Samacheer Kalvi Class 12 Maths Solutions Chapter 1 Applications of Matrices and Determinants Exercise 1.1

Question 1. Find the adjoint of the following:
(i) \( \begin{bmatrix} -3 & 4 \\ 6 & 2 \end{bmatrix} \)
(ii) \( \begin{bmatrix} 2 & 3 & 1 \\ 3 & 4 & 1 \\ 3 & 7 & 2 \end{bmatrix} \)
(iii) \( \begin{bmatrix} 2 & 2 & 1 \\ 1 & -2 & 2 \\ 1 & -2 & 2 \end{bmatrix} \)
Answer:
(i) For \( A = \begin{bmatrix} -3 & 4 \\ 6 & 2 \end{bmatrix} \), the adjoint matrix is found by swapping the diagonal elements and changing the sign of the off-diagonal elements.
\( \text{adj } A = \begin{bmatrix} 2 & -4 \\ -6 & -3 \end{bmatrix} \)
(ii) For \( A = \begin{bmatrix} 2 & 3 & 1 \\ 3 & 4 & 1 \\ 3 & 7 & 2 \end{bmatrix} \), we first find the cofactor matrix. Each cofactor is the determinant of a smaller matrix, with alternating signs. Then we transpose this matrix to get the adjoint.
\( \text{adj } A = \begin{bmatrix} +(8-7) & -(6-3) & +(21-12) \\ -(6-7) & +(4-3) & -(14-9) \\ +(3-4) & -(2-3) & +(8-9) \end{bmatrix}^T \)
\( = \begin{bmatrix} 1 & -3 & 9 \\ 1 & 1 & -5 \\ -1 & -1 & 1 \end{bmatrix}^T = \begin{bmatrix} 1 & 1 & -1 \\ -3 & 1 & -1 \\ 9 & -5 & 1 \end{bmatrix} \)
(iii) For \( A = \frac{1}{3} \begin{bmatrix} 2 & 2 & 1 \\ 1 & -2 & 2 \\ 1 & -2 & 2 \end{bmatrix} \), to find the adjoint, we use the property \( \text{adj}(kA) = k^{n-1} \text{adj } A \). Here \( k = \frac{1}{3} \) and \( n=3 \). So, \( \text{adj } A = (\frac{1}{3})^{3-1} \text{adj } \begin{bmatrix} 2 & 2 & 1 \\ 1 & -2 & 2 \\ 1 & -2 & 2 \end{bmatrix} = \frac{1}{9} \text{adj } \begin{bmatrix} 2 & 2 & 1 \\ 1 & -2 & 2 \\ 1 & -2 & 2 \end{bmatrix} \).
Let \( B = \begin{bmatrix} 2 & 2 & 1 \\ 1 & -2 & 2 \\ 1 & -2 & 2 \end{bmatrix} \). We calculate \( \text{adj } B \):
\( \text{adj } B = \begin{bmatrix} +(2+4) & -(-4-2) & +(4-1) \\ -(4+2) & +(4-1) & -(-4-2) \\ +(4-1) & -(4+2) & +(2+4) \end{bmatrix}^T \)
\( = \begin{bmatrix} 6 & 6 & 3 \\ -6 & 3 & 6 \\ 3 & -6 & 6 \end{bmatrix}^T = \begin{bmatrix} 6 & -6 & 3 \\ 6 & 3 & -6 \\ 3 & 6 & 6 \end{bmatrix} \)
Therefore, \( \text{adj } A = \frac{1}{9} \begin{bmatrix} 6 & -6 & 3 \\ 6 & 3 & -6 \\ 3 & 6 & 6 \end{bmatrix} = \frac{1}{3} \begin{bmatrix} 2 & -2 & 1 \\ 2 & 1 & -2 \\ 1 & 2 & 2 \end{bmatrix} \)
In simple words: The adjoint of a matrix is found by creating a matrix of cofactors and then transposing it. For a 2x2 matrix, you swap diagonal numbers and change signs of off-diagonal numbers. For a 3x3 matrix, it's a longer process of finding many small determinants and then flipping the result.

🎯 Exam Tip: Remember to apply the correct signs for the cofactor matrix, as mistakes there are common. For larger matrices, organize your calculations carefully to avoid errors.

Question 2. Find the inverse (if it exists) of the following.
(i) \( \begin{bmatrix} -2 & 4 \\ 1 & -3 \end{bmatrix} \)
(ii) \( \begin{bmatrix} 5 & 1 & 1 \\ 1 & 5 & 1 \\ 1 & 1 & 5 \end{bmatrix} \)
(iii) \( \begin{bmatrix} 2 & 3 & 1 \\ 3 & 4 & 1 \\ 3 & 7 & 2 \end{bmatrix} \)
Answer:
The inverse of a matrix \( A \) is given by \( A^{-1} = \frac{1}{|A|} \text{adj } A \), provided that the determinant \( |A| \neq 0 \).
(i) For \( A = \begin{bmatrix} -2 & 4 \\ 1 & -3 \end{bmatrix} \):
First, calculate the determinant: \( |A| = (-2)(-3) - (4)(1) = 6 - 4 = 2 \). Since \( |A| \neq 0 \), the inverse exists.
Next, find the adjoint matrix: \( \text{adj } A = \begin{bmatrix} -3 & -4 \\ -1 & -2 \end{bmatrix} \).
Finally, calculate the inverse: \( A^{-1} = \frac{1}{2} \begin{bmatrix} -3 & -4 \\ -1 & -2 \end{bmatrix} \).
(ii) For \( A = \begin{bmatrix} 5 & 1 & 1 \\ 1 & 5 & 1 \\ 1 & 1 & 5 \end{bmatrix} \):
First, calculate the determinant: \( |A| = 5(25-1) - 1(5-1) + 1(1-5) = 5(24) - 1(4) + 1(-4) = 120 - 4 - 4 = 112 \). Since \( |A| \neq 0 \), the inverse exists.
Next, find the adjoint matrix:
\( \text{adj } A = \begin{bmatrix} +(25-1) & -(5-1) & +(1-5) \\ -(5-1) & +(25-1) & -(5-1) \\ +(1-5) & -(5-1) & +(25-1) \end{bmatrix}^T = \begin{bmatrix} 24 & -4 & -4 \\ -4 & 24 & -4 \\ -4 & -4 & 24 \end{bmatrix} \).
Finally, calculate the inverse: \( A^{-1} = \frac{1}{112} \begin{bmatrix} 24 & -4 & -4 \\ -4 & 24 & -4 \\ -4 & -4 & 24 \end{bmatrix} = \frac{1}{28} \begin{bmatrix} 6 & -1 & -1 \\ -1 & 6 & -1 \\ -1 & -1 & 6 \end{bmatrix} \).
(iii) For \( A = \begin{bmatrix} 2 & 3 & 1 \\ 3 & 4 & 1 \\ 3 & 7 & 2 \end{bmatrix} \):
First, calculate the determinant: \( |A| = 2(8-7) - 3(6-3) + 1(21-12) = 2(1) - 3(3) + 1(9) = 2 - 9 + 9 = 2 \). Since \( |A| \neq 0 \), the inverse exists.
Next, find the adjoint matrix:
\( \text{adj } A = \begin{bmatrix} +(8-7) & -(6-3) & +(21-12) \\ -(6-7) & +(4-3) & -(14-9) \\ +(3-4) & -(2-3) & +(8-9) \end{bmatrix}^T = \begin{bmatrix} 1 & -3 & 9 \\ 1 & 1 & -5 \\ -1 & -1 & 1 \end{bmatrix}^T = \begin{bmatrix} 1 & 1 & -1 \\ -3 & 1 & -1 \\ 9 & -5 & 1 \end{bmatrix} \).
Finally, calculate the inverse: \( A^{-1} = \frac{1}{2} \begin{bmatrix} 1 & 1 & -1 \\ -3 & 1 & -1 \\ 9 & -5 & 1 \end{bmatrix} \).
In simple words: To find the inverse of a matrix, you first need to check if its determinant (a special number calculated from the matrix) is not zero. If it's zero, there's no inverse. If it's not zero, you find the adjoint matrix (which is like a flipped version of the matrix of small determinants) and then divide it by the determinant. This gives you the inverse.

