RBSE Solutions Class 12 Maths Chapter 5 Inverse of a Matrix and Linear Equations Exercise 5.1

Question 1. For which value of x, matrix \[ \begin{bmatrix} 1 & -2 & 3 \\ 1 & 2 & 1 \\ x & 2 & -3 \end{bmatrix} \] is singular ?
Answer: A matrix is singular if its determinant is zero. We need to find the value of x that makes the determinant of the given matrix equal to zero. To do this, we expand the determinant along the first row. We multiply the elements of the first row by their respective cofactors. After performing the multiplication and subtraction steps, we get a simple linear equation. Solving this equation for x, we find its value to be -1. The determinant of a matrix tells us important properties about the matrix, like whether its inverse exists.
In simple words: The matrix is singular when its determinant is zero. We find the value of x that makes the determinant zero by solving the equation.

🎯 Exam Tip: Remember that a matrix is singular if and only if its determinant is zero. Always set up the determinant equal to zero and solve for the unknown variable.

Question 2. If matrix \[ A = \begin{bmatrix} 1 & -1 & 2 \\ 3 & 0 & -2 \\ 1 & 0 & 3 \end{bmatrix} \] , then find adj.A, Also prove that, A(adj.A) = \(|A|\) I\(_3\) = (adj.A)A.
Answer: First, we calculate the determinant of matrix A, which is 11. Then, we find the cofactor for each element of matrix A. The cofactor is found by calculating the determinant of the smaller matrix that remains after removing the row and column of that element, with an alternating sign pattern. After finding all cofactors, we arrange them into a cofactor matrix. The adjoint of A (adj.A) is the transpose of this cofactor matrix, meaning its rows become columns and vice versa. Finally, we multiply matrix A by its adjoint and then multiply the adjoint by A. We show that both results are equal to the determinant of A (which is 11) multiplied by the identity matrix of order 3. The relationship \( A(adj.A) = |A| I \) is a fundamental property that helps us find the inverse of a matrix.
In simple words: We find a special matrix called the 'adjoint' of A. Then we check if multiplying A by its adjoint (in both orders) gives the same result as multiplying the determinant of A by the identity matrix. It helps us understand how matrix multiplication works with special matrices.

🎯 Exam Tip: When proving matrix identities, calculate each part (determinant, cofactors, adjoint, matrix products) carefully step-by-step. Double-check all calculations, especially signs for cofactors and matrix multiplication entries.

Question 3. Find the inverse matrix of the following matrix:
(i) \( \begin{bmatrix} 1 & 2 & 5 \\ -1 & -1 & -1 \\ 2 & 3 & -1 \end{bmatrix} \)
(ii) \( \begin{bmatrix} 1 & 3 & 3 \\ 1 & 4 & 3 \\ 1 & 3 & 4 \end{bmatrix} \)
(iii) \( \begin{bmatrix} 0 & 1 & -1 \\ 4 & -3 & 4 \\ 3 & -3 & 4 \end{bmatrix} \)
Answer: To find the inverse of a matrix, we first calculate its determinant. If the determinant is not zero, the inverse exists. Then, we find the cofactor for each element of the matrix. These cofactors form the cofactor matrix. The adjoint of the matrix is the transpose of this cofactor matrix (its rows become columns). Finally, the inverse matrix is found by dividing the adjoint matrix by the determinant. This process is repeated for each of the given matrices. Finding the inverse of a matrix is crucial for solving systems of linear equations and in many areas of engineering and computer graphics.
(i) Let \( A = \begin{bmatrix} 1 & 2 & 5 \\ -1 & -1 & -1 \\ 2 & 3 & -1 \end{bmatrix} \)
Determinant \( |A| = -7 \)
Cofactor matrix \( C = \begin{bmatrix} 4 & -3 & -1 \\ 17 & -11 & 1 \\ 3 & -4 & 1 \end{bmatrix} \)
Adjoint \( adj.A = C^T = \begin{bmatrix} 4 & 17 & 3 \\ -3 & -11 & -4 \\ -1 & 1 & 1 \end{bmatrix} \)
Inverse \( A^{-1} = \frac{1}{|A|} adj.A = \frac{1}{-7} \begin{bmatrix} 4 & 17 & 3 \\ -3 & -11 & -4 \\ -1 & 1 & 1 \end{bmatrix} = \begin{bmatrix} -\frac{4}{7} & -\frac{17}{7} & -\frac{3}{7} \\ \frac{3}{7} & \frac{11}{7} & \frac{4}{7} \\ \frac{1}{7} & -\frac{1}{7} & -\frac{1}{7} \end{bmatrix} \)
(ii) Let \( A = \begin{bmatrix} 1 & 3 & 3 \\ 1 & 4 & 3 \\ 1 & 3 & 4 \end{bmatrix} \)
Determinant \( |A| = 1 \)
Cofactor matrix \( C = \begin{bmatrix} 7 & -1 & -1 \\ -3 & 1 & 0 \\ -3 & 0 & 1 \end{bmatrix} \)
Adjoint \( adj.A = C^T = \begin{bmatrix} 7 & -3 & -3 \\ -1 & 1 & 0 \\ -1 & 0 & 1 \end{bmatrix} \)
Inverse \( A^{-1} = \frac{1}{|A|} adj.A = \frac{1}{1} \begin{bmatrix} 7 & -3 & -3 \\ -1 & 1 & 0 \\ -1 & 0 & 1 \end{bmatrix} = \begin{bmatrix} 7 & -3 & -3 \\ -1 & 1 & 0 \\ -1 & 0 & 1 \end{bmatrix} \)
(iii) Let \( A = \begin{bmatrix} 0 & 1 & -1 \\ 4 & -3 & 4 \\ 3 & -3 & 4 \end{bmatrix} \)
Determinant \( |A| = -1 \)
Cofactor matrix \( C = \begin{bmatrix} 0 & -4 & -3 \\ -1 & 3 & 3 \\ 1 & -4 & -4 \end{bmatrix} \)
Adjoint \( adj.A = C^T = \begin{bmatrix} 0 & -1 & 1 \\ -4 & 3 & -4 \\ -3 & 3 & -4 \end{bmatrix} \)
Inverse \( A^{-1} = \frac{1}{|A|} adj.A = \frac{1}{-1} \begin{bmatrix} 0 & -1 & 1 \\ -4 & 3 & -4 \\ -3 & 3 & -4 \end{bmatrix} = \begin{bmatrix} 0 & 1 & -1 \\ 4 & -3 & 4 \\ 3 & -3 & 4 \end{bmatrix} \)
In simple words: We find the inverse of each matrix. This involves three main steps: first, find a special number called the 'determinant'. Second, find all 'cofactors' and make a new matrix from them. Third, flip this new matrix (transpose it) to get the 'adjoint', and then divide it by the determinant.

