} 3 & 0 \\ 0 & 3 \end{array}\right] \right) \) \( A^{-1} = \frac{1}{4} \left[\begin{array}{cc} 1-3 & 1-0 \\ -2-0 & 2-3 \end{array}\right] \) \( A^{-1} = \frac{1}{4} \left[\begin{array}{rr} -2 & 1 \\ -2 & -1 \end{array}\right] \) This is the inverse of A. We can also write it as \( -\frac{1}{4} \left[\begin{array}{rr} 2 & -1 \\ 2 & 1 \end{array}\right] \).In simple words: First, we proved that the given matrix A satisfies a specific equation by calculating \( A^2 \) and then adding \( 3A \) and \( 4I \) (identity matrix). The result was a zero matrix, confirming the equation. Then, using this equation, we found the inverse of A by multiplying the whole equation by \( A^{-1} \), simplifying it, and rearranging terms to solve for \( A^{-1} \). This method is often quicker than finding the inverse directly by determinants and adjoints.

🎯 Exam Tip: When asked to find \( A^{-1} \) after showing that A satisfies a polynomial equation, always use the equation to find the inverse. Multiplying the equation by \( A^{-1} \) and isolating \( A^{-1} \) is generally faster and less error-prone than calculating the determinant and adjoint directly for larger matrices. Remember that \( A^{-1}A = I \) and \( AI = A \).

Question 13. If \( A^2 - A + I = 0, \) then show that \( A^{-1} = I - A \).
Answer:Given the matrix equation \( A^2 - A + I = O \) ... (1) To show that \( A^{-1} = I - A \), we can multiply equation (1) by \( A^{-1} \). Pre-multiply both sides of equation (1) by \( A^{-1} \): \( A^{-1}(A^2 - A + I) = A^{-1}O \) Distribute \( A^{-1} \): \( A^{-1}A^2 - A^{-1}A + A^{-1}I = O \) Use the properties: \( A^{-1}A = I \), \( A^2 = A \cdot A \), and \( A^{-1}I = A^{-1} \). So, \( (A^{-1}A)A - I + A^{-1} = O \) This simplifies to: \( IA - I + A^{-1} = O \) We know that \( IA = A \). So, \( A - I + A^{-1} = O \) To isolate \( A^{-1} \), move A and I to the other side: \( A^{-1} = I - A \) This shows the desired result. The identity matrix I acts like the number 1 in basic algebra.In simple words: We started with a given matrix equation. To prove the inverse of A is \( I - A \), we multiplied every part of the original equation by \( A^{-1} \). Using basic matrix rules like \( A^{-1}A = I \) (the identity matrix) and \( A^{-1}I = A^{-1} \), we simplified the equation. After moving terms around, we successfully showed that \( A^{-1} \) is equal to \( I - A \).

🎯 Exam Tip: For problems proving a relationship involving \( A^{-1} \) from a given polynomial equation, always multiply the equation by \( A^{-1} \) (either pre- or post-multiplying, ensuring consistency) and use the fundamental properties \( A^{-1}A = I \) and \( AI = A \). This method avoids direct computation of \( A^{-1} \).

Question 14. For the matrix \( A = \left[\begin{array}{rr} 2 & -3 \\ 3 & 4 \end{array}\right], \) show that \( A^2 - 6A + 17I = O \). Hence find \( A^{-1} \).
Answer:Given matrix \( A = \left[\begin{array}{rr} 2 & -3 \\ 3 & 4 \end{array}\right] \). First, calculate \( A^2 \): \( A^2 = A \times A = \left[\begin{array}{rr} 2 & -3 \\ 3 & 4 \end{array}\right] \left[\begin{array}{rr} 2 & -3 \\ 3 & 4 \end{array}\right] \) \( A^2 = \left[\begin{array}{cc} (2)(2)+(-3)(3) & (2)(-3)+(-3)(4) \\ (3)(2)+(4)(3) & (3)(-3)+(4)(4) \end{array}\right] \) \( A^2 = \left[\begin{array}{cc} 4-9 & -6-12 \\ 6+12 & -9+16 \end{array}\right] = \left[\begin{array}{rr} -5 & -18 \\ 18 & 7 \end{array}\right] \) Next, calculate \( 6A \): \( 6A = 6 \left[\begin{array}{rr} 2 & -3 \\ 3 & 4 \end{array}\right] = \left[\begin{array}{rr} 12 & -18 \\ 18 & 24 \end{array}\right] \) And for \( 17I \), where I is the identity matrix of order 2: \( 17I = 17 \left[\begin{array}{rr} 1 & 0 \\ 0 & 1 \end{array}\right] = \left[\begin{array}{rr} 17 & 0 \\ 0 & 17 \end{array}\right] \) Now, substitute these into the equation \( A^2 - 6A + 17I \): \( A^2 - 6A + 17I = \left[\begin{array}{rr} -5 & -18 \\ 18 & 7 \end{array}\right] - \left[\begin{array}{rr} 12 & -18 \\ 18 & 24 \end{array}\right] + \left[\begin{array}{rr} 17 & 0 \\ 0 & 17 \end{array}\right] \) \( A^2 - 6A + 17I = \left[\begin{array}{cc} -5-12+17 & -18-(-18)+0 \\ 18-18+0 & 7-24+17 \end{array}\right] \) \( A^2 - 6A + 17I = \left[\begin{array}{cc} -17+17 & -18+18 \\ 0 & -17+17 \end{array}\right] = \left[\begin{array}{cc} 0 & 0 \\ 0 & 0 \end{array}\right] = O \) So, we have shown that \( A^2 - 6A + 17I = O \). This is often called the Cayley-Hamilton theorem. To find \( A^{-1} \), we can use this equation. From \( A^2 - 6A + 17I = O \), we can write: \( 17I = 6A - A^2 \) Pre-multiply both sides by \( A^{-1} \): \( A^{-1}(17I) = A^{-1}(6A - A^2) \) \( 17(A^{-1}I) = 6(A^{-1}A) - (A^{-1}A)A \) We know that \( A^{-1}I = A^{-1} \) and \( A^{-1}A = I \). \( 17A^{-1} = 6I - IA \) \( 17A^{-1} = 6I - A \) Now, solve for \( A^{-1} \): \( A^{-1} = \frac{1}{17}(6I - A) \) \( A^{-1} = \frac{1}{17} \left( 6\left[\begin{array}{rr} 1 & 0 \\ 0 & 1 \end{array}\right] - \left[\begin{array}{rr} 2 & -3 \\ 3 & 4 \end{array}\right] \right) \) \( A^{-1} = \frac{1}{17} \left( \left[\begin{array}{rr} 6 & 0 \\ 0 & 6 \end{array}\right] - \left[\begin{array}{rr} 2 & -3 \\ 3 & 4 \end{array}\right] \right) \) \( A^{-1} = \frac{1}{17} \left[\begin{array}{cc} 6-2 & 0-(-3) \\ 0-3 & 6-4 \end{array}\right] \) \( A^{-1} = \frac{1}{17} \left[\begin{array}{rr} 4 & 3 \\ -3 & 2 \end{array}\right] \)In simple words: First, we checked if matrix A fit a certain algebraic rule, \( A^2 - 6A + 17I = O \), by calculating \( A^2 \), \( 6A \), and \( 17I \) separately and adding them up. Once confirmed, we used this same rule to find the inverse of A. We multiplied the entire equation by \( A^{-1} \) and rearranged the terms to easily figure out what \( A^{-1} \) was, which saved us from using the longer determinant and adjoint method.

🎯 Exam Tip: For these types of problems, showing that the matrix satisfies a given polynomial equation is usually the first part. The second part, finding the inverse, should then be done by algebraic manipulation of that polynomial equation (by multiplying with \( A^{-1} \) and isolating it), not by directly calculating the determinant and adjoint. This is a common and efficient method.