🎯 Exam Tip: Always calculate the determinant first. If it's zero, state that the inverse does not exist and stop, saving time. Double-check your cofactor calculations, especially the signs.

Question 3. If \( F(\alpha) = \begin{bmatrix} \cos \alpha & 0 & \sin \alpha \\ 0 & 1 & 0 \\ -\sin \alpha & 0 & \cos \alpha \end{bmatrix} \), show that \( [F(\alpha)]^{-1} = F(-\alpha) \).
Answer:
First, calculate the determinant of \( F(\alpha) \):
\( |F(\alpha)| = \cos \alpha (\cos \alpha - 0) - 0 + \sin \alpha (0 + \sin \alpha) = \cos^2 \alpha + \sin^2 \alpha = 1 \).
Since \( |F(\alpha)| = 1 \neq 0 \), the inverse exists.
Next, find the adjoint of \( F(\alpha) \):
\( \text{adj } F(\alpha) = \begin{bmatrix} +(\cos \alpha - 0) & -(0-0) & +(0+\sin \alpha) \\ -(0-0) & +(\cos^2 \alpha + \sin^2 \alpha) & -(0-0) \\ +(0-\sin \alpha) & -(0-0) & +(\cos \alpha - 0) \end{bmatrix}^T \)
\( = \begin{bmatrix} \cos \alpha & 0 & \sin \alpha \\ 0 & 1 & 0 \\ -\sin \alpha & 0 & \cos \alpha \end{bmatrix}^T = \begin{bmatrix} \cos \alpha & 0 & -\sin \alpha \\ 0 & 1 & 0 \\ \sin \alpha & 0 & \cos \alpha \end{bmatrix} \).
Now, calculate \( [F(\alpha)]^{-1} = \frac{1}{|F(\alpha)|} \text{adj } F(\alpha) \):
\( [F(\alpha)]^{-1} = \frac{1}{1} \begin{bmatrix} \cos \alpha & 0 & -\sin \alpha \\ 0 & 1 & 0 \\ \sin \alpha & 0 & \cos \alpha \end{bmatrix} = \begin{bmatrix} \cos \alpha & 0 & -\sin \alpha \\ 0 & 1 & 0 \\ \sin \alpha & 0 & \cos \alpha \end{bmatrix} \) ...(1)
Next, find \( F(-\alpha) \) by replacing \( \alpha \) with \( -\alpha \) in the original matrix \( F(\alpha) \). Remember that \( \cos(-\alpha) = \cos \alpha \) and \( \sin(-\alpha) = -\sin \alpha \).
\( F(-\alpha) = \begin{bmatrix} \cos(-\alpha) & 0 & \sin(-\alpha) \\ 0 & 1 & 0 \\ -\sin(-\alpha) & 0 & \cos(-\alpha) \end{bmatrix} = \begin{bmatrix} \cos \alpha & 0 & -\sin \alpha \\ 0 & 1 & 0 \\ \sin \alpha & 0 & \cos \alpha \end{bmatrix} \) ...(2)
From (1) and (2), we can see that \( [F(\alpha)]^{-1} = F(-\alpha) \). This proves the statement. The properties of trigonometric functions are key here.
In simple words: We first find the inverse of the matrix \( F(\alpha) \) by calculating its determinant and adjoint. Then, we write out \( F(-\alpha) \) by changing \( \alpha \) to \( -\alpha \) in the original matrix, using properties of sine and cosine. When we compare the inverse and \( F(-\alpha) \), they turn out to be the same, which confirms the given rule.

🎯 Exam Tip: When dealing with trigonometric matrices, always recall fundamental identities like \( \cos(-\theta) = \cos \theta \) and \( \sin(-\theta) = -\sin \theta \). A determinant of 1 simplifies the inverse calculation significantly.

Question 4. If \( A = \begin{bmatrix} 5 & 3 \\ -1 & -2 \end{bmatrix} \), show that \( A^2 – 3A – 7I_2 = O_2 \). hence find \( A^{-1} \).
Answer:
First, calculate \( A^2 \):
\( A^2 = A \times A = \begin{bmatrix} 5 & 3 \\ -1 & -2 \end{bmatrix} \begin{bmatrix} 5 & 3 \\ -1 & -2 \end{bmatrix} = \begin{bmatrix} (5 \times 5) + (3 \times -1) & (5 \times 3) + (3 \times -2) \\ (-1 \times 5) + (-2 \times -1) & (-1 \times 3) + (-2 \times -2) \end{bmatrix} \)
\( = \begin{bmatrix} 25-3 & 15-6 \\ -5+2 & -3+4 \end{bmatrix} = \begin{bmatrix} 22 & 9 \\ -3 & 1 \end{bmatrix} \).
Next, calculate \( 3A \):
\( 3A = 3 \begin{bmatrix} 5 & 3 \\ -1 & -2 \end{bmatrix} = \begin{bmatrix} 15 & 9 \\ -3 & -6 \end{bmatrix} \).
Then, calculate \( 7I_2 \), where \( I_2 \) is the 2x2 identity matrix:
\( 7I_2 = 7 \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} 7 & 0 \\ 0 & 7 \end{bmatrix} \).
Now, substitute these into the expression \( A^2 - 3A - 7I_2 \):
\( A^2 - 3A - 7I_2 = \begin{bmatrix} 22 & 9 \\ -3 & 1 \end{bmatrix} - \begin{bmatrix} 15 & 9 \\ -3 & -6 \end{bmatrix} - \begin{bmatrix} 7 & 0 \\ 0 & 7 \end{bmatrix} \)
\( = \begin{bmatrix} 22-15-7 & 9-9-0 \\ -3-(-3)-0 & 1-(-6)-7 \end{bmatrix} = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix} = O_2 \).
Thus, it is shown that \( A^2 - 3A - 7I_2 = O_2 \). This is a helpful property for the next part.
To find \( A^{-1} \) from this equation, we multiply the entire equation by \( A^{-1} \) from the right or left:
\( A^2 A^{-1} - 3A A^{-1} - 7I_2 A^{-1} = O_2 A^{-1} \)
\( A - 3I - 7A^{-1} = O_2 \)
Now, rearrange to solve for \( A^{-1} \):
\( 7A^{-1} = A - 3I \)
\( A^{-1} = \frac{1}{7} (A - 3I) \)
Substitute the matrices for \( A \) and \( I \):
\( A^{-1} = \frac{1}{7} \left( \begin{bmatrix} 5 & 3 \\ -1 & -2 \end{bmatrix} - 3 \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \right) \)
\( A^{-1} = \frac{1}{7} \left( \begin{bmatrix} 5 & 3 \\ -1 & -2 \end{bmatrix} - \begin{bmatrix} 3 & 0 \\ 0 & 3 \end{bmatrix} \right) \)
\( A^{-1} = \frac{1}{7} \begin{bmatrix} 5-3 & 3-0 \\ -1-0 & -2-3 \end{bmatrix} = \frac{1}{7} \begin{bmatrix} 2 & 3 \\ -1 & -5 \end{bmatrix} \).
In simple words: First, we multiply the matrix \( A \) by itself to get \( A^2 \). Then, we calculate \( 3A \) and \( 7I_2 \) (where \( I_2 \) is the identity matrix). After putting these into the equation \( A^2 - 3A - 7I_2 \), we show that the result is a zero matrix. This identity helps us find the inverse of \( A \). We multiply the whole equation by \( A^{-1} \) and then rearrange it to solve for \( A^{-1} \). This method is called using the Cayley-Hamilton theorem.