🎯 Exam Tip: Always check that the determinant is non-zero before proceeding, as an inverse only exists for non-singular matrices. Be very careful with signs when calculating cofactors.

Question 4. If matrix \[ A = F(\alpha) = \begin{bmatrix} \cos \alpha & -\sin \alpha & 0 \\ \sin \alpha & \cos \alpha & 0 \\ 0 & 0 & 1 \end{bmatrix} \] then find \( A^{-1} \) and prove that: (i) \( A^{-1}A = I_3 \) (ii) \( A^{-1} = F(-\alpha) \) (iii) \( A(adj.A) = |A|I = (adj.A)A \)
Answer: First, we find the determinant of matrix A, which simplifies to 1 (using the identity \( \cos^2 \alpha + \sin^2 \alpha = 1 \)). Then, we find the cofactor for each element and arrange them into a cofactor matrix. The adjoint of A is the transpose of this cofactor matrix. Since the determinant is 1, the inverse matrix \( A^{-1} \) is simply equal to the adjoint matrix. This type of matrix, called a rotation matrix, is fundamental in geometry for rotating points and objects around an axis.
Determinant \( |A| = 1 \)
Cofactor matrix \( C = \begin{bmatrix} \cos \alpha & -\sin \alpha & 0 \\ \sin \alpha & \cos \alpha & 0 \\ 0 & 0 & 1 \end{bmatrix} \)
Adjoint \( adj.A = C^T = \begin{bmatrix} \cos \alpha & \sin \alpha & 0 \\ -\sin \alpha & \cos \alpha & 0 \\ 0 & 0 & 1 \end{bmatrix} \)
Inverse \( A^{-1} = \frac{1}{|A|} adj.A = \frac{1}{1} \begin{bmatrix} \cos \alpha & \sin \alpha & 0 \\ -\sin \alpha & \cos \alpha & 0 \\ 0 & 0 & 1 \end{bmatrix} = \begin{bmatrix} \cos \alpha & \sin \alpha & 0 \\ -\sin \alpha & \cos \alpha & 0 \\ 0 & 0 & 1 \end{bmatrix} \)
(i) To prove \( A^{-1}A = I_3 \): We multiply the calculated \( A^{-1} \) by the original matrix A. This matrix multiplication results in the identity matrix \( I_3 \), which confirms the property. \[ A^{-1}A = \begin{bmatrix} \cos \alpha & \sin \alpha & 0 \\ -\sin \alpha & \cos \alpha & 0 \\ 0 & 0 & 1 \end{bmatrix} \begin{bmatrix} \cos \alpha & -\sin \alpha & 0 \\ \sin \alpha & \cos \alpha & 0 \\ 0 & 0 & 1 \end{bmatrix} \] \( = \begin{bmatrix} \cos^2 \alpha + \sin^2 \alpha & -\cos \alpha \sin \alpha + \sin \alpha \cos \alpha & 0 \\ -\sin \alpha \cos \alpha + \cos \alpha \sin \alpha & \sin^2 \alpha + \cos^2 \alpha & 0 \\ 0 & 0 & 1 \end{bmatrix} \) \( = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} = I_3 \)
(ii) To prove \( A^{-1} = F(-\alpha) \): We replace \( \alpha \) with \( -\alpha \) in the original matrix definition \( F(\alpha) \). Using trigonometric rules \( \cos(-\alpha) = \cos \alpha \) and \( \sin(-\alpha) = -\sin \alpha \), we find that \( F(-\alpha) \) exactly matches the calculated \( A^{-1} \). \[ F(-\alpha) = \begin{bmatrix} \cos (-\alpha) & -\sin (-\alpha) & 0 \\ \sin (-\alpha) & \cos (-\alpha) & 0 \\ 0 & 0 & 1 \end{bmatrix} = \begin{bmatrix} \cos \alpha & \sin \alpha & 0 \\ -\sin \alpha & \cos \alpha & 0 \\ 0 & 0 & 1 \end{bmatrix} = A^{-1} \]
(iii) To prove \( A(adj.A) = |A|I = (adj.A)A \): We substitute the value of \( |A| \) (which is 1) and multiply A by its adjoint (adj.A) and vice versa. Both products give the identity matrix \( I_3 \), thus confirming the identity.
We have \( |A| = 1 \), so \( |A|I_3 = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} \)
\( A(adj.A) = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} = |A|I_3 \)
\( (adj.A)A = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} = |A|I_3 \)
In simple words: We are given a special matrix A that depends on an angle \( \alpha \). First, we find its inverse matrix. Then we prove three things: (i) When we multiply A by its inverse, we get the identity matrix. (ii) The inverse matrix is the same as the original matrix but with the angle \( \alpha \) changed to \( -\alpha \). (iii) Multiplying A by its adjoint (a special related matrix) gives the same result as multiplying its determinant by the identity matrix, and this works both ways.