Question 15. Show that \( A = \left[\begin{array}{rrr} 1 & 0 & -2 \\ -2 & -1 & 2 \\ 3 & 4 & 1 \end{array}\right] \) satisfies the equation \( A^3 - A^2 - 3A - I = 0 \). Hence, find \( A^{-1} \).
Answer:Given matrix \( A = \left[\begin{array}{rrr} 1 & 0 & -2 \\ -2 & -1 & 2 \\ 3 & 4 & 1 \end{array}\right] \). First, calculate \( A^2 \): \( A^2 = A \times A = \left[\begin{array}{rrr} 1 & 0 & -2 \\ -2 & -1 & 2 \\ 3 & 4 & 1 \end{array}\right] \left[\begin{array}{rrr} 1 & 0 & -2 \\ -2 & -1 & 2 \\ 3 & 4 & 1 \end{array}\right] \) \( A^2 = \left[\begin{array}{ccc} (1)(1)+(0)(-2)+(-2)(3) & (1)(0)+(0)(-1)+(-2)(4) & (1)(-2)+(0)(2)+(-2)(1) \\ (-2)(1)+(-1)(-2)+(2)(3) & (-2)(0)+(-1)(-1)+(2)(4) & (-2)(-2)+(-1)(2)+(2)(1) \\ (3)(1)+(4)(-2)+(1)(3) & (3)(0)+(4)(-1)+(1)(4) & (3)(-2)+(4)(2)+(1)(1) \end{array}\right] \) \( A^2 = \left[\begin{array}{ccc} 1+0-6 & 0+0-8 & -2+0-2 \\ -2+2+6 & 0+1+8 & 4-2+2 \\ 3-8+3 & 0-4+4 & -6+8+1 \end{array}\right] = \left[\begin{array}{rrr} -5 & -8 & -4 \\ 6 & 9 & 4 \\ -2 & 0 & 3 \end{array}\right] \) Next, calculate \( A^3 \): \( A^3 = A^2 \times A = \left[\begin{array}{rrr} -5 & -8 & -4 \\ 6 & 9 & 4 \\ -2 & 0 & 3 \end{array}\right] \left[\begin{array}{rrr} 1 & 0 & -2 \\ -2 & -1 & 2 \\ 3 & 4 & 1 \end{array}\right] \) \( A^3 = \left[\begin{array}{ccc} (-5)(1)+(-8)(-2)+(-4)(3) & (-5)(0)+(-8)(-1)+(-4)(4) & (-5)(-2)+(-8)(2)+(-4)(1) \\ (6)(1)+(9)(-2)+(4)(3) & (6)(0)+(9)(-1)+(4)(4) & (6)(-2)+(9)(2)+(4)(1) \\ (-2)(1)+(0)(-2)+(3)(3) & (-2)(0)+(0)(-1)+(3)(4) & (-2)(-2)+(0)(2)+(3)(1) \end{array}\right] \) \( A^3 = \left[\begin{array}{ccc} -5+16-12 & 0+8-16 & 10-16-4 \\ 6-18+12 & 0-9+16 & -12+18+4 \\ -2+0+9 & 0+0+12 & 4+0+3 \end{array}\right] = \left[\begin{array}{rrr} -1 & -8 & -10 \\ 0 & 7 & 10 \\ 7 & 12 & 7 \end{array}\right] \) Now, we calculate \( A^3 - A^2 - 3A - I \). For \( 3A \): \( 3A = 3 \left[\begin{array}{rrr} 1 & 0 & -2 \\ -2 & -1 & 2 \\ 3 & 4 & 1 \end{array}\right] = \left[\begin{array}{rrr} 3 & 0 & -6 \\ -6 & -3 & 6 \\ 9 & 12 & 3 \end{array}\right] \) For \( I \), the identity matrix of order 3: \( I = \left[\begin{array}{rrr} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right] \) Substitute into \( A^3 - A^2 - 3A - I \): \( = \left[\begin{array}{rrr} -1 & -8 & -10 \\ 0 & 7 & 10 \\ 7 & 12 & 7 \end{array}\right] - \left[\begin{array}{rrr} -5 & -8 & -4 \\ 6 & 9 & 4 \\ -2 & 0 & 3 \end{array}\right] - \left[\begin{array}{rrr} 3 & 0 & -6 \\ -6 & -3 & 6 \\ 9 & 12 & 3 \end{array}\right] - \left[\begin{array}{rrr} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right] \) \( = \left[\begin{array}{ccc} -1-(-5)-3-1 & -8-(-8)-0-0 & -10-(-4)-(-6)-0 \\ 0-6-(-6)-0 & 7-9-(-3)-1 & 10-4-6-0 \\ 7-(-2)-9-0 & 12-0-12-0 & 7-3-3-1 \end{array}\right] \) \( = \left[\begin{array}{ccc} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{array}\right] = O \) Thus, \( A^3 - A^2 - 3A - I = O \). This shows the Cayley-Hamilton theorem in action for a 3x3 matrix. To find \( A^{-1} \), we use the equation \( A^3 - A^2 - 3A - I = O \). Pre-multiply by \( A^{-1} \): \( A^{-1}(A^3 - A^2 - 3A - I) = A^{-1}O \) \( A^{-1}A^3 - A^{-1}A^2 - A^{-1}(3A) - A^{-1}I = O \) \( (A^{-1}A)A^2 - (A^{-1}A)A - 3(A^{-1}A) - A^{-1} = O \) \( IA^2 - IA - 3I - A^{-1} = O \) \( A^2 - A - 3I - A^{-1} = O \) Solve for \( A^{-1} \): \( A^{-1} = A^2 - A - 3I \) Substitute the matrices: \( A^{-1} = \left[\begin{array}{rrr} -5 & -8 & -4 \\ 6 & 9 & 4 \\ -2 & 0 & 3 \end{array}\right] - \left[\begin{array}{rrr} 1 & 0 & -2 \\ -2 & -1 & 2 \\ 3 & 4 & 1 \end{array}\right] - 3\left[\begin{array}{rrr} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right] \) \( A^{-1} = \left[\begin{array}{rrr} -5 & -8 & -4 \\ 6 & 9 & 4 \\ -2 & 0 & 3 \end{array}\right] - \left[\begin{array}{rrr} 1 & 0 & -2 \\ -2 & -1 & 2 \\ 3 & 4 & 1 \end{array}\right] - \left[\begin{array}{rrr} 3 & 0 & 0 \\ 0 & 3 & 0 \\ 0 & 0 & 3 \end{array}\right] \) \( A^{-1} = \left[\begin{array}{ccc} -5-1-3 & -8-0-0 & -4-(-2)-0 \\ 6-(-2)-0 & 9-(-1)-3 & 4-2-0 \\ -2-3-0 & 0-4-0 & 3-1-3 \end{array}\right] \) \( A^{-1} = \left[\begin{array}{ccc} -9 & -8 & -2 \\ 8 & 7 & 2 \\ -5 & -4 & -1 \end{array}\right] \)In simple words: First, we performed several matrix multiplications to find \( A^2 \) and \( A^3 \). Then, we substituted these, along with 3A and the identity matrix I, into the given equation \( A^3 - A^2 - 3A - I \) to show that it equals the zero matrix. After confirming the equation, we used it to find \( A^{-1} \) by multiplying the entire equation by \( A^{-1} \) and rearranging the terms. This algebraic method makes finding the inverse much simpler than direct calculation.

🎯 Exam Tip: When proving a matrix equation for a 3x3 matrix, break down the calculations into steps (e.g., \( A^2 \), \( A^3 \), \( kA \)) to avoid errors. When finding \( A^{-1} \) from the characteristic equation, always multiply the equation by \( A^{-1} \) to obtain an expression for \( A^{-1} \) in terms of A and I. This method is efficient and less prone to numerical mistakes.