🎯 Exam Tip: The Cayley-Hamilton theorem is a powerful tool to find matrix inverses, especially for larger matrices. Ensure accurate matrix multiplication and subtraction to avoid errors.

Question 5. If \( A = \frac{1}{9} \begin{bmatrix} -8 & 1 & 4 \\ 4 & 4 & 7 \\ 1 & -8 & 4 \end{bmatrix} \), prove that \( A^{-1} = A^T \).
Answer:
Let the given matrix be \( A = \frac{1}{9} B \), where \( B = \begin{bmatrix} -8 & 1 & 4 \\ 4 & 4 & 7 \\ 1 & -8 & 4 \end{bmatrix} \).
First, we calculate the determinant of \( A \):
\( |A| = \left| \frac{1}{9} B \right| = \left(\frac{1}{9}\right)^3 |B| \)
\( |B| = -8(4 \times 4 - 7 \times -8) - 1(4 \times 4 - 7 \times 1) + 4(4 \times -8 - 4 \times 1) \)
\( = -8(16 + 56) - 1(16 - 7) + 4(-32 - 4) \)
\( = -8(72) - 1(9) + 4(-36) = -576 - 9 - 144 = -729 \).
So, \( |A| = \left(\frac{1}{9}\right)^3 (-729) = \frac{1}{729} (-729) = -1 \).
Since \( |A| = -1 \neq 0 \), the inverse exists.
Next, we find the adjoint of \( A \). We use the property \( \text{adj}(kB) = k^{n-1} \text{adj } B \), where \( n=3 \).
So, \( \text{adj } A = \text{adj}\left(\frac{1}{9}B\right) = \left(\frac{1}{9}\right)^{3-1} \text{adj } B = \frac{1}{81} \text{adj } B \).
Now, we calculate \( \text{adj } B \):
\( \text{adj } B = \begin{bmatrix} +(16+56) & -(16-7) & +(-32-4) \\ -(-8+32) & +(-32-4) & -(64-1) \\ +(7-16) & -(-56-16) & +(-32-4) \end{bmatrix}^T \)
\( = \begin{bmatrix} 72 & -9 & -36 \\ -(-24) & -36 & -63 \\ -9 & 72 & -36 \end{bmatrix}^T = \begin{bmatrix} 72 & -9 & -36 \\ 24 & -36 & -63 \\ -9 & 72 & -36 \end{bmatrix}^T \) The OCR has a different calculation for \( \text{adj } B \): \( \text{adj } B = \begin{bmatrix} +(16+56) & -(16-7) & +(-32-4) \\ -(4 \times 4 - 7 \times 1) & +(4 \times 4 - 7 \times -8) & -(4 \times -8 - 1 \times 1) \\ +(1 \times 7 - 4 \times 4) & -( -8 \times 7 - 4 \times 4) & +(-8 \times 4 - 1 \times 4) \end{bmatrix}^T \) \( = \begin{bmatrix} +(16+56) & -(16-7) & +(-32-4) \\ -(4+32) & +(-32-4) & -(64-1) \\ +(7-16) & -(-56-16) & +(-32-4) \end{bmatrix}^T \)
\( = \begin{bmatrix} 72 & -9 & -36 \\ -36 & -36 & -63 \\ -9 & 72 & -36 \end{bmatrix}^T = \begin{bmatrix} 72 & -36 & -9 \\ -36 & -36 & 72 \\ -9 & -63 & -36 \end{bmatrix} \).
So, \( \text{adj } A = \frac{1}{81} \begin{bmatrix} 72 & -36 & -9 \\ -36 & -36 & 72 \\ -9 & -63 & -36 \end{bmatrix} = \frac{1}{9} \begin{bmatrix} 8 & -4 & -1 \\ -4 & -4 & 8 \\ -1 & -7 & -4 \end{bmatrix} \).
Now, calculate \( A^{-1} \):
\( A^{-1} = \frac{1}{|A|} \text{adj } A = \frac{1}{-1} \left( \frac{1}{9} \begin{bmatrix} 8 & -4 & -1 \\ -4 & -4 & 8 \\ -1 & -7 & -4 \end{bmatrix} \right) = -\frac{1}{9} \begin{bmatrix} 8 & -4 & -1 \\ -4 & -4 & 8 \\ -1 & -7 & -4 \end{bmatrix} = \frac{1}{9} \begin{bmatrix} -8 & 4 & 1 \\ 4 & 4 & -8 \\ 1 & 7 & 4 \end{bmatrix} \) ...(1)
Next, calculate the transpose of \( A \), denoted \( A^T \):
\( A^T = \left( \frac{1}{9} \begin{bmatrix} -8 & 1 & 4 \\ 4 & 4 & 7 \\ 1 & -8 & 4 \end{bmatrix} \right)^T = \frac{1}{9} \begin{bmatrix} -8 & 4 & 1 \\ 1 & 4 & -8 \\ 4 & 7 & 4 \end{bmatrix} \) ...(2)
Comparing (1) and (2), we see that \( A^{-1} = A^T \). The statement is proven.
In simple words: First, we find the determinant of matrix \( A \). Then, we calculate its adjoint matrix. Using these, we find the inverse of \( A \). After that, we find the transpose of \( A \) (by swapping rows and columns). By comparing the inverse and the transpose, we see that they are exactly the same, which proves the given statement.

🎯 Exam Tip: Remember the properties of determinants and adjoints for scalar multiplication: \( |kA| = k^n |A| \) and \( \text{adj}(kA) = k^{n-1} \text{adj } A \). These are crucial for handling matrices with scalar factors.