🎯 Exam Tip: When dealing with trigonometric matrices, always simplify expressions using identities like \( \sin^2 \alpha + \cos^2 \alpha = 1 \). Be careful with signs when substituting negative angles.

Question 6. If matrix \[ A = \frac{1}{9} \begin{bmatrix} -8 & 1 & 4 \\ 4 & 4 & 7 \\ 1 & -8 & 4 \end{bmatrix} \] , then prove that : \( A^{-1} = A^T \).
Answer: We are given matrix A, which is a scalar factor \( \frac{1}{9} \) multiplied by another matrix, X. To prove \( A^{-1} = A^T \), we first calculate \( A^{-1} \). This involves finding the determinant of X and then its adjoint. The adjoint of X is the transpose of its cofactor matrix. After finding \( X^{-1} \), we scale it by 9 to get \( A^{-1} \). Next, we find the transpose of the original matrix A, which is \( A^T \), by swapping its rows and columns while keeping the scalar factor. By comparing the calculated \( A^{-1} \) and \( A^T \), we show that they are indeed equal. A matrix whose inverse is equal to its transpose is called an orthogonal matrix, which is very common in rotations and transformations in geometry.
Let \( A = \frac{1}{9}X \), where \( X = \begin{bmatrix} -8 & 1 & 4 \\ 4 & 4 & 7 \\ 1 & -8 & 4 \end{bmatrix} \)
Determinant \( |X| = -729 \)
Cofactor matrix of X is \( C_X = \begin{bmatrix} 72 & -9 & -36 \\ -36 & -36 & -63 \\ -9 & 72 & -36 \end{bmatrix} \)
Adjoint \( adj.X = C_X^T = \begin{bmatrix} 72 & -36 & -9 \\ -9 & -36 & 72 \\ -36 & -63 & -36 \end{bmatrix} \)
Inverse \( X^{-1} = \frac{1}{|X|} adj.X = \frac{1}{-729} \begin{bmatrix} 72 & -36 & -9 \\ -9 & -36 & 72 \\ -36 & -63 & -36 \end{bmatrix} \)
Now, \( A^{-1} = 9 X^{-1} = 9 \cdot \frac{1}{-729} \begin{bmatrix} 72 & -36 & -9 \\ -9 & -36 & 72 \\ -36 & -63 & -36 \end{bmatrix} \)
\( A^{-1} = \frac{1}{-81} \begin{bmatrix} 72 & -36 & -9 \\ -9 & -36 & 72 \\ -36 & -63 & -36 \end{bmatrix} = \frac{1}{9} \begin{bmatrix} -8 & 4 & 1 \\ 1 & 4 & -8 \\ 4 & 7 & 4 \end{bmatrix} \)
Now, find the transpose of A:
\( A^T = \left( \frac{1}{9} \begin{bmatrix} -8 & 1 & 4 \\ 4 & 4 & 7 \\ 1 & -8 & 4 \end{bmatrix} \right)^T = \frac{1}{9} \begin{bmatrix} -8 & 4 & 1 \\ 1 & 4 & -8 \\ 4 & 7 & 4 \end{bmatrix} \)
Comparing \( A^{-1} \) and \( A^T \), we see that they are equal.
Hence, \( A^{-1} = A^T \) is proved.
In simple words: We have a matrix A. We need to show that its inverse is the same as its transpose. First, we find the inverse of A by calculating its determinant and adjoint. Then, we find the transpose of A by swapping its rows and columns. When we compare the inverse and the transpose, they turn out to be the same, which proves the statement.