Question 16. Let \( A = \left[\begin{array}{rrr} 1 & 2 & 2 \\ 2 & 1 & 2 \\ 2 & 2 & 1 \end{array}\right], \) Prove that \( A^2 - 4A - 5I = 0 \). Hence obtain \( A^{-1} \).
Answer:Given matrix \( A = \left[\begin{array}{rrr} 1 & 2 & 2 \\ 2 & 1 & 2 \\ 2 & 2 & 1 \end{array}\right] \). First, calculate \( A^2 \): \( A^2 = A \times A = \left[\begin{array}{rrr} 1 & 2 & 2 \\ 2 & 1 & 2 \\ 2 & 2 & 1 \end{array}\right] \left[\begin{array}{rrr} 1 & 2 & 2 \\ 2 & 1 & 2 \\ 2 & 2 & 1 \end{array}\right] \) \( A^2 = \left[\begin{array}{ccc} (1)(1)+(2)(2)+(2)(2) & (1)(2)+(2)(1)+(2)(2) & (1)(2)+(2)(2)+(2)(1) \\ (2)(1)+(1)(2)+(2)(2) & (2)(2)+(1)(1)+(2)(2) & (2)(2)+(1)(2)+(2)(1) \\ (2)(1)+(2)(2)+(1)(2) & (2)(2)+(2)(1)+(1)(2) & (2)(2)+(2)(2)+(1)(1) \end{array}\right] \) \( A^2 = \left[\begin{array}{ccc} 1+4+4 & 2+2+4 & 2+4+2 \\ 2+2+4 & 4+1+4 & 4+2+2 \\ 2+4+2 & 4+2+2 & 4+4+1 \end{array}\right] = \left[\begin{array}{rrr} 9 & 8 & 8 \\ 8 & 9 & 8 \\ 8 & 8 & 9 \end{array}\right] \) Next, calculate \( 4A \): \( 4A = 4 \left[\begin{array}{rrr} 1 & 2 & 2 \\ 2 & 1 & 2 \\ 2 & 2 & 1 \end{array}\right] = \left[\begin{array}{rrr} 4 & 8 & 8 \\ 8 & 4 & 8 \\ 8 & 8 & 4 \end{array}\right] \) And for \( 5I \), where I is the identity matrix of order 3: \( 5I = 5 \left[\begin{array}{rrr} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right] = \left[\begin{array}{rrr} 5 & 0 & 0 \\ 0 & 5 & 0 \\ 0 & 0 & 5 \end{array}\right] \) Now, substitute these into the equation \( A^2 - 4A - 5I \): \( A^2 - 4A - 5I = \left[\begin{array}{rrr} 9 & 8 & 8 \\ 8 & 9 & 8 \\ 8 & 8 & 9 \end{array}\right] - \left[\begin{array}{rrr} 4 & 8 & 8 \\ 8 & 4 & 8 \\ 8 & 8 & 4 \end{array}\right] - \left[\begin{array}{rrr} 5 & 0 & 0 \\ 0 & 5 & 0 \\ 0 & 0 & 5 \end{array}\right] \) \( A^2 - 4A - 5I = \left[\begin{array}{ccc} 9-4-5 & 8-8-0 & 8-8-0 \\ 8-8-0 & 9-4-5 & 8-8-0 \\ 8-8-0 & 8-8-0 & 9-4-5 \end{array}\right] \) \( A^2 - 4A - 5I = \left[\begin{array}{ccc} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{array}\right] = O \) Thus, \( A^2 - 4A - 5I = O \). To obtain \( A^{-1} \), we use the equation \( A^2 - 4A - 5I = O \). From this equation, we can write: \( 5I = A^2 - 4A \) Pre-multiply both sides by \( A^{-1} \): \( A^{-1}(5I) = A^{-1}(A^2 - 4A) \) \( 5(A^{-1}I) = A^{-1}A^2 - A^{-1}(4A) \) \( 5A^{-1} = (A^{-1}A)A - 4(A^{-1}A) \) \( 5A^{-1} = IA - 4I \) \( 5A^{-1} = A - 4I \) Solve for \( A^{-1} \): \( A^{-1} = \frac{1}{5}(A - 4I) \) Substitute the matrices: \( A^{-1} = \frac{1}{5} \left( \left[\begin{array}{rrr} 1 & 2 & 2 \\ 2 & 1 & 2 \\ 2 & 2 & 1 \end{array}\right] - 4\left[\begin{array}{rrr} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right] \right) \) \( A^{-1} = \frac{1}{5} \left( \left[\begin{array}{rrr} 1 & 2 & 2 \\ 2 & 1 & 2 \\ 2 & 2 & 1 \end{array}\right] - \left[\begin{array}{rrr} 4 & 0 & 0 \\ 0 & 4 & 0 \\ 0 & 0 & 4 \end{array}\right] \right) \) \( A^{-1} = \frac{1}{5} \left[\begin{array}{ccc} 1-4 & 2-0 & 2-0 \\ 2-0 & 1-4 & 2-0 \\ 2-0 & 2-0 & 1-4 \end{array}\right] \) \( A^{-1} = \frac{1}{5} \left[\begin{array}{rrr} -3 & 2 & 2 \\ 2 & -3 & 2 \\ 2 & 2 & -3 \end{array}\right] \)In simple words: First, we calculated \( A^2 \) by multiplying matrix A by itself. Then, we found \( 4A \) and \( 5I \), where I is the identity matrix. By putting these values into the equation \( A^2 - 4A - 5I \), we showed that it results in a zero matrix. Finally, we used this proven equation to find the inverse of A. We multiplied the equation by \( A^{-1} \) and rearranged it to find a simple expression for \( A^{-1} \). This algebraic method is very effective for finding matrix inverses.

🎯 Exam Tip: When using the characteristic equation to find the inverse, remember to isolate the term with \( A^{-1} \) after multiplying by \( A^{-1} \). Ensure that all matrix operations (addition, subtraction, scalar multiplication) are performed correctly. This is often the most efficient way to find \( A^{-1} \) when a polynomial relation is given.

Question 17. If \( A = \frac{1}{9}\left[\begin{array}{rrr} 4 & -8 & 1 \\ 4 & 4 & 7 \\ 1 & -8 & 8 \end{array}\right], \) find \( (A')^{-1} \).
Answer:Given matrix \( A = \frac{1}{9}\left[\begin{array}{rrr} 4 & -8 & 1 \\ 4 & 4 & 7 \\ 1 & -8 & 8 \end{array}\right] \). First, find the transpose of A, which is \( A' \). \( A' = (\frac{1}{9}\left[\begin{array}{rrr} 4 & -8 & 1 \\ 4 & 4 & 7 \\ 1 & -8 & 8 \end{array}\right])' = \frac{1}{9}\left[\begin{array}{rrr} 4 & 4 & 1 \\ -8 & 4 & -8 \\ 1 & 7 & 8 \end{array}\right] \) Now, let \( K = \left[\begin{array}{rrr} 4 & 4 & 1 \\ -8 & 4 & -8 \\ 1 & 7 & 8 \end{array}\right] \), so \( A' = \frac{1}{9}K \). Then \( (A')^{-1} = ( \frac{1}{9}K )^{-1} = 9 K^{-1} \). First, find the determinant of K: \( |K| = \left|\begin{array}{rrr} 4 & 4 & 1 \\ -8 & 4 & -8 \\ 1 & 7 & 8 \end{array}\right| \) Expanding along R₁: \( |K| = 4 \left|\begin{array}{cc} 4 & -8 \\ 7 & 8 \end{array}\right| - 4 \left|\begin{array}{cc} -8 & -8 \\ 1 & 8 \end{array}\right| + 1 \left|\begin{array}{cc} -8 & 4 \\ 1 & 7 \end{array}\right| \) \( |K| = 4(32 - (-56)) - 4(-64 - (-8)) + 1(-56 - 4) \) \( |K| = 4(88) - 4(-56) + 1(-60) \) \( |K| = 352 + 224 - 60 = 576 - 60 = 516 \) Since \( |K| = 516 \neq 0 \), \( K^{-1} \) exists. Next, find the cofactors of K: Cofactor of \( k_{11} = \left|\begin{array}{cc} 4 & -8 \\ 7 & 8 \end{array}\right| = 32+56 = 88 \) Cofactor of \( k_{12} = -\left|\begin{array}{cc} -8 & -8 \\ 1 & 8 \end{array}\right| = -(-64+8) = 56 \) Cofactor of \( k_{13} = \left|\begin{array}{cc} -8 & 4 \\ 1 & 7 \end{array}\right| = -56-4 = -60 \) Cofactor of \( k_{21} = -\left|\begin{array}{cc} 4 & 1 \\ 7 & 8 \end{array}\right| = -(32-7) = -25 \) Cofactor of \( k_{22} = \left|\begin{array}{cc} 4 & 1 \\ 1 & 8 \end{array}\right| = 32-1 = 31 \) Cofactor of \( k_{23} = -\left|\begin{array}{cc} 4 & 4 \\ 1 & 7 \end{array}\right| = -(28-4) = -24 \) Cofactor of \( k_{31} = \left|\begin{array}{cc} 4 & 1 \\ 4 & -8 \end{array}\right| = -32-4 = -36 \) Cofactor of \( k_{32} = -\left|\begin{array}{cc} 4 & 1 \\ -8 & -8 \end{array}\right| = -(-32-(-8)) = 24 \) Cofactor of \( k_{33} = \left|\begin{array}{cc} 4 & 4 \\ -8 & 4 \end{array}\right| = 16-(-32) = 48 \) The cofactor matrix of K is: \( C_K = \left[\begin{array}{rrr} 88 & 56 & -60 \\ -25 & 31 & -24 \\ -36 & 24 & 48 \end{array}\right] \) The adjoint of K is the transpose of its cofactor matrix: \( \text{adj } K = C_K^T = \left[\begin{array}{rrr} 88 & -25 & -36 \\ 56 & 31 & 24 \\ -60 & -24 & 48 \end{array}\right] \) Now, calculate \( K^{-1} = \frac{1}{|K|} \text{adj } K \): \( K^{-1} = \frac{1}{516} \left[\begin{array}{rrr} 88 & -25 & -36 \\ 56 & 31 & 24 \\ -60 & -24 & 48 \end{array}\right] \) Finally, calculate \( (A')^{-1} = 9 K^{-1} \): \( (A')^{-1} = 9 \times \frac{1}{516} \left[\begin{array}{rrr} 88 & -25 & -36 \\ 56 & 31 & 24 \\ -60 & -24 & 48 \end{array}\right] \) \( (A')^{-1} = \frac{9}{516} \left[\begin{array}{rrr} 88 & -25 & -36 \\ 56 & 31 & 24 \\ -60 & -24 & 48 \end{array}\right] \) Simplify the fraction \( \frac{9}{516} \). Both are divisible by 3: \( \frac{3}{172} \). \( (A')^{-1} = \frac{3}{172} \left[\begin{array}{rrr} 88 & -25 & -36 \\ 56 & 31 & 24 \\ -60 & -24 & 48 \end{array}\right] \)In simple words: We were asked to find the inverse of the transpose of matrix A. First, we found the transpose of A by swapping its rows and columns. Since A had a scalar part (1/9), we separated it to simplify calculations. We then found the determinant, cofactors, and adjoint of the inner matrix. Finally, we used these to calculate its inverse, then multiplied by 9 to get the inverse of \( A' \). This involves a series of standard matrix operations.