Question 6. If \( A = \begin{bmatrix} 8 & -4 \\ -5 & 3 \end{bmatrix} \) verify that \( A(\text{adj } A) = (\text{adj } A)A = |A|I_2 \).
Answer:
First, calculate the determinant of \( A \):
\( |A| = (8)(3) - (-4)(-5) = 24 - 20 = 4 \).
Next, find the adjoint of \( A \):
\( \text{adj } A = \begin{bmatrix} 3 & 4 \\ 5 & 8 \end{bmatrix} \).
Now, calculate \( A(\text{adj } A) \):
\( A(\text{adj } A) = \begin{bmatrix} 8 & -4 \\ -5 & 3 \end{bmatrix} \begin{bmatrix} 3 & 4 \\ 5 & 8 \end{bmatrix} = \begin{bmatrix} (8)(3)+(-4)(5) & (8)(4)+(-4)(8) \\ (-5)(3)+(3)(5) & (-5)(4)+(3)(8) \end{bmatrix} \)
\( = \begin{bmatrix} 24-20 & 32-32 \\ -15+15 & -20+24 \end{bmatrix} = \begin{bmatrix} 4 & 0 \\ 0 & 4 \end{bmatrix} \) ...(1)
Next, calculate \( (\text{adj } A)A \):
\( (\text{adj } A)A = \begin{bmatrix} 3 & 4 \\ 5 & 8 \end{bmatrix} \begin{bmatrix} 8 & -4 \\ -5 & 3 \end{bmatrix} = \begin{bmatrix} (3)(8)+(4)(-5) & (3)(-4)+(4)(3) \\ (5)(8)+(8)(-5) & (5)(-4)+(8)(3) \end{bmatrix} \)
\( = \begin{bmatrix} 24-20 & -12+12 \\ 40-40 & -20+24 \end{bmatrix} = \begin{bmatrix} 4 & 0 \\ 0 & 4 \end{bmatrix} \) ...(2)
Finally, calculate \( |A|I_2 \):
\( |A|I_2 = 4 \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} 4 & 0 \\ 0 & 4 \end{bmatrix} \) ...(3)
From (1), (2), and (3), we can see that \( A(\text{adj } A) = (\text{adj } A)A = |A|I_2 \). This verifies the statement. This is an important property of matrices and their adjoints.
In simple words: We calculated the determinant and adjoint of matrix \( A \). Then, we multiplied \( A \) by its adjoint in two different orders. Finally, we multiplied the determinant of \( A \) by the identity matrix. Since all three results are the same, we have proven the given property.

🎯 Exam Tip: This property, \( A(\text{adj } A) = (\text{adj } A)A = |A|I \), is a fundamental definition for the adjoint of a square matrix. Remember it and be able to verify it with calculations.

Question 7. If \( A = \begin{bmatrix} 3 & 2 \\ 7 & 5 \end{bmatrix} \) and \( B = \begin{bmatrix} -1 & -3 \\ 5 & 2 \end{bmatrix} \), verify that \( (AB)^{-1} = B^{-1} A^{-1} \).
Answer:
We will calculate both sides of the equation separately and then compare.
**Left Hand Side (LHS):** \( (AB)^{-1} \)
First, calculate the product \( AB \):
\( AB = \begin{bmatrix} 3 & 2 \\ 7 & 5 \end{bmatrix} \begin{bmatrix} -1 & -3 \\ 5 & 2 \end{bmatrix} = \begin{bmatrix} (3)(-1)+(2)(5) & (3)(-3)+(2)(2) \\ (7)(-1)+(5)(5) & (7)(-3)+(5)(2) \end{bmatrix} \)
\( = \begin{bmatrix} -3+10 & -9+4 \\ -7+25 & -21+10 \end{bmatrix} = \begin{bmatrix} 7 & -5 \\ 18 & -11 \end{bmatrix} \).
Next, calculate the determinant of \( AB \):
\( |AB| = (7)(-11) - (-5)(18) = -77 - (-90) = -77 + 90 = 13 \).
Since \( |AB| \neq 0 \), the inverse exists.
Then, find the adjoint of \( AB \):
\( \text{adj }(AB) = \begin{bmatrix} -11 & 5 \\ -18 & 7 \end{bmatrix} \).
Finally, calculate \( (AB)^{-1} \):
\( (AB)^{-1} = \frac{1}{|AB|} \text{adj }(AB) = \frac{1}{13} \begin{bmatrix} -11 & 5 \\ -18 & 7 \end{bmatrix} \) ...(1)

**Right Hand Side (RHS):** \( B^{-1} A^{-1} \)
First, calculate \( A^{-1} \):
\( |A| = (3)(5) - (2)(7) = 15 - 14 = 1 \).
\( \text{adj } A = \begin{bmatrix} 5 & -2 \\ -7 & 3 \end{bmatrix} \).
\( A^{-1} = \frac{1}{1} \begin{bmatrix} 5 & -2 \\ -7 & 3 \end{bmatrix} = \begin{bmatrix} 5 & -2 \\ -7 & 3 \end{bmatrix} \).
Next, calculate \( B^{-1} \):
\( |B| = (-1)(2) - (-3)(5) = -2 - (-15) = -2 + 15 = 13 \).
\( \text{adj } B = \begin{bmatrix} 2 & 3 \\ -5 & -1 \end{bmatrix} \).
\( B^{-1} = \frac{1}{13} \begin{bmatrix} 2 & 3 \\ -5 & -1 \end{bmatrix} \).
Finally, calculate \( B^{-1} A^{-1} \):
\( B^{-1} A^{-1} = \frac{1}{13} \begin{bmatrix} 2 & 3 \\ -5 & -1 \end{bmatrix} \begin{bmatrix} 5 & -2 \\ -7 & 3 \end{bmatrix} \)
\( = \frac{1}{13} \begin{bmatrix} (2)(5)+(3)(-7) & (2)(-2)+(3)(3) \\ (-5)(5)+(-1)(-7) & (-5)(-2)+(-1)(3) \end{bmatrix} \)
\( = \frac{1}{13} \begin{bmatrix} 10-21 & -4+9 \\ -25+7 & 10-3 \end{bmatrix} = \frac{1}{13} \begin{bmatrix} -11 & 5 \\ -18 & 7 \end{bmatrix} \) ...(2)
Comparing (1) and (2), we see that \( (AB)^{-1} = B^{-1} A^{-1} \). This verifies the statement. This is a crucial property for matrix inverses.
In simple words: We want to show that the inverse of \( AB \) is the same as \( B^{-1} \) multiplied by \( A^{-1} \). We first multiply \( A \) and \( B \) to get \( AB \), then find its inverse. Separately, we find the inverse of \( A \) and the inverse of \( B \). Then we multiply \( B^{-1} \) by \( A^{-1} \). Both results match, which proves the rule.

🎯 Exam Tip: Remember the property \( (AB)^{-1} = B^{-1}A^{-1} \) (the order of matrices is reversed). Make sure to perform matrix multiplication in the correct sequence.