🎯 Exam Tip: When a matrix has a scalar multiplier, remember that \( (kA)^{-1} = \frac{1}{k} A^{-1} \) and \( (kA)^T = k A^T \). This simplifies calculations by handling the scalar separately.

Question 5. If matrix \( A = \frac{1}{9} \begin{bmatrix} -8 & 1 & 4 \\ 4 & 4 & 7 \\ 1 & -8 & 4 \end{bmatrix} \), then prove that \( A^{-1} = A^T \).
Answer:
Let the given matrix be \( A \). To prove \( A^{-1} = A^T \), we first find \( A^{-1} \) and \( A^T \).
Let \( X = \begin{bmatrix} -8 & 1 & 4 \\ 4 & 4 & 7 \\ 1 & -8 & 4 \end{bmatrix} \).
First, find the determinant of X:
\( |X| = -8(4 \times 4 - 7 \times (-8)) - 1(4 \times 4 - 7 \times 1) + 4(4 \times (-8) - 4 \times 1) \)
\( = -8(16 + 56) - 1(16 - 7) + 4(-32 - 4) \)
\( = -8(72) - 1(9) + 4(-36) \)
\( = -576 - 9 - 144 = -729 \)
Next, we find the cofactors of matrix X:
\( F_{11} = +(16 - (-56)) = 72 \)
\( F_{12} = -(16 - 7) = -9 \)
\( F_{13} = +(-32 - 4) = -36 \)
\( F_{21} = -(4 - (-32)) = -36 \)
\( F_{22} = +(-32 - 4) = -36 \)
\( F_{23} = -(-64 - 1) = 63 \)
\( F_{31} = +(7 - 16) = -9 \)
\( F_{32} = -(-56 - 16) = 72 \)
\( F_{33} = +(-32 - 4) = -36 \)
The cofactor matrix is \( C = \begin{bmatrix} 72 & -9 & -36 \\ -36 & -36 & 63 \\ -9 & 72 & -36 \end{bmatrix} \).
The adjoint of X is the transpose of the cofactor matrix:
\( \text{adj.X} = C^T = \begin{bmatrix} 72 & -36 & -9 \\ -9 & -36 & 72 \\ -36 & 63 & -36 \end{bmatrix} \)
Now, calculate \( X^{-1} \):
\( X^{-1} = \frac{1}{|X|} \text{adj.X} = \frac{1}{-729} \begin{bmatrix} 72 & -36 & -9 \\ -9 & -36 & 72 \\ -36 & 63 & -36 \end{bmatrix} = \frac{1}{81} \begin{bmatrix} -8 & 4 & 1 \\ 1 & 4 & -8 \\ 4 & -7 & 4 \end{bmatrix} \)
Given \( A = \frac{1}{9} X \). Then, using the property \( (kM)^{-1} = \frac{1}{k} M^{-1} \):
\( A^{-1} = \left(\frac{1}{9}X\right)^{-1} = 9X^{-1} \)
\( A^{-1} = 9 \times \frac{1}{81} \begin{bmatrix} -8 & 4 & 1 \\ 1 & 4 & -8 \\ 4 & -7 & 4 \end{bmatrix} = \frac{1}{9} \begin{bmatrix} -8 & 4 & 1 \\ 1 & 4 & -8 \\ 4 & -7 & 4 \end{bmatrix} \)
Next, we find \( A^T \). Using the property \( (kM)^T = kM^T \):
\( A^T = \left(\frac{1}{9}X\right)^T = \frac{1}{9}X^T \)
\( A^T = \frac{1}{9} \begin{bmatrix} -8 & 1 & 4 \\ 4 & 4 & 7 \\ 1 & -8 & 4 \end{bmatrix}^T = \frac{1}{9} \begin{bmatrix} -8 & 4 & 1 \\ 1 & 4 & -8 \\ 4 & 7 & 4 \end{bmatrix} \)
Comparing \( A^{-1} \) and \( A^T \), we see that they are not equal due to \( -7 \) vs \( 7 \). The source calculations for cofactors seem to have a sign error for F23 or F32. Let's recheck F23 from OCR: `F23=- (-8*-8 - 1*1) = -(64-1) = -63`. My F23 is `-(5*2 - 0*1) = -(10-0)=-10` for Q7. Re-calculating F23 for Q5 with the source values: \( F_{23} = -((-8)\times (-8) - 1 \times 1) = -(64 - 1) = -63 \). So, the cofactor matrix from the source should be: \( C = \begin{bmatrix} 72 & -9 & -36 \\ -36 & -36 & -63 \\ -9 & 72 & -36 \end{bmatrix} \) Then, \( \text{adj.X} = C^T = \begin{bmatrix} 72 & -36 & -9 \\ -9 & -36 & 72 \\ -36 & -63 & -36 \end{bmatrix} \) And \( X^{-1} = \frac{1}{-729} \begin{bmatrix} 72 & -36 & -9 \\ -9 & -36 & 72 \\ -36 & -63 & -36 \end{bmatrix} = \frac{1}{81} \begin{bmatrix} -8 & 4 & 1 \\ 1 & 4 & -8 \\ 4 & 7 & 4 \end{bmatrix} \). This calculation is correct. The \( A^T \) is \( \frac{1}{9} \begin{bmatrix} -8 & 4 & 1 \\ 1 & 4 & -8 \\ 4 & 7 & 4 \end{bmatrix} \). Since \( A^{-1} = \frac{1}{9} \begin{bmatrix} -8 & 4 & 1 \\ 1 & 4 & -8 \\ 4 & 7 & 4 \end{bmatrix} \) and \( A^T = \frac{1}{9} \begin{bmatrix} -8 & 4 & 1 \\ 1 & 4 & -8 \\ 4 & 7 & 4 \end{bmatrix} \), we have \( A^{-1} = A^T \). Thus, it is proved.
In simple words: We first found the inverse of matrix A and its transpose. After calculating both, we showed that the final matrices for \( A^{-1} \) and \( A^T \) are exactly the same. This proves the given statement.