🎯 Exam Tip: When a matrix is given with a scalar multiple, like \( A = kM \), remember that \( A' = kM' \) and \( (A')^{-1} = \frac{1}{k} (M')^{-1} \). It's often easier to first find the inverse of the unscaled matrix and then apply the scalar factor. Be extremely careful with sign changes for cofactors, as they are a common source of error.

Question 18. Show that \( A = \left[\begin{array}{rr} -8 & 5 \\ 2 & 4 \end{array}\right] \) satisfies the equation \( x^2 + 4x - 42 = 0 \). Hence find \( A^{-1} \).
Answer:Given matrix \( A = \left[\begin{array}{rr} -8 & 5 \\ 2 & 4 \end{array}\right] \). First, calculate \( A^2 \): \( A^2 = A \times A = \left[\begin{array}{rr} -8 & 5 \\ 2 & 4 \end{array}\right] \left[\begin{array}{rr} -8 & 5 \\ 2 & 4 \end{array}\right] \) \( A^2 = \left[\begin{array}{cc} (-8)(-8)+(5)(2) & (-8)(5)+(5)(4) \\ (2)(-8)+(4)(2) & (2)(5)+(4)(4) \end{array}\right] \) \( A^2 = \left[\begin{array}{cc} 64+10 & -40+20 \\ -16+8 & 10+16 \end{array}\right] = \left[\begin{array}{rr} 74 & -20 \\ -8 & 26 \end{array}\right] \) Now, we need to show that \( A^2 + 4A - 42I = O \). Calculate \( 4A \): \( 4A = 4 \left[\begin{array}{rr} -8 & 5 \\ 2 & 4 \end{array}\right] = \left[\begin{array}{rr} -32 & 20 \\ 8 & 16 \end{array}\right] \) And for \( 42I \), where I is the identity matrix of order 2: \( 42I = 42 \left[\begin{array}{rr} 1 & 0 \\ 0 & 1 \end{array}\right] = \left[\begin{array}{rr} 42 & 0 \\ 0 & 42 \end{array}\right] \) Substitute these into the equation \( A^2 + 4A - 42I \): \( A^2 + 4A - 42I = \left[\begin{array}{rr} 74 & -20 \\ -8 & 26 \end{array}\right] + \left[\begin{array}{rr} -32 & 20 \\ 8 & 16 \end{array}\right] - \left[\begin{array}{rr} 42 & 0 \\ 0 & 42 \end{array}\right] \) \( A^2 + 4A - 42I = \left[\begin{array}{cc} 74-32-42 & -20+20-0 \\ -8+8-0 & 26+16-42 \end{array}\right] \) \( A^2 + 4A - 42I = \left[\begin{array}{cc} 42-42 & 0 \\ 0 & 42-42 \end{array}\right] = \left[\begin{array}{cc} 0 & 0 \\ 0 & 0 \end{array}\right] = O \) Thus, \( A^2 + 4A - 42I = O \). This confirms the matrix satisfies the given equation. To find \( A^{-1} \), we use the equation \( A^2 + 4A - 42I = O \). From this equation, we can write: \( 42I = A^2 + 4A \) Pre-multiply both sides by \( A^{-1} \): \( A^{-1}(42I) = A^{-1}(A^2 + 4A) \) \( 42(A^{-1}I) = A^{-1}A^2 + A^{-1}(4A) \) \( 42A^{-1} = (A^{-1}A)A + 4(A^{-1}A) \) \( 42A^{-1} = IA + 4I \) \( 42A^{-1} = A + 4I \) Solve for \( A^{-1} \): \( A^{-1} = \frac{1}{42}(A + 4I) \) Substitute the matrices: \( A^{-1} = \frac{1}{42} \left( \left[\begin{array}{rr} -8 & 5 \\ 2 & 4 \end{array}\right] + 4\left[\begin{array}{rr} 1 & 0 \\ 0 & 1 \end{array}\right] \right) \) \( A^{-1} = \frac{1}{42} \left( \left[\begin{array}{rr} -8 & 5 \\ 2 & 4 \end{array}\right] + \left[\begin{array}{rr} 4 & 0 \\ 0 & 4 \end{array}\right] \right) \) \( A^{-1} = \frac{1}{42} \left[\begin{array}{cc} -8+4 & 5+0 \\ 2+0 & 4+4 \end{array}\right] \) \( A^{-1} = \frac{1}{42} \left[\begin{array}{rr} -4 & 5 \\ 2 & 8 \end{array}\right] \)In simple words: We proved that matrix A satisfies the given algebraic equation by calculating its square (\( A^2 \)) and then combining it with \( 4A \) and \( -42I \). The result was a zero matrix, confirming the equation is true for A. Then, to find the inverse of A, we used this same equation. By multiplying the equation by \( A^{-1} \) and rearranging the terms, we found a simple way to express \( A^{-1} \), which is more efficient than the usual method for inverses.

🎯 Exam Tip: When a matrix is said to satisfy a polynomial equation, you can use that equation to find its inverse. The steps involve calculating powers of A, scalar multiples, and then multiplying the entire equation by \( A^{-1} \). This method is generally faster and less complex than finding the determinant and adjoint, especially for matrices of higher order.