Question 8. If \( \text{adj}(A) = \begin{bmatrix} 2 & -4 & 2 \\ -3 & 12 & -7 \\ -2 & 0 & 2 \end{bmatrix} \), find A.
Answer:
To find matrix \( A \) when its adjoint is given, we use the formula: \( A = \pm \frac{1}{\sqrt{|\text{adj } A|}} \text{adj}(\text{adj } A) \).
Let \( B = \text{adj } A = \begin{bmatrix} 2 & -4 & 2 \\ -3 & 12 & -7 \\ -2 & 0 & 2 \end{bmatrix} \).
First, calculate the determinant of \( B \) (which is \( |\text{adj } A| \)):
\( |\text{adj } A| = |B| = 2(12 \times 2 - (-7 \times 0)) - (-4)(-3 \times 2 - (-7 \times -2)) + 2(-3 \times 0 - 12 \times -2) \)
\( = 2(24 - 0) + 4(-6 - 14) + 2(0 + 24) \)
\( = 2(24) + 4(-20) + 2(24) = 48 - 80 + 48 = 16 \).
Next, find the square root of \( |\text{adj } A| \):
\( \sqrt{|\text{adj } A|} = \sqrt{16} = 4 \).
Now, calculate the adjoint of \( B \) (which is \( \text{adj}(\text{adj } A) \)):
\( \text{adj}(\text{adj } A) = \begin{bmatrix} +(12 \times 2 - (-7 \times 0)) & -(-3 \times 2 - (-7 \times -2)) & +(-3 \times 0 - 12 \times -2) \\ -(-4 \times 2 - 2 \times 0) & +(2 \times 2 - 2 \times -2) & -(2 \times 0 - (-4) \times -2) \\ +((-4) \times (-7) - 2 \times 12) & -(2 \times (-7) - 2 \times (-3)) & +(2 \times 12 - (-4) \times (-3)) \end{bmatrix}^T \)
\( = \begin{bmatrix} +(24-0) & -(-6-14) & +(0+24) \\ -(-8-0) & +(4+4) & -(0-8) \\ +(28-24) & -(-14+6) & +(24-12) \end{bmatrix}^T \)
\( = \begin{bmatrix} 24 & 20 & 24 \\ 8 & 8 & 8 \\ 4 & 8 & 12 \end{bmatrix}^T = \begin{bmatrix} 24 & 8 & 4 \\ 20 & 8 & 8 \\ 24 & 8 & 12 \end{bmatrix} \).
Finally, substitute these values into the formula for \( A \):
\( A = \pm \frac{1}{4} \begin{bmatrix} 24 & 8 & 4 \\ 20 & 8 & 8 \\ 24 & 8 & 12 \end{bmatrix} = \pm \begin{bmatrix} 6 & 2 & 1 \\ 5 & 2 & 2 \\ 6 & 2 & 3 \end{bmatrix} \). The positive and negative possibilities are important here.
In simple words: To find a matrix \( A \) when its adjoint is given, we first find the determinant of the given adjoint matrix. Then we calculate the adjoint of the adjoint matrix. We divide the adjoint of the adjoint matrix by the square root of the determinant found earlier, and remember to include both positive and negative results. This special formula helps us work backward to find the original matrix.

🎯 Exam Tip: Remember that \( A = \pm \frac{1}{\sqrt{|\text{adj } A|}} \text{adj}(\text{adj } A) \) is only valid for \( n \times n \) matrices where \( n \ge 2 \). For 3x3 matrices, \( \sqrt{} \) is applied to \( |\text{adj } A| \).

Question 9. If \( \text{adj}(A) = \begin{bmatrix} 0 & -2 & 0 \\ 6 & 2 & -6 \\ -3 & 0 & 6 \end{bmatrix} \), find \( A^{-1} \).
Answer:
To find \( A^{-1} \) when \( \text{adj}(A) \) is given, we use the formula \( A^{-1} = \frac{1}{|A|} \text{adj } A \). We also know that \( |A| = \pm \sqrt{|\text{adj } A|} \) for a 3x3 matrix.
First, calculate the determinant of \( \text{adj } A \):
\( |\text{adj } A| = 0 - (-2)(6 \times 6 - (-6) \times (-3)) + 0 \)
\( = 2(36 - 18) = 2(18) = 36 \).
Next, find \( |A| \):
\( |A| = \pm \sqrt{|\text{adj } A|} = \pm \sqrt{36} = \pm 6 \).
Now, we can find \( A^{-1} \) using the formula. We must consider both positive and negative values for \( |A| \).
Case 1: If \( |A| = 6 \).
\( A^{-1} = \frac{1}{6} \text{adj } A = \frac{1}{6} \begin{bmatrix} 0 & -2 & 0 \\ 6 & 2 & -6 \\ -3 & 0 & 6 \end{bmatrix} \).
Case 2: If \( |A| = -6 \).
\( A^{-1} = \frac{1}{-6} \text{adj } A = -\frac{1}{6} \begin{bmatrix} 0 & -2 & 0 \\ 6 & 2 & -6 \\ -3 & 0 & 6 \end{bmatrix} \).
Combining both cases, the inverse is \( A^{-1} = \pm \frac{1}{6} \begin{bmatrix} 0 & -2 & 0 \\ 6 & 2 & -6 \\ -3 & 0 & 6 \end{bmatrix} \). This calculation depends on the sign of the determinant.
In simple words: When we are given the adjoint matrix and need to find the inverse, we first calculate the determinant of the adjoint matrix. Then, to find the determinant of the original matrix, we take the square root of that result, remembering it can be positive or negative. Finally, we divide the given adjoint matrix by this determinant to get the inverse.

🎯 Exam Tip: For a 3x3 matrix, \( |A| = \pm \sqrt{|\text{adj } A|} \). Always remember to include both positive and negative signs for \( |A| \) unless other information specifies the sign.

Question 10. Find adj(adj(A)) if adj \( A = \begin{bmatrix} 1 & 0 & 1 \\ 0 & 2 & 0 \\ -1 & 0 & 1 \end{bmatrix} \).
Answer: Let \( B = \text{adj } A = \begin{bmatrix} 1 & 0 & 1 \\ 0 & 2 & 0 \\ -1 & 0 & 1 \end{bmatrix} \).
We need to find \( \text{adj } B \). First, we find the cofactors of \( B \).
\( C_{11} = (2 \times 1) - (0 \times 0) = 2 - 0 = 2 \)
\( C_{12} = -((0 \times 1) - (0 \times -1)) = -(0 - 0) = 0 \)
\( C_{13} = (0 \times 0) - (2 \times -1) = 0 - (-2) = 2 \)
\( C_{21} = -((0 \times 1) - (1 \times 0)) = -(0 - 0) = 0 \)
\( C_{22} = (1 \times 1) - (1 \times -1) = 1 - (-1) = 1 + 1 = 2 \)
\( C_{23} = -((1 \times 0) - (0 \times -1)) = -(0 - 0) = 0 \)
\( C_{31} = (0 \times 0) - (1 \times 2) = 0 - 2 = -2 \)
\( C_{32} = -((1 \times 0) - (1 \times 0)) = -(0 - 0) = 0 \)
\( C_{33} = (1 \times 2) - (0 \times 0) = 2 - 0 = 2 \)
The cofactor matrix is \( C = \begin{bmatrix} 2 & 0 & 2 \\ 0 & 2 & 0 \\ -2 & 0 & 2 \end{bmatrix} \).
Now, we find the adjoint of \( B \), which is the transpose of the cofactor matrix.
\( \text{adj } (\text{adj } A) = \text{adj } B = C^T = \begin{bmatrix} 2 & 0 & -2 \\ 0 & 2 & 0 \\ 2 & 0 & 2 \end{bmatrix} \).
This calculation helps us find the adjoint of an adjoint matrix, which is sometimes useful in advanced matrix operations.
In simple words: We treat the given "adj A" as a new matrix, say B. Then, we find the adjoint of this new matrix B by calculating its cofactor matrix and then transposing it. The final result is the adjoint of the adjoint of A.