🎯 Exam Tip: When proving matrix identities, always calculate each side of the equation separately and show that they result in identical matrices. Watch out for sign errors when calculating cofactors and transposes.

Question 6. If matrix \( A = \begin{bmatrix} 1 & -1 \\ 2 & -1 \end{bmatrix} \), then prove that \( A^{-1} = A^3 \).
Answer:
Given matrix \( A = \begin{bmatrix} 1 & -1 \\ 2 & -1 \end{bmatrix} \).
First, we find the determinant of A:
\( |A| = (1)(-1) - (-1)(2) = -1 + 2 = 1 \).
Since \( |A| = 1 \neq 0 \), the inverse \( A^{-1} \) exists.
Next, we find the cofactors of A:
\( A_{11} = -1 \)
\( A_{12} = -2 \)
\( A_{21} = -(-1) = 1 \)
\( A_{22} = 1 \)
The cofactor matrix is \( C = \begin{bmatrix} -1 & -2 \\ 1 & 1 \end{bmatrix} \).
The adjoint of A is the transpose of the cofactor matrix:
\( \text{adj.A} = C^T = \begin{bmatrix} -1 & 1 \\ -2 & 1 \end{bmatrix} \).
Now, we calculate \( A^{-1} \):
\( A^{-1} = \frac{1}{|A|} \text{adj.A} = \frac{1}{1} \begin{bmatrix} -1 & 1 \\ -2 & 1 \end{bmatrix} = \begin{bmatrix} -1 & 1 \\ -2 & 1 \end{bmatrix} \).
Next, we calculate \( A^2 \) and \( A^3 \):
\( A^2 = A \times A = \begin{bmatrix} 1 & -1 \\ 2 & -1 \end{bmatrix} \begin{bmatrix} 1 & -1 \\ 2 & -1 \end{bmatrix} \)
\( = \begin{bmatrix} (1)(1)+(-1)(2) & (1)(-1)+(-1)(-1) \\ (2)(1)+(-1)(2) & (2)(-1)+(-1)(-1) \end{bmatrix} \)
\( = \begin{bmatrix} 1-2 & -1+1 \\ 2-2 & -2+1 \end{bmatrix} = \begin{bmatrix} -1 & 0 \\ 0 & -1 \end{bmatrix} \).
Now, calculate \( A^3 \):
\( A^3 = A^2 \times A = \begin{bmatrix} -1 & 0 \\ 0 & -1 \end{bmatrix} \begin{bmatrix} 1 & -1 \\ 2 & -1 \end{bmatrix} \)
\( = \begin{bmatrix} (-1)(1)+(0)(2) & (-1)(-1)+(0)(-1) \\ (0)(1)+(-1)(2) & (0)(-1)+(-1)(-1) \end{bmatrix} \)
\( = \begin{bmatrix} -1 & 1 \\ -2 & 1 \end{bmatrix} \).
Comparing \( A^{-1} \) and \( A^3 \), we see that:
\( A^{-1} = \begin{bmatrix} -1 & 1 \\ -2 & 1 \end{bmatrix} \) and \( A^3 = \begin{bmatrix} -1 & 1 \\ -2 & 1 \end{bmatrix} \).
Therefore, \( A^{-1} = A^3 \). The statement is proved.
In simple words: We first found the inverse of matrix A. Then, we multiplied A by itself three times to find A cubed. Since both results were exactly the same, we proved that \( A^{-1} \) equals \( A^3 \).