Question 19. If \( A = \left[\begin{array}{ll} 4 & 3 \\ 2 & 5 \end{array}\right], \) find x and y such that \( A^2 - xA + yI = 0 \).
Answer:Given matrix \( A = \left[\begin{array}{ll} 4 & 3 \\ 2 & 5 \end{array}\right] \). We are given the equation \( A^2 - xA + yI = O \). First, calculate \( A^2 \): \( A^2 = A \times A = \left[\begin{array}{ll} 4 & 3 \\ 2 & 5 \end{array}\right] \left[\begin{array}{ll} 4 & 3 \\ 2 & 5 \end{array}\right] \) \( A^2 = \left[\begin{array}{cc} (4)(4)+(3)(2) & (4)(3)+(3)(5) \\ (2)(4)+(5)(2) & (2)(3)+(5)(5) \end{array}\right] \) \( A^2 = \left[\begin{array}{cc} 16+6 & 12+15 \\ 8+10 & 6+25 \end{array}\right] = \left[\begin{array}{cc} 22 & 27 \\ 18 & 31 \end{array}\right] \) Now, calculate \( xA \): \( xA = x \left[\begin{array}{ll} 4 & 3 \\ 2 & 5 \end{array}\right] = \left[\begin{array}{cc} 4x & 3x \\ 2x & 5x \end{array}\right] \) And for \( yI \), where I is the identity matrix of order 2: \( yI = y \left[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right] = \left[\begin{array}{cc} y & 0 \\ 0 & y \end{array}\right] \) Substitute these into the equation \( A^2 - xA + yI = O \): \( \left[\begin{array}{cc} 22 & 27 \\ 18 & 31 \end{array}\right] - \left[\begin{array}{cc} 4x & 3x \\ 2x & 5x \end{array}\right] + \left[\begin{array}{cc} y & 0 \\ 0 & y \end{array}\right] = \left[\begin{array}{cc} 0 & 0 \\ 0 & 0 \end{array}\right] \) Perform the matrix operations: \( \left[\begin{array}{cc} 22-4x+y & 27-3x \\ 18-2x & 31-5x+y \end{array}\right] = \left[\begin{array}{cc} 0 & 0 \\ 0 & 0 \end{array}\right] \) For two matrices to be equal, their corresponding elements must be equal. This gives us a system of equations: 1. \( 22 - 4x + y = 0 \) 2. \( 27 - 3x = 0 \) 3. \( 18 - 2x = 0 \) 4. \( 31 - 5x + y = 0 \) From equation (2): \( 27 - 3x = 0 \) \( 3x = 27 \) \( x = 9 \) From equation (3): \( 18 - 2x = 0 \) \( 2x = 18 \) \( x = 9 \) Both equations (2) and (3) give \( x = 9 \), which is consistent. Now, substitute \( x = 9 \) into equation (1): \( 22 - 4(9) + y = 0 \) \( 22 - 36 + y = 0 \) \( -14 + y = 0 \) \( y = 14 \) Now, verify these values using equation (4): \( 31 - 5x + y = 0 \) \( 31 - 5(9) + 14 = 0 \) \( 31 - 45 + 14 = 0 \) \( -14 + 14 = 0 \) \( 0 = 0 \) The values \( x = 9 \) and \( y = 14 \) satisfy all equations. Therefore, \( x = 9 \) and \( y = 14 \). This process is an application of the Cayley-Hamilton theorem.In simple words: We were given a matrix A and an equation involving A, x, and y. We first calculated \( A^2 \) by multiplying A by itself. Then, we wrote out \( xA \) and \( yI \) (where I is the identity matrix) as matrices. We put all these into the given equation and added/subtracted the matrices. Since the final result was a zero matrix, we set each element of the resulting matrix equal to zero, which gave us a set of simple equations. Solving these equations helped us find the values of x and y.

🎯 Exam Tip: When given a matrix and a polynomial equation to find unknown scalars, calculate each matrix term (like \( A^2 \), \( xA \), \( yI \)) separately. Combine these terms into a single matrix equation. Then, equate corresponding elements of this matrix to the zero matrix to form a system of linear equations. Solve this system to find the unknown scalar values, ensuring consistency across all equations.

Question 20. If \( A = \left[\begin{array}{rrr} 1 & -1 & 1 \\ 2 & -1 & 0 \\ 1 & 0 & 0 \end{array}\right], \) find \( A^2 \) and show that \( A^2 = A^{-1} \).
Answer:Given matrix \( A = \left[\begin{array}{rrr} 1 & -1 & 1 \\ 2 & -1 & 0 \\ 1 & 0 & 0 \end{array}\right] \). First, find \( A^2 \): \( A^2 = A \times A = \left[\begin{array}{rrr} 1 & -1 & 1 \\ 2 & -1 & 0 \\ 1 & 0 & 0 \end{array}\right] \left[\begin{array}{rrr} 1 & -1 & 1 \\ 2 & -1 & 0 \\ 1 & 0 & 0 \end{array}\right] \) \( A^2 = \left[\begin{array}{ccc} (1)(1)+(-1)(2)+(1)(1) & (1)(-1)+(-1)(-1)+(1)(0) & (1)(1)+(-1)(0)+(1)(0) \\ (2)(1)+(-1)(2)+(0)(1) & (2)(-1)+(-1)(-1)+(0)(0) & (2)(1)+(-1)(0)+(0)(0) \\ (1)(1)+(0)(2)+(0)(1) & (1)(-1)+(0)(-1)+(0)(0) & (1)(1)+(0)(0)+(0)(0) \end{array}\right] \) \( A^2 = \left[\begin{array}{ccc} 1-2+1 & -1+1+0 & 1+0+0 \\ 2-2+0 & -2+1+0 & 2+0+0 \\ 1+0+0 & -1+0+0 & 1+0+0 \end{array}\right] \) \( A^2 = \left[\begin{array}{rrr} 0 & 0 & 1 \\ 0 & -1 & 2 \\ 1 & -1 & 1 \end{array}\right] \) Next, we need to show that \( A^2 = A^{-1} \). This means that if we multiply \( A^2 \) by A, we should get the identity matrix I, because \( A^{-1}A = I \). So, we need to show \( A^2 A = I \). We already have \( A^2 \), so let's compute \( A^2 A \): \( A^2 A = \left[\begin{array}{rrr} 0 & 0 & 1 \\ 0 & -1 & 2 \\ 1 & -1 & 1 \end{array}\right] \left[\begin{array}{rrr} 1 & -1 & 1 \\ 2 & -1 & 0 \\ 1 & 0 & 0 \end{array}\right] \) \( A^2 A = \left[\begin{array}{ccc} (0)(1)+(0)(2)+(1)(1) & (0)(-1)+(0)(-1)+(1)(0) & (0)(1)+(0)(0)+(1)(0) \\ (0)(1)+(-1)(2)+(2)(1) & (0)(-1)+(-1)(-1)+(2)(0) & (0)(1)+(-1)(0)+(2)(0) \\ (1)(1)+(-1)(2)+(1)(1) & (1)(-1)+(-1)(-1)+(1)(0) & (1)(1)+(-1)(0)+(1)(0) \end{array}\right] \) \( A^2 A = \left[\begin{array}{ccc} 0+0+1 & 0+0+0 & 0+0+0 \\ 0-2+2 & 0+1+0 & 0+0+0 \\ 1-2+1 & -1+1+0 & 1+0+0 \end{array}\right] \) \( A^2 A = \left[\begin{array}{rrr} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right] = I \) Since \( A^2 A = I \), it implies that \( A^{-1} = A^2 \). This is a property of inverse matrices. Alternatively, we can find \( A^{-1} \) directly and then compare. First, find the determinant of A: \( |A| = \left|\begin{array}{rrr} 1 & -1 & 1 \\ 2 & -1 & 0 \\ 1 & 0 & 0 \end{array}\right| \) Expanding along C₃ (since it has two zeros): \( |A| = 1 \left|\begin{array}{cc} 2 & -1 \\ 1 & 0 \end{array}\right| - 0 + 0 \) \( |A| = 1((2)(0) - (-1)(1)) = 1(0 - (-1)) = 1(1) = 1 \) Since \( |A| = 1 \neq 0 \), \( A^{-1} \) exists. Next, find the cofactors of A: Cofactor of \( a_{11} = \left|\begin{array}{cc} -1 & 0 \\ 0 & 0 \end{array}\right| = 0-0 = 0 \) Cofactor of \( a_{12} = -\left|\begin{array}{cc} 2 & 0 \\ 1 & 0 \end{array}\right| = -(0-0) = 0 \) Cofactor of \( a_{13} = \left|\begin{array}{cc} 2 & -1 \\ 1 & 0 \end{array}\right| = 0-(-1) = 1 \) Cofactor of \( a_{21} = -\left|\begin{array}{cc} -1 & 1 \\ 0 & 0 \end{array}\right| = -(0-0) = 0 \) Cofactor of \( a_{22} = \left|\begin{array}{cc} 1 & 1 \\ 1 & 0 \end{array}\right| = 0-1 = -1 \) Cofactor of \( a_{23} = -\left|\begin{array}{cc} 1 & -1 \\ 1 & 0 \end{array}\right| = -(0-(-1)) = -1 \) Cofactor of \( a_{31} = \left|\begin{array}{cc} -1 & 1 \\ -1 & 0 \end{array}\right| = 0-(-1) = 1 \) Cofactor of \( a_{32} = -\left|\begin{array}{cc} 1 & 1 \\ 2 & 0 \end{array}\right| = -(0-2) = 2 \) Cofactor of \( a_{33} = \left|\begin{array}{cc} 1 & -1 \\ 2 & -1 \end{array}\right| = -1-(-2) = -1+2 = 1 \) The cofactor matrix is: \( C = \left[\begin{array}{rrr} 0 & 0 & 1 \\ 0 & -1 & -1 \\ 1 & 2 & 1 \end{array}\right] \) The adjoint of A is the transpose of the cofactor matrix: \( \text{adj } A = C^T = \left[\begin{array}{rrr} 0 & 0 & 1 \\ 0 & -1 & 2 \\ 1 & -1 & 1 \end{array}\right] \) Now, calculate \( A^{-1} = \frac{1}{|A|} \text{adj } A \): \( A^{-1} = \frac{1}{1} \left[\begin{array}{rrr} 0 & 0 & 1 \\ 0 & -1 & 2 \\ 1 & -1 & 1 \end{array}\right] = \left[\begin{array}{rrr} 0 & 0 & 1 \\ 0 & -1 & 2 \\ 1 & -1 & 1 \end{array}\right] \) Comparing \( A^2 \) with \( A^{-1} \), we see they are equal: \( A^2 = \left[\begin{array}{rrr} 0 & 0 & 1 \\ 0 & -1 & 2 \\ 1 & -1 & 1 \end{array}\right] \) and \( A^{-1} = \left[\begin{array}{rrr} 0 & 0 & 1 \\ 0 & -1 & 2 \\ 1 & -1 & 1 \end{array}\right] \) Thus, \( A^2 = A^{-1} \).In simple words: First, we multiplied matrix A by itself to find \( A^2 \). Then, to show that \( A^2 \) is also the inverse of A, we checked if multiplying \( A^2 \) by A would give us the identity matrix (I). Since it did, this proved that \( A^2 \) is indeed the inverse. We also calculated the inverse directly using the determinant and adjoint method, and confirmed that the result was the same as \( A^2 \), which is a unique property for some matrices.