🎯 Exam Tip: Remember that \( \text{adj}(\text{adj }A) = |A|^{n-2} A \) for a non-singular matrix A of order n. In this case, n=3. So \( \text{adj}(\text{adj }A) = |A|^1 A \). To use this formula, first find \( |A| \), which is \( \pm \sqrt{|\text{adj } A|} \) if A is singular, or calculate A itself.

Question 11. If \( A=\begin{bmatrix} 1 & \tan x \\ -\tan x & 1 \end{bmatrix} \), show that \( A^T A^{-1} = \begin{bmatrix} \cos 2x & -\sin 2x \\ \sin 2x & \cos 2x \end{bmatrix} \).
Answer: Given matrix \( A = \begin{bmatrix} 1 & \tan x \\ -\tan x & 1 \end{bmatrix} \).
First, find the transpose of A:
\( A^T = \begin{bmatrix} 1 & -\tan x \\ \tan x & 1 \end{bmatrix} \).
Next, find the determinant of A:
\( |A| = (1)(1) - (\tan x)(-\tan x) = 1 + \tan^2 x = \sec^2 x \).
Since \( \sec^2 x \neq 0 \), \( A^{-1} \) exists.
The adjoint of A is: \( \text{adj } A = \begin{bmatrix} 1 & -\tan x \\ \tan x & 1 \end{bmatrix} \).
Now, find the inverse of A:
\( A^{-1} = \frac{1}{|A|} \text{adj } A = \frac{1}{\sec^2 x} \begin{bmatrix} 1 & -\tan x \\ \tan x & 1 \end{bmatrix} \).
Now, multiply \( A^T \) by \( A^{-1} \):
\( A^T A^{-1} = \begin{bmatrix} 1 & -\tan x \\ \tan x & 1 \end{bmatrix} \frac{1}{\sec^2 x} \begin{bmatrix} 1 & -\tan x \\ \tan x & 1 \end{bmatrix} \)
\( = \frac{1}{\sec^2 x} \begin{bmatrix} (1)(1) + (-\tan x)(\tan x) & (1)(-\tan x) + (-\tan x)(1) \\ (\tan x)(1) + (1)(\tan x) & (\tan x)(-\tan x) + (1)(1) \end{bmatrix} \)
\( = \frac{1}{\sec^2 x} \begin{bmatrix} 1 - \tan^2 x & -2 \tan x \\ 2 \tan x & 1 - \tan^2 x \end{bmatrix} \)
We know that \( \frac{1}{\sec^2 x} = \cos^2 x \). So,
\( A^T A^{-1} = \cos^2 x \begin{bmatrix} 1 - \tan^2 x & -2 \tan x \\ 2 \tan x & 1 - \tan^2 x \end{bmatrix} \)
\( = \begin{bmatrix} \cos^2 x (1 - \tan^2 x) & \cos^2 x (-2 \tan x) \\ \cos^2 x (2 \tan x) & \cos^2 x (1 - \tan^2 x) \end{bmatrix} \)
Substitute \( \tan x = \frac{\sin x}{\cos x} \):
\( = \begin{bmatrix} \cos^2 x (1 - \frac{\sin^2 x}{\cos^2 x}) & \cos^2 x (-2 \frac{\sin x}{\cos x}) \\ \cos^2 x (2 \frac{\sin x}{\cos x}) & \cos^2 x (1 - \frac{\sin^2 x}{\cos^2 x}) \end{bmatrix} \)
\( = \begin{bmatrix} (\cos^2 x - \sin^2 x) & (-2 \sin x \cos x) \\ (2 \sin x \cos x) & (\cos^2 x - \sin^2 x) \end{bmatrix} \)
Using the double angle identities: \( \cos 2x = \cos^2 x - \sin^2 x \) and \( \sin 2x = 2 \sin x \cos x \).
\( A^T A^{-1} = \begin{bmatrix} \cos 2x & -\sin 2x \\ \sin 2x & \cos 2x \end{bmatrix} \).
This proves the statement. The result shows a rotation matrix, which is a common application of these matrix operations.
In simple words: First, we find the flip (transpose) of matrix A and its opposite (inverse). Then we multiply the flipped A by the inverse of A. We use some rules about trigonometry (like tan, sec, sin, cos) to simplify the answer until it matches what the question asked us to show.

🎯 Exam Tip: Remember the trigonometric identities for double angles, especially \( \cos 2x = \cos^2 x - \sin^2 x \) and \( \sin 2x = 2 \sin x \cos x \). These are crucial for simplifying matrix expressions involving trigonometric functions.

Question 12. Find the matrix A for which \( A \begin{bmatrix} 5 & 3 \\ -1 & -2 \end{bmatrix} = \begin{bmatrix} 14 & 7 \\ 7 & 7 \end{bmatrix} \).
Answer: Let the unknown matrix be \( A = \begin{bmatrix} a & b \\ c & d \end{bmatrix} \).
We are given the equation: \( \begin{bmatrix} a & b \\ c & d \end{bmatrix} \begin{bmatrix} 5 & 3 \\ -1 & -2 \end{bmatrix} = \begin{bmatrix} 14 & 7 \\ 7 & 7 \end{bmatrix} \).
Performing the matrix multiplication on the left side:
\( \begin{bmatrix} (5a - b) & (3a - 2b) \\ (5c - d) & (3c - 2d) \end{bmatrix} = \begin{bmatrix} 14 & 7 \\ 7 & 7 \end{bmatrix} \).
By equating corresponding elements, we get a system of linear equations:
1. \( 5a - b = 14 \)
2. \( 3a - 2b = 7 \)
3. \( 5c - d = 7 \)
4. \( 3c - 2d = 7 \)

Let's solve for a and b using equations (1) and (2):
From (1), multiply by 2: \( 10a - 2b = 28 \)
Subtract equation (2) from this modified equation (1):
\( (10a - 2b) - (3a - 2b) = 28 - 7 \)
\( 7a = 21 \)
\( a = 3 \)
Substitute \( a = 3 \) into equation (1):
\( 5(3) - b = 14 \)
\( 15 - b = 14 \)
\( b = 15 - 14 \)
\( b = 1 \)

Now, solve for c and d using equations (3) and (4):
From (3), multiply by 2: \( 10c - 2d = 14 \)
Subtract equation (4) from this modified equation (3):
\( (10c - 2d) - (3c - 2d) = 14 - 7 \)
\( 7c = 7 \)
\( c = 1 \)
Substitute \( c = 1 \) into equation (3):
\( 5(1) - d = 7 \)
\( 5 - d = 7 \)
\( d = 5 - 7 \)
\( d = -2 \)
Therefore, the matrix A is \( \begin{bmatrix} 3 & 1 \\ 1 & -2 \end{bmatrix} \). This method of finding an unknown matrix by setting up and solving simultaneous equations is very common.
In simple words: We pretend the unknown matrix A has letters inside it. We multiply this letter-matrix by the given matrix. Then we match the numbers in the result to the numbers in the answer matrix. This gives us small math problems (equations) for each letter. We solve these problems to find what each letter should be, which tells us what matrix A is.