🎯 Exam Tip: For proving matrix powers or inverses, compute each side of the equation separately and show that they produce identical matrices. This method helps verify each step clearly.

Question 7. If \( A = \begin{bmatrix} 5 & 0 & 4 \\ 2 & 3 & 2 \\ 1 & 2 & 1 \end{bmatrix} \) and \( B^{-1} = \begin{bmatrix} 1 & 3 & 3 \\ 1 & 4 & 3 \\ 1 & 3 & 4 \end{bmatrix} \), then find \( (AB)^{-1} \).
Answer:
We need to find \( (AB)^{-1} \). We know the property \( (AB)^{-1} = B^{-1} A^{-1} \).
We are given \( B^{-1} \), so we need to find \( A^{-1} \).
Given \( A = \begin{bmatrix} 5 & 0 & 4 \\ 2 & 3 & 2 \\ 1 & 2 & 1 \end{bmatrix} \).
First, find the determinant of A:
\( |A| = 5(3 \times 1 - 2 \times 2) - 0(2 \times 1 - 2 \times 1) + 4(2 \times 2 - 3 \times 1) \)
\( = 5(3 - 4) - 0 + 4(4 - 3) \)
\( = 5(-1) + 4(1) = -5 + 4 = -1 \).
Since \( |A| = -1 \neq 0 \), \( A^{-1} \) exists.
Next, find the cofactors of A:
\( F_{11} = +(3 \times 1 - 2 \times 2) = 3 - 4 = -1 \)
\( F_{12} = -(2 \times 1 - 2 \times 1) = -(2 - 2) = 0 \)
\( F_{13} = +(2 \times 2 - 3 \times 1) = 4 - 3 = 1 \)
\( F_{21} = -(0 \times 1 - 4 \times 2) = -(0 - 8) = 8 \)
\( F_{22} = +(5 \times 1 - 4 \times 1) = 5 - 4 = 1 \)
\( F_{23} = -(5 \times 2 - 0 \times 1) = -(10 - 0) = -10 \)
\( F_{31} = +(0 \times 2 - 4 \times 3) = 0 - 12 = -12 \)
\( F_{32} = -(5 \times 2 - 4 \times 2) = -(10 - 8) = -2 \)
\( F_{33} = +(5 \times 3 - 0 \times 2) = 15 - 0 = 15 \)
The cofactor matrix is \( C = \begin{bmatrix} -1 & 0 & 1 \\ 8 & 1 & -10 \\ -12 & -2 & 15 \end{bmatrix} \).
The adjoint of A is the transpose of the cofactor matrix:
\( \text{adj.A} = C^T = \begin{bmatrix} -1 & 8 & -12 \\ 0 & 1 & -2 \\ 1 & -10 & 15 \end{bmatrix} \).
Now, calculate \( A^{-1} \):
\( A^{-1} = \frac{1}{|A|} \text{adj.A} = \frac{1}{-1} \begin{bmatrix} -1 & 8 & -12 \\ 0 & 1 & -2 \\ 1 & -10 & 15 \end{bmatrix} = \begin{bmatrix} 1 & -8 & 12 \\ 0 & -1 & 2 \\ -1 & 10 & -15 \end{bmatrix} \).
Finally, calculate \( (AB)^{-1} = B^{-1} A^{-1} \):
\( (AB)^{-1} = \begin{bmatrix} 1 & 3 & 3 \\ 1 & 4 & 3 \\ 1 & 3 & 4 \end{bmatrix} \begin{bmatrix} 1 & -8 & 12 \\ 0 & -1 & 2 \\ -1 & 10 & -15 \end{bmatrix} \)
\( = \begin{bmatrix} 1+0-3 & -8-3+30 & 12+6-45 \\ 1+0-3 & -8-4+30 & 12+8-45 \\ 1+0-4 & -8-3+40 & 12+6-60 \end{bmatrix} \)
\( = \begin{bmatrix} -2 & 19 & -27 \\ -2 & 18 & -25 \\ -3 & 29 & -42 \end{bmatrix} \).
In simple words: To find the inverse of the product of two matrices (AB), we used the rule that it equals the inverse of B multiplied by the inverse of A. We already had \( B^{-1} \), so we calculated \( A^{-1} \) first. Then, we multiplied \( B^{-1} \) by \( A^{-1} \) to get the final answer.

🎯 Exam Tip: Remember the property \( (AB)^{-1} = B^{-1} A^{-1} \) for matrix products. Always calculate inverses accurately by finding the determinant and adjoint first.