🎯 Exam Tip: When asked to show \( A^2 = A^{-1} \), the most direct way is to verify that \( A^3 = I \). If this is true, then multiplying \( A^3 = I \) by \( A^{-1} \) gives \( A^2 = A^{-1}I \), which simplifies to \( A^2 = A^{-1} \). Alternatively, calculate \( A^{-1} \) using the adjoint method and directly compare it with \( A^2 \).

Question 20. If \( A = \begin{bmatrix} 1 & -1 & 1 \\ 2 & -1 & 0 \\ 1 & 0 & 0 \end{bmatrix} \), find \( A^2 \) and show that \( A^2 = A^{-1} \).
Answer: First, we will find \( A^2 \) by multiplying matrix A by itself.
\( A^2 = A \cdot A = \begin{bmatrix} 1 & -1 & 1 \\ 2 & -1 & 0 \\ 1 & 0 & 0 \end{bmatrix} \begin{bmatrix} 1 & -1 & 1 \\ 2 & -1 & 0 \\ 1 & 0 & 0 \end{bmatrix} \)
\( = \begin{bmatrix} (1)(1)+(-1)(2)+(1)(1) & (1)(-1)+(-1)(-1)+(1)(0) & (1)(1)+(-1)(0)+(1)(0) \\ (2)(1)+(-1)(2)+(0)(1) & (2)(-1)+(-1)(-1)+(0)(0) & (2)(1)+(-1)(0)+(0)(0) \\ (1)(1)+(0)(2)+(0)(1) & (1)(-1)+(0)(-1)+(0)(0) & (1)(1)+(0)(0)+(0)(0) \end{bmatrix} \)
\( = \begin{bmatrix} 1-2+1 & -1+1+0 & 1+0+0 \\ 2-2+0 & -2+1+0 & 2+0+0 \\ 1+0+0 & -1+0+0 & 1+0+0 \end{bmatrix} \)
\( \implies A^2 = \begin{bmatrix} 0 & 0 & 1 \\ 0 & -1 & 2 \\ 1 & -1 & 1 \end{bmatrix} \) (Equation 1)

Next, we will find the inverse of A, \( A^{-1} \). For this, we first need to find the determinant of A, \( |A| \).
\( |A| = \begin{vmatrix} 1 & -1 & 1 \\ 2 & -1 & 0 \\ 1 & 0 & 0 \end{vmatrix} \)
We expand along the third column (C3) because it has two zeros, making the calculation simpler.
\( |A| = 1 \cdot \begin{vmatrix} 2 & -1 \\ 1 & 0 \end{vmatrix} - 0 \cdot \begin{vmatrix} 1 & -1 \\ 1 & 0 \end{vmatrix} + 0 \cdot \begin{vmatrix} 1 & -1 \\ 2 & -1 \end{vmatrix} \)
\( = 1((2)(0) - (-1)(1)) = 1(0+1) = 1 \)
Since \( |A| = 1 \neq 0 \), the inverse of A exists. The determinant tells us if a matrix can be inverted.

Now, we calculate the cofactors of each element in matrix A to find the adjoint of A.
Cofactors for Row 1:
\( C_{11} = \begin{vmatrix} -1 & 0 \\ 0 & 0 \end{vmatrix} = 0 \)
\( C_{12} = - \begin{vmatrix} 2 & 0 \\ 1 & 0 \end{vmatrix} = 0 \)
\( C_{13} = \begin{vmatrix} 2 & -1 \\ 1 & 0 \end{vmatrix} = 1 \)
Cofactors for Row 2:
\( C_{21} = - \begin{vmatrix} -1 & 1 \\ 0 & 0 \end{vmatrix} = 0 \)
\( C_{22} = \begin{vmatrix} 1 & 1 \\ 1 & 0 \end{vmatrix} = -1 \)
\( C_{23} = - \begin{vmatrix} 1 & -1 \\ 1 & 0 \end{vmatrix} = -1 \)
Cofactors for Row 3:
\( C_{31} = \begin{vmatrix} -1 & 1 \\ -1 & 0 \end{vmatrix} = 1 \)
\( C_{32} = - \begin{vmatrix} 1 & 1 \\ 2 & 0 \end{vmatrix} = 2 \)
\( C_{33} = \begin{vmatrix} 1 & -1 \\ 2 & -1 \end{vmatrix} = 1 \)

The matrix of cofactors is: \( \begin{bmatrix} 0 & 0 & 1 \\ 0 & -1 & -1 \\ 1 & 2 & 1 \end{bmatrix} \)
The adjoint of A (adj A) is the transpose of the cofactor matrix.
\( \text{adj } A = \begin{bmatrix} 0 & 0 & 1 \\ 0 & -1 & 2 \\ 1 & -1 & 1 \end{bmatrix} \)

Now we can find \( A^{-1} \).
\( A^{-1} = \frac{1}{|A|} \text{adj } A = \frac{1}{1} \begin{bmatrix} 0 & 0 & 1 \\ 0 & -1 & 2 \\ 1 & -1 & 1 \end{bmatrix} \)
\( \implies A^{-1} = \begin{bmatrix} 0 & 0 & 1 \\ 0 & -1 & 2 \\ 1 & -1 & 1 \end{bmatrix} \) (Equation 2)

By comparing Equation 1 (\( A^2 \)) and Equation 2 (\( A^{-1} \)), we can see that both matrices are identical. Thus, \( A^2 = A^{-1} \) is shown.
In simple words: First, we multiplied matrix A by itself to get \( A^2 \). Then, we found the inverse of A, \( A^{-1} \), by calculating its determinant and cofactors. When we looked at the final matrices for \( A^2 \) and \( A^{-1} \), they were exactly the same, proving the statement.

🎯 Exam Tip: When showing \( A^2 = A^{-1} \), always calculate both independently and then compare the final matrices. Expanding along a row or column with more zeros simplifies determinant calculations for 3x3 matrices.

Question 21. If \( A = \begin{bmatrix} 1 & 1 & 2 \\ 1 & 9 & 3 \\ 1 & 4 & 2 \end{bmatrix} \) and \( B = \begin{bmatrix} 1 & 2 & 0 \\ 2 & 3 & -1 \\ 1 & -1 & 3 \end{bmatrix} \), Verify that \( (AB)^{-1} = B^{-1}A^{-1} \).
Answer: We need to verify the property \( (AB)^{-1} = B^{-1}A^{-1} \) for the given matrices A and B. We will calculate both sides of the equation.