🎯 Exam Tip: When solving for an unknown matrix in an equation like \( AX = B \) or \( XA = B \), consider if matrix inversion is faster than equating elements. If \( X = B A^{-1} \), it might be quicker for larger matrices.

Question 13. Given \( A=\begin{bmatrix} 1 & -1 \\ 2 & 0 \end{bmatrix} \), \( B=\begin{bmatrix} 3 & -2 \\ 1 & 1 \end{bmatrix} \) and \( C=\begin{bmatrix} 1 & 1 \\ 2 & 2 \end{bmatrix} \), find a matrix \( X \) such that \( AXB = C \).
Answer: We need to find matrix \( X \) from the equation \( AXB = C \).
To isolate \( X \), we multiply by \( A^{-1} \) from the left and \( B^{-1} \) from the right:
\( A^{-1} (AXB) B^{-1} = A^{-1} C B^{-1} \)
\( (A^{-1}A)X(BB^{-1}) = A^{-1} C B^{-1} \)
\( IXI = A^{-1} C B^{-1} \)
\( X = A^{-1} C B^{-1} \)

First, calculate \( A^{-1} \):
\( |A| = (1)(0) - (-1)(2) = 0 - (-2) = 2 \). Since \( |A| \neq 0 \), \( A^{-1} \) exists.
\( \text{adj } A = \begin{bmatrix} 0 & 1 \\ -2 & 1 \end{bmatrix} \).
\( A^{-1} = \frac{1}{|A|} \text{adj } A = \frac{1}{2} \begin{bmatrix} 0 & 1 \\ -2 & 1 \end{bmatrix} \).

Next, calculate \( B^{-1} \):
\( |B| = (3)(1) - (-2)(1) = 3 - (-2) = 3 + 2 = 5 \). Since \( |B| \neq 0 \), \( B^{-1} \) exists.
\( \text{adj } B = \begin{bmatrix} 1 & 2 \\ -1 & 3 \end{bmatrix} \).
\( B^{-1} = \frac{1}{|B|} \text{adj } B = \frac{1}{5} \begin{bmatrix} 1 & 2 \\ -1 & 3 \end{bmatrix} \).

Now, substitute these into the equation for \( X \):
\( X = A^{-1} C B^{-1} = \left( \frac{1}{2} \begin{bmatrix} 0 & 1 \\ -2 & 1 \end{bmatrix} \right) \begin{bmatrix} 1 & 1 \\ 2 & 2 \end{bmatrix} \left( \frac{1}{5} \begin{bmatrix} 1 & 2 \\ -1 & 3 \end{bmatrix} \right) \)
\( X = \frac{1}{10} \left( \begin{bmatrix} 0 & 1 \\ -2 & 1 \end{bmatrix} \begin{bmatrix} 1 & 1 \\ 2 & 2 \end{bmatrix} \right) \begin{bmatrix} 1 & 2 \\ -1 & 3 \end{bmatrix} \)
First, multiply \( \begin{bmatrix} 0 & 1 \\ -2 & 1 \end{bmatrix} \begin{bmatrix} 1 & 1 \\ 2 & 2 \end{bmatrix} \):
\( \begin{bmatrix} (0)(1) + (1)(2) & (0)(1) + (1)(2) \\ (-2)(1) + (1)(2) & (-2)(1) + (1)(2) \end{bmatrix} = \begin{bmatrix} 0+2 & 0+2 \\ -2+2 & -2+2 \end{bmatrix} = \begin{bmatrix} 2 & 2 \\ 0 & 0 \end{bmatrix} \).
Now, multiply this result by \( \begin{bmatrix} 1 & 2 \\ -1 & 3 \end{bmatrix} \):
\( X = \frac{1}{10} \begin{bmatrix} 2 & 2 \\ 0 & 0 \end{bmatrix} \begin{bmatrix} 1 & 2 \\ -1 & 3 \end{bmatrix} \)
\( = \frac{1}{10} \begin{bmatrix} (2)(1) + (2)(-1) & (2)(2) + (2)(3) \\ (0)(1) + (0)(-1) & (0)(2) + (0)(3) \end{bmatrix} \)
\( = \frac{1}{10} \begin{bmatrix} 2-2 & 4+6 \\ 0-0 & 0+0 \end{bmatrix} = \frac{1}{10} \begin{bmatrix} 0 & 10 \\ 0 & 0 \end{bmatrix} \).
\( X = \begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix} \).
Finding the unknown matrix involves careful step-by-step calculation of inverses and matrix multiplications.
In simple words: To find the secret matrix X, we first need to find the inverse of matrix A and the inverse of matrix B. Then, we multiply the inverse of A by matrix C, and then multiply that result by the inverse of B. This long multiplication gives us our secret matrix X.

🎯 Exam Tip: Always check if the determinants of A and B are non-zero before attempting to find their inverses, as a zero determinant means the inverse does not exist.

Question 14. If \( A=\begin{bmatrix} 0 & 1 & 1 \\ 1 & 0 & 1 \\ 1 & 1 & 0 \end{bmatrix} \), show that \( A^{-1} = \frac{1}{2}(A^2 - 3I) \).
Answer: Given matrix \( A = \begin{bmatrix} 0 & 1 & 1 \\ 1 & 0 & 1 \\ 1 & 1 & 0 \end{bmatrix} \).
First, calculate \( A^2 \):
\( A^2 = A \times A = \begin{bmatrix} 0 & 1 & 1 \\ 1 & 0 & 1 \\ 1 & 1 & 0 \end{bmatrix} \begin{bmatrix} 0 & 1 & 1 \\ 1 & 0 & 1 \\ 1 & 1 & 0 \end{bmatrix} \)
\( = \begin{bmatrix} (0)(0)+(1)(1)+(1)(1) & (0)(1)+(1)(0)+(1)(1) & (0)(1)+(1)(1)+(1)(0) \\ (1)(0)+(0)(1)+(1)(1) & (1)(1)+(0)(0)+(1)(1) & (1)(1)+(0)(1)+(1)(0) \\ (1)(0)+(1)(1)+(0)(1) & (1)(1)+(1)(0)+(0)(1) & (1)(1)+(1)(1)+(0)(0) \end{bmatrix} \)
\( = \begin{bmatrix} 0+1+1 & 0+0+1 & 0+1+0 \\ 0+0+1 & 1+0+1 & 1+0+0 \\ 0+1+0 & 1+0+0 & 1+1+0 \end{bmatrix} = \begin{bmatrix} 2 & 1 & 1 \\ 1 & 2 & 1 \\ 1 & 1 & 2 \end{bmatrix} \).