Question 8. If \( A = \begin{bmatrix} 1 & \tan \alpha \\ -\tan \alpha & 1 \end{bmatrix} \), then prove that : \( A^T A^{-1} = \begin{bmatrix} \cos 2\alpha & -\sin 2\alpha \\ \sin 2\alpha & \cos 2\alpha \end{bmatrix} \).
Answer:
Given matrix \( A = \begin{bmatrix} 1 & \tan \alpha \\ -\tan \alpha & 1 \end{bmatrix} \).
First, find the transpose of A:
\( A^T = \begin{bmatrix} 1 & -\tan \alpha \\ \tan \alpha & 1 \end{bmatrix} \).
Next, find the inverse of A, \( A^{-1} \).
Calculate the determinant of A:
\( |A| = (1)(1) - (\tan \alpha)(-\tan \alpha) = 1 + \tan^2 \alpha \).
We know that \( 1 + \tan^2 \alpha = \sec^2 \alpha \). So, \( |A| = \sec^2 \alpha \).
Since \( |A| \neq 0 \), \( A^{-1} \) exists.
The cofactor matrix of A is \( C = \begin{bmatrix} 1 & -(-\tan \alpha) \\ -(\tan \alpha) & 1 \end{bmatrix} = \begin{bmatrix} 1 & \tan \alpha \\ -\tan \alpha & 1 \end{bmatrix} \).
The adjoint of A is the transpose of the cofactor matrix:
\( \text{adj.A} = C^T = \begin{bmatrix} 1 & -\tan \alpha \\ \tan \alpha & 1 \end{bmatrix} \).
Now, calculate \( A^{-1} \):
\( A^{-1} = \frac{1}{|A|} \text{adj.A} = \frac{1}{\sec^2 \alpha} \begin{bmatrix} 1 & -\tan \alpha \\ \tan \alpha & 1 \end{bmatrix} = \cos^2 \alpha \begin{bmatrix} 1 & -\tan \alpha \\ \tan \alpha & 1 \end{bmatrix} \).
Finally, calculate \( A^T A^{-1} \):
\( A^T A^{-1} = \begin{bmatrix} 1 & -\tan \alpha \\ \tan \alpha & 1 \end{bmatrix} \times \cos^2 \alpha \begin{bmatrix} 1 & -\tan \alpha \\ \tan \alpha & 1 \end{bmatrix} \)
\( = \cos^2 \alpha \begin{bmatrix} 1 & -\tan \alpha \\ \tan \alpha & 1 \end{bmatrix} \begin{bmatrix} 1 & -\tan \alpha \\ \tan \alpha & 1 \end{bmatrix} \)
\( = \cos^2 \alpha \begin{bmatrix} (1)(1)+(-\tan \alpha)(\tan \alpha) & (1)(-\tan \alpha)+(-\tan \alpha)(1) \\ (\tan \alpha)(1)+(1)(\tan \alpha) & (\tan \alpha)(-\tan \alpha)+(1)(1) \end{bmatrix} \)
\( = \cos^2 \alpha \begin{bmatrix} 1-\tan^2 \alpha & -2\tan \alpha \\ 2\tan \alpha & 1-\tan^2 \alpha \end{bmatrix} \).
Distribute \( \cos^2 \alpha \) and use trigonometric identities \( \cos 2\alpha = \cos^2 \alpha - \sin^2 \alpha = \cos^2 \alpha (1-\tan^2 \alpha) \) and \( \sin 2\alpha = 2\sin \alpha \cos \alpha = 2\tan \alpha \cos^2 \alpha \):
\( = \begin{bmatrix} \cos^2 \alpha (1-\tan^2 \alpha) & -2\tan \alpha \cos^2 \alpha \\ 2\tan \alpha \cos^2 \alpha & \cos^2 \alpha (1-\tan^2 \alpha) \end{bmatrix} \)
\( = \begin{bmatrix} \cos 2\alpha & -\sin 2\alpha \\ \sin 2\alpha & \cos 2\alpha \end{bmatrix} \).
Thus, it is proved that \( A^T A^{-1} = \begin{bmatrix} \cos 2\alpha & -\sin 2\alpha \\ \sin 2\alpha & \cos 2\alpha \end{bmatrix} \).
In simple words: We calculated the transpose of A and the inverse of A separately. Then, we multiplied these two results together. By using basic trigonometry rules, we showed that the final matrix matches the target matrix with \( \cos 2\alpha \) and \( \sin 2\alpha \) terms.

🎯 Exam Tip: When dealing with trigonometric functions in matrices, remember to use identities like \( \sec^2 \alpha = 1 + \tan^2 \alpha \), \( \cos 2\alpha = \cos^2 \alpha - \sin^2 \alpha \), and \( \sin 2\alpha = 2\sin \alpha \cos \alpha \). These simplify the final expression.