First, let's find \( A^{-1} \). We start by finding the determinant of A.
\( |A| = \begin{vmatrix} 1 & 1 & 2 \\ 1 & 9 & 3 \\ 1 & 4 & 2 \end{vmatrix} \)
Expanding along Row 1 (R1):
\( |A| = 1(9 \cdot 2 - 3 \cdot 4) - 1(1 \cdot 2 - 3 \cdot 1) + 2(1 \cdot 4 - 9 \cdot 1) \)
\( = 1(18 - 12) - 1(2 - 3) + 2(4 - 9) \)
\( = 1(6) - 1(-1) + 2(-5) \)
\( = 6 + 1 - 10 = -3 \)
Since \( |A| = -3 \neq 0 \), the inverse of A exists.

Now we find the cofactors of A:
Cofactors for Row 1:
\( C_{11} = \begin{vmatrix} 9 & 3 \\ 4 & 2 \end{vmatrix} = 18 - 12 = 6 \)
\( C_{12} = - \begin{vmatrix} 1 & 3 \\ 1 & 2 \end{vmatrix} = - (2 - 3) = 1 \)
\( C_{13} = \begin{vmatrix} 1 & 9 \\ 1 & 4 \end{vmatrix} = 4 - 9 = -5 \)
Cofactors for Row 2:
\( C_{21} = - \begin{vmatrix} 1 & 2 \\ 4 & 2 \end{vmatrix} = - (2 - 8) = 6 \)
\( C_{22} = \begin{vmatrix} 1 & 2 \\ 1 & 2 \end{vmatrix} = 2 - 2 = 0 \)
\( C_{23} = - \begin{vmatrix} 1 & 1 \\ 1 & 4 \end{vmatrix} = - (4 - 1) = -3 \)
Cofactors for Row 3:
\( C_{31} = \begin{vmatrix} 1 & 2 \\ 9 & 3 \end{vmatrix} = 3 - 18 = -15 \)
\( C_{32} = - \begin{vmatrix} 1 & 2 \\ 1 & 3 \end{vmatrix} = - (3 - 2) = -1 \)
\( C_{33} = \begin{vmatrix} 1 & 1 \\ 1 & 9 \end{vmatrix} = 9 - 1 = 8 \)

The adjoint of A is the transpose of the cofactor matrix:
\( \text{adj } A = \begin{bmatrix} 6 & 6 & -15 \\ 1 & 0 & -1 \\ -5 & -3 & 8 \end{bmatrix} \)
So, \( A^{-1} = \frac{1}{|A|} \text{adj } A = \frac{1}{-3} \begin{bmatrix} 6 & 6 & -15 \\ 1 & 0 & -1 \\ -5 & -3 & 8 \end{bmatrix} \)

Next, let's find \( B^{-1} \). We start by finding the determinant of B.
\( |B| = \begin{vmatrix} 1 & 2 & 0 \\ 2 & 3 & -1 \\ 1 & -1 & 3 \end{vmatrix} \)
Expanding along Row 1 (R1):
\( |B| = 1(3 \cdot 3 - (-1)(-1)) - 2(2 \cdot 3 - (-1)(1)) + 0(2(-1) - 3(1)) \)
\( = 1(9 - 1) - 2(6 - (-1)) + 0 \)
\( = 1(8) - 2(7) = 8 - 14 = -6 \)
Since \( |B| = -6 \neq 0 \), the inverse of B exists.

Now we find the cofactors of B:
Cofactors for Row 1:
\( C_{11} = \begin{vmatrix} 3 & -1 \\ -1 & 3 \end{vmatrix} = 9 - 1 = 8 \)
\( C_{12} = - \begin{vmatrix} 2 & -1 \\ 1 & 3 \end{vmatrix} = - (6 - (-1)) = -7 \)
\( C_{13} = \begin{vmatrix} 2 & 3 \\ 1 & -1 \end{vmatrix} = -2 - 3 = -5 \)
Cofactors for Row 2:
\( C_{21} = - \begin{vmatrix} 2 & 0 \\ -1 & 3 \end{vmatrix} = - (6 - 0) = -6 \)
\( C_{22} = \begin{vmatrix} 1 & 0 \\ 1 & 3 \end{vmatrix} = 3 - 0 = 3 \)
\( C_{23} = - \begin{vmatrix} 1 & 2 \\ 1 & -1 \end{vmatrix} = - (-1 - 2) = 3 \)
Cofactors for Row 3:
\( C_{31} = \begin{vmatrix} 2 & 0 \\ 3 & -1 \end{vmatrix} = -2 - 0 = -2 \)
\( C_{32} = - \begin{vmatrix} 1 & 0 \\ 2 & -1 \end{vmatrix} = - (-1 - 0) = 1 \)
\( C_{33} = \begin{vmatrix} 1 & 2 \\ 2 & 3 \end{vmatrix} = 3 - 4 = -1 \)

The adjoint of B is the transpose of the cofactor matrix:
\( \text{adj } B = \begin{bmatrix} 8 & -6 & -2 \\ -7 & 3 & 1 \\ -5 & 3 & -1 \end{bmatrix} \)
So, \( B^{-1} = \frac{1}{|B|} \text{adj } B = \frac{1}{-6} \begin{bmatrix} 8 & -6 & -2 \\ -7 & 3 & 1 \\ -5 & 3 & -1 \end{bmatrix} \)

Now, let's calculate the right-hand side of the equation: \( B^{-1}A^{-1} \).
\( B^{-1}A^{-1} = \frac{1}{-6} \begin{bmatrix} 8 & -6 & -2 \\ -7 & 3 & 1 \\ -5 & 3 & -1 \end{bmatrix} \cdot \frac{1}{-3} \begin{bmatrix} 6 & 6 & -15 \\ 1 & 0 & -1 \\ -5 & -3 & 8 \end{bmatrix} \)
\( = \frac{1}{18} \begin{bmatrix} 48-6+10 & 48+0+6 & -120+6-16 \\ -42+3-5 & -42+0-3 & 105-3+8 \\ -30+3+5 & -30+0+3 & 75-3-8 \end{bmatrix} \)
\( \implies B^{-1}A^{-1} = \frac{1}{18} \begin{bmatrix} 52 & 54 & -130 \\ -44 & -45 & 110 \\ -22 & -27 & 64 \end{bmatrix} \) (Equation 1)

Next, we calculate the left-hand side of the equation: \( (AB)^{-1} \). First, we find the product AB.
\( AB = \begin{bmatrix} 1 & 1 & 2 \\ 1 & 9 & 3 \\ 1 & 4 & 2 \end{bmatrix} \begin{bmatrix} 1 & 2 & 0 \\ 2 & 3 & -1 \\ 1 & -1 & 3 \end{bmatrix} \)
\( = \begin{bmatrix} 1(1)+1(2)+2(1) & 1(2)+1(3)+2(-1) & 1(0)+1(-1)+2(3) \\ 1(1)+9(2)+3(1) & 1(2)+9(3)+3(-1) & 1(0)+9(-1)+3(3) \\ 1(1)+4(2)+2(1) & 1(2)+4(3)+2(-1) & 1(0)+4(-1)+2(3) \end{bmatrix} \)
\( = \begin{bmatrix} 1+2+2 & 2+3-2 & 0-1+6 \\ 1+18+3 & 2+27-3 & 0-9+9 \\ 1+8+2 & 2+12-2 & 0-4+6 \end{bmatrix} \)
\( \implies AB = \begin{bmatrix} 5 & 3 & 5 \\ 22 & 26 & 0 \\ 11 & 12 & 2 \end{bmatrix} \)

Now, we find the determinant of AB.
\( |AB| = \begin{vmatrix} 5 & 3 & 5 \\ 22 & 26 & 0 \\ 11 & 12 & 2 \end{vmatrix} \)
Expanding along Row 1 (R1):
\( |AB| = 5(26 \cdot 2 - 0 \cdot 12) - 3(22 \cdot 2 - 0 \cdot 11) + 5(22 \cdot 12 - 26 \cdot 11) \)
\( = 5(52) - 3(44) + 5(264 - 286) \)
\( = 260 - 132 + 5(-22) \)
\( = 260 - 132 - 110 = 18 \)
Since \( |AB| = 18 \neq 0 \), the inverse of AB exists.