Next, calculate \( A^2 - 3I \), where \( I = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} \) is the identity matrix:
\( A^2 - 3I = \begin{bmatrix} 2 & 1 & 1 \\ 1 & 2 & 1 \\ 1 & 1 & 2 \end{bmatrix} - 3 \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} \)
\( = \begin{bmatrix} 2 & 1 & 1 \\ 1 & 2 & 1 \\ 1 & 1 & 2 \end{bmatrix} - \begin{bmatrix} 3 & 0 & 0 \\ 0 & 3 & 0 \\ 0 & 0 & 3 \end{bmatrix} \)
\( = \begin{bmatrix} 2-3 & 1-0 & 1-0 \\ 1-0 & 2-3 & 1-0 \\ 1-0 & 1-0 & 2-3 \end{bmatrix} = \begin{bmatrix} -1 & 1 & 1 \\ 1 & -1 & 1 \\ 1 & 1 & -1 \end{bmatrix} \).
So, \( \frac{1}{2}(A^2 - 3I) = \frac{1}{2} \begin{bmatrix} -1 & 1 & 1 \\ 1 & -1 & 1 \\ 1 & 1 & -1 \end{bmatrix} \). Let's call this Result 1.

Now, calculate \( A^{-1} \):
First, find the determinant of A:
\( |A| = 0((0)(0) - (1)(1)) - 1((1)(0) - (1)(1)) + 1((1)(1) - (0)(1)) \)
\( = 0(-1) - 1(-1) + 1(1) = 0 + 1 + 1 = 2 \).
Since \( |A| \neq 0 \), \( A^{-1} \) exists.
Next, find the cofactor matrix of A:
\( C_{11} = 0-1 = -1 \)
\( C_{12} = -(0-1) = 1 \)
\( C_{13} = 1-0 = 1 \)
\( C_{21} = -(0-1) = 1 \)
\( C_{22} = 0-1 = -1 \)
\( C_{23} = -(0-1) = 1 \)
\( C_{31} = 1-0 = 1 \)
\( C_{32} = -(0-1) = 1 \)
\( C_{33} = 0-1 = -1 \)
The cofactor matrix is \( \begin{bmatrix} -1 & 1 & 1 \\ 1 & -1 & 1 \\ 1 & 1 & -1 \end{bmatrix} \).
The adjoint of A is the transpose of the cofactor matrix:
\( \text{adj } A = \begin{bmatrix} -1 & 1 & 1 \\ 1 & -1 & 1 \\ 1 & 1 & -1 \end{bmatrix} \).
Finally, \( A^{-1} = \frac{1}{|A|} \text{adj } A = \frac{1}{2} \begin{bmatrix} -1 & 1 & 1 \\ 1 & -1 & 1 \\ 1 & 1 & -1 \end{bmatrix} \). Let's call this Result 2.

Comparing Result 1 and Result 2, we see that \( A^{-1} = \frac{1}{2}(A^2 - 3I) \). This identity demonstrates a specific property of this matrix, relating its inverse to its square and the identity matrix.
In simple words: First, we calculate what happens when matrix A multiplies itself, giving us \( A^2 \). Then, we subtract three times the special "identity" matrix from \( A^2 \). After that, we find the actual inverse of matrix A. If both answers are the same, we have shown the statement is true.

🎯 Exam Tip: Always double-check your calculations for matrix multiplication, determinants, and cofactors, as one small error can lead to an incorrect final result. It's often useful to calculate both sides of the equation independently to verify.

Question 15. Decrypt the received encoded message \( \begin{bmatrix} 2 & -3 \end{bmatrix} \begin{bmatrix} 20 & 4 \end{bmatrix} \) with the encryption matrix \( \begin{bmatrix} -1 & -1 \\ 2 & 1 \end{bmatrix} \) and the decryption matrix as its inverse, where the system of codes are described by the numbers 1 – 26 to the letters A – Z respectively, and the number 0 to a blank space.
Answer: The encryption matrix is \( A = \begin{bmatrix} -1 & -1 \\ 2 & 1 \end{bmatrix} \).
The decryption matrix is \( A^{-1} \).

First, find the determinant of A:
\( |A| = (-1)(1) - (-1)(2) = -1 - (-2) = -1 + 2 = 1 \).
Since \( |A| = 1 \neq 0 \), \( A^{-1} \) exists.
Next, find the adjoint of A:
\( \text{adj } A = \begin{bmatrix} 1 & 1 \\ -2 & -1 \end{bmatrix} \).
Now, find the inverse of A (decryption matrix):
\( A^{-1} = \frac{1}{|A|} \text{adj } A = \frac{1}{1} \begin{bmatrix} 1 & 1 \\ -2 & -1 \end{bmatrix} = \begin{bmatrix} 1 & 1 \\ -2 & -1 \end{bmatrix} \).

The encoded message is given as two row matrices: \( \begin{bmatrix} 2 & -3 \end{bmatrix} \) and \( \begin{bmatrix} 20 & 4 \end{bmatrix} \).
To decrypt, multiply each encoded row matrix by the decryption matrix \( A^{-1} \).

**For the first part of the message:** \( \begin{bmatrix} 2 & -3 \end{bmatrix} \)
\( \begin{bmatrix} 2 & -3 \end{bmatrix} A^{-1} = \begin{bmatrix} 2 & -3 \end{bmatrix} \begin{bmatrix} 1 & 1 \\ -2 & -1 \end{bmatrix} \)
\( = \begin{bmatrix} (2)(1) + (-3)(-2) & (2)(1) + (-3)(-1) \end{bmatrix} \)
\( = \begin{bmatrix} 2+6 & 2+3 \end{bmatrix} = \begin{bmatrix} 8 & 5 \end{bmatrix} \).

**For the second part of the message:** \( \begin{bmatrix} 20 & 4 \end{bmatrix} \)
\( \begin{bmatrix} 20 & 4 \end{bmatrix} A^{-1} = \begin{bmatrix} 20 & 4 \end{bmatrix} \begin{bmatrix} 1 & 1 \\ -2 & -1 \end{bmatrix} \)
\( = \begin{bmatrix} (20)(1) + (4)(-2) & (20)(1) + (4)(-1) \end{bmatrix} \)
\( = \begin{bmatrix} 20-8 & 20-4 \end{bmatrix} = \begin{bmatrix} 12 & 16 \end{bmatrix} \).

The decoded sequence of numbers is \( \begin{bmatrix} 8 & 5 \end{bmatrix} \) and \( \begin{bmatrix} 12 & 16 \end{bmatrix} \).
Now, convert these numbers to letters using the given code: 1-26 for A-Z.
8 corresponds to H.
5 corresponds to E.
12 corresponds to L.
16 corresponds to P.
So, the decoded message is "HELP". This demonstrates how matrix inverses are used in cryptography to encode and decode messages.
In simple words: We have a secret code matrix. To break the code, we first find its opposite matrix (inverse). Then, we take each part of the secret message (which is in numbers) and multiply it by this opposite matrix. The new numbers we get match letters of the alphabet, and when put together, they spell out the secret word.

🎯 Exam Tip: In cryptography problems, ensure you correctly calculate the inverse matrix, as any error will lead to an unreadable decrypted message. Also, pay close attention to the order of multiplication (row matrix times inverse vs. inverse times row matrix).