Now we find the cofactors of AB:
Cofactors for Row 1:
\( C_{11} = \begin{vmatrix} 26 & 0 \\ 12 & 2 \end{vmatrix} = 52 - 0 = 52 \)
\( C_{12} = - \begin{vmatrix} 22 & 0 \\ 11 & 2 \end{vmatrix} = - (44 - 0) = -44 \)
\( C_{13} = \begin{vmatrix} 22 & 26 \\ 11 & 12 \end{vmatrix} = 264 - 286 = -22 \)
Cofactors for Row 2:
\( C_{21} = - \begin{vmatrix} 3 & 5 \\ 12 & 2 \end{vmatrix} = - (6 - 60) = 54 \)
\( C_{22} = \begin{vmatrix} 5 & 5 \\ 11 & 2 \end{vmatrix} = 10 - 55 = -45 \)
\( C_{23} = - \begin{vmatrix} 5 & 3 \\ 11 & 12 \end{vmatrix} = - (60 - 33) = -27 \)
Cofactors for Row 3:
\( C_{31} = \begin{vmatrix} 3 & 5 \\ 26 & 0 \end{vmatrix} = 0 - 130 = -130 \)
\( C_{32} = - \begin{vmatrix} 5 & 5 \\ 22 & 0 \end{vmatrix} = - (0 - 110) = 110 \)
\( C_{33} = \begin{vmatrix} 5 & 3 \\ 22 & 26 \end{vmatrix} = 130 - 66 = 64 \)

The adjoint of AB is the transpose of the cofactor matrix:
\( \text{adj } (AB) = \begin{bmatrix} 52 & 54 & -130 \\ -44 & -45 & 110 \\ -22 & -27 & 64 \end{bmatrix} \)
So, \( (AB)^{-1} = \frac{1}{|AB|} \text{adj } (AB) = \frac{1}{18} \begin{bmatrix} 52 & 54 & -130 \\ -44 & -45 & 110 \\ -22 & -27 & 64 \end{bmatrix} \) (Equation 2)

By comparing Equation 1 and Equation 2, we see that \( (AB)^{-1} \) is equal to \( B^{-1}A^{-1} \). This verifies the property.
In simple words: We took two matrices, A and B. We calculated the inverse of A and the inverse of B separately. Then we multiplied those inverses together in reverse order to get \( B^{-1}A^{-1} \). After that, we first multiplied A and B to get AB, and then found the inverse of that product, \( (AB)^{-1} \). Both final results were exactly the same, which confirms the property.

🎯 Exam Tip: Remember to always perform matrix multiplication and inversion steps carefully. A common error is mixing up the order of multiplication for \( B^{-1}A^{-1} \) versus \( A^{-1}B^{-1} \).

Question 22. If \( A = \frac{1}{9}\begin{bmatrix} -8 & 1 & 4 \\ 4 & 4 & 7 \\ 1 & -8 & 4 \end{bmatrix} \), Prove that \( A^{-1} = A' \).
Answer: We are given matrix A and need to prove that its inverse \( A^{-1} \) is equal to its transpose \( A' \).

First, let's find the determinant of A. For a matrix of the form \( kM \), its determinant is \( |kM| = k^n|M| \), where n is the order of the matrix (here, n=3).
Let \( M = \begin{bmatrix} -8 & 1 & 4 \\ 4 & 4 & 7 \\ 1 & -8 & 4 \end{bmatrix} \)
\( |M| = -8(4 \cdot 4 - 7(-8)) - 1(4 \cdot 4 - 7 \cdot 1) + 4(4(-8) - 4 \cdot 1) \)
\( = -8(16 + 56) - 1(16 - 7) + 4(-32 - 4) \)
\( = -8(72) - 1(9) + 4(-36) \)
\( = -576 - 9 - 144 = -729 \)
Now, \( |A| = (\frac{1}{9})^3 |M| = \frac{1}{729} (-729) = -1 \)
Since \( |A| = -1 \neq 0 \), the inverse \( A^{-1} \) exists.

Next, we find the cofactors of matrix M:
Cofactors for Row 1:
\( M_{11} = \begin{vmatrix} 4 & 7 \\ -8 & 4 \end{vmatrix} = 16 - (-56) = 72 \)
\( M_{12} = - \begin{vmatrix} 4 & 7 \\ 1 & 4 \end{vmatrix} = - (16 - 7) = -9 \)
\( M_{13} = \begin{vmatrix} 4 & 4 \\ 1 & -8 \end{vmatrix} = -32 - 4 = -36 \)
Cofactors for Row 2:
\( M_{21} = - \begin{vmatrix} 1 & 4 \\ -8 & 4 \end{vmatrix} = - (4 - (-32)) = -36 \)
\( M_{22} = \begin{vmatrix} -8 & 4 \\ 1 & 4 \end{vmatrix} = -32 - 4 = -36 \)
\( M_{23} = - \begin{vmatrix} -8 & 1 \\ 1 & -8 \end{vmatrix} = - (64 - 1) = -63 \)
Cofactors for Row 3:
\( M_{31} = \begin{vmatrix} 1 & 4 \\ 4 & 7 \end{vmatrix} = 7 - 16 = -9 \)
\( M_{32} = - \begin{vmatrix} -8 & 4 \\ 4 & 7 \end{vmatrix} = - (-56 - 16) = 72 \)
\( M_{33} = \begin{vmatrix} -8 & 1 \\ 4 & 4 \end{vmatrix} = -32 - 4 = -36 \)

The adjoint of M (adj M) is the transpose of its cofactor matrix:
\( \text{adj } M = \begin{bmatrix} 72 & -36 & -9 \\ -9 & -36 & 72 \\ -36 & -63 & -36 \end{bmatrix} \)

Now, we find \( A^{-1} \). Since \( A = \frac{1}{9}M \), its inverse \( A^{-1} \) can be found using the formula \( A^{-1} = \frac{1}{|A|} (\frac{1}{9})^{3-1} \text{adj } M \). This simplifies to \( A^{-1} = \frac{1}{|A|} \frac{1}{81} \text{adj } M \).
\( A^{-1} = \frac{1}{-1} \cdot \frac{1}{81} \begin{bmatrix} 72 & -36 & -9 \\ -9 & -36 & 72 \\ -36 & -63 & -36 \end{bmatrix} \)
\( = -\frac{1}{81} \begin{bmatrix} 72 & -36 & -9 \\ -9 & -36 & 72 \\ -36 & -63 & -36 \end{bmatrix} \)
\( \implies A^{-1} = \frac{1}{81} \begin{bmatrix} -72 & 36 & 9 \\ 9 & 36 & -72 \\ 36 & 63 & 36 \end{bmatrix} \)
We can factor out 9 from each element in the matrix inside the bracket:
\( A^{-1} = \frac{9}{81} \begin{bmatrix} -8 & 4 & 1 \\ 1 & 4 & -8 \\ 4 & 7 & 4 \end{bmatrix} = \frac{1}{9} \begin{bmatrix} -8 & 4 & 1 \\ 1 & 4 & -8 \\ 4 & 7 & 4 \end{bmatrix} \) (Equation 1)

Next, let's find the transpose of A, \( A' \). The transpose is obtained by interchanging rows and columns of A.
\( A' = \left( \frac{1}{9}\begin{bmatrix} -8 & 1 & 4 \\ 4 & 4 & 7 \\ 1 & -8 & 4 \end{bmatrix} \right)' \)
\( \implies A' = \frac{1}{9}\begin{bmatrix} -8 & 4 & 1 \\ 1 & 4 & -8 \\ 4 & 7 & 4 \end{bmatrix} \) (Equation 2)

By comparing Equation 1 and Equation 2, we can see that \( A^{-1} = A' \). This proves the given statement.
In simple words: We had to show that the inverse of matrix A is the same as its transpose. We calculated the inverse of A using its determinant and cofactors. Then we simply flipped the rows and columns of A to get its transpose. Both results matched perfectly, proving the statement. This kind of matrix is special because it's its own inverse, but with a sign change, and its transpose.

🎯 Exam Tip: When proving \( A^{-1} = A' \) for a matrix with a scalar multiple, remember to apply the scalar property \( |kA| = k^n|A| \) for the determinant and \( \text{adj}(kA) = k^{n-1} \text{adj}(A) \) for the adjoint. Alternatively, compute \( A^{-1} \) and \( A' \) separately and compare the final matrices.