Question 9. Show that matrix \( A = \begin{bmatrix} 2 & -3 \\ 3 & 4 \end{bmatrix} \) satisfies the equation \( A^2 – 6A + 17I = O \). Thus find \( A^{-1} \).
Answer:
Given matrix \( A = \begin{bmatrix} 2 & -3 \\ 3 & 4 \end{bmatrix} \).
First, calculate \( A^2 \):
\( A^2 = A \times A = \begin{bmatrix} 2 & -3 \\ 3 & 4 \end{bmatrix} \begin{bmatrix} 2 & -3 \\ 3 & 4 \end{bmatrix} \)
\( = \begin{bmatrix} (2)(2)+(-3)(3) & (2)(-3)+(-3)(4) \\ (3)(2)+(4)(3) & (3)(-3)+(4)(4) \end{bmatrix} \)
\( = \begin{bmatrix} 4-9 & -6-12 \\ 6+12 & -9+16 \end{bmatrix} = \begin{bmatrix} -5 & -18 \\ 18 & 7 \end{bmatrix} \).
Next, calculate \( 6A \):
\( 6A = 6 \begin{bmatrix} 2 & -3 \\ 3 & 4 \end{bmatrix} = \begin{bmatrix} 12 & -18 \\ 18 & 24 \end{bmatrix} \).
Next, calculate \( 17I \), where \( I = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \) is the identity matrix:
\( 17I = 17 \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} 17 & 0 \\ 0 & 17 \end{bmatrix} \).
Now, substitute these into the equation \( A^2 – 6A + 17I \):
\( A^2 – 6A + 17I = \begin{bmatrix} -5 & -18 \\ 18 & 7 \end{bmatrix} - \begin{bmatrix} 12 & -18 \\ 18 & 24 \end{bmatrix} + \begin{bmatrix} 17 & 0 \\ 0 & 17 \end{bmatrix} \)
\( = \begin{bmatrix} -5-12+17 & -18-(-18)+0 \\ 18-18+0 & 7-24+17 \end{bmatrix} \)
\( = \begin{bmatrix} -17+17 & -18+18 \\ 0 & 24-24 \end{bmatrix} = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix} = O \).
Thus, the matrix A satisfies the equation \( A^2 – 6A + 17I = O \).
To find \( A^{-1} \), we use the satisfied equation. Multiply the equation by \( A^{-1} \):
\( A^{-1} (A^2 – 6A + 17I) = A^{-1} O \)
\( A^{-1} A^2 – 6A^{-1} A + 17A^{-1} I = O \)
\( A – 6I + 17A^{-1} = O \)
\( 17A^{-1} = 6I – A \)
\( A^{-1} = \frac{1}{17} (6I – A) \).
Now, calculate \( 6I - A \):
\( 6I - A = \begin{bmatrix} 6 & 0 \\ 0 & 6 \end{bmatrix} - \begin{bmatrix} 2 & -3 \\ 3 & 4 \end{bmatrix} = \begin{bmatrix} 6-2 & 0-(-3) \\ 0-3 & 6-4 \end{bmatrix} = \begin{bmatrix} 4 & 3 \\ -3 & 2 \end{bmatrix} \).
Therefore, \( A^{-1} = \frac{1}{17} \begin{bmatrix} 4 & 3 \\ -3 & 2 \end{bmatrix} \).
In simple words: We first checked if the matrix A fits the given equation by calculating A squared, 6 times A, and 17 times the identity matrix. When we added and subtracted these, we got a matrix of all zeros, which proves the equation. Then, we used this equation to find the inverse of A.

🎯 Exam Tip: For problems like this, first calculate all the terms in the equation (like \( A^2 \), \( kA \), \( cI \)) separately before combining them. To find \( A^{-1} \) using the characteristic equation, remember to multiply by \( A^{-1} \) and rearrange the terms.

Question 10. If matrix \( A = \begin{bmatrix} 2 & -8 \\ 3 & 4 \end{bmatrix} \), then show that \( A^2 - 6A + 32I = O \). Hence find \( A^{-1} \).
Answer:
Given matrix \( A = \begin{bmatrix} 2 & -8 \\ 3 & 4 \end{bmatrix} \).
First, calculate \( A^2 \):
\( A^2 = A \times A = \begin{bmatrix} 2 & -8 \\ 3 & 4 \end{bmatrix} \begin{bmatrix} 2 & -8 \\ 3 & 4 \end{bmatrix} \)
\( = \begin{bmatrix} (2)(2)+(-8)(3) & (2)(-8)+(-8)(4) \\ (3)(2)+(4)(3) & (3)(-8)+(4)(4) \end{bmatrix} \)
\( = \begin{bmatrix} 4-24 & -16-32 \\ 6+12 & -24+16 \end{bmatrix} = \begin{bmatrix} -20 & -48 \\ 18 & -8 \end{bmatrix} \).
Next, calculate \( 6A \):
\( 6A = 6 \begin{bmatrix} 2 & -8 \\ 3 & 4 \end{bmatrix} = \begin{bmatrix} 12 & -48 \\ 18 & 24 \end{bmatrix} \).
Next, calculate \( 32I \), where \( I = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \) is the identity matrix:
\( 32I = 32 \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} 32 & 0 \\ 0 & 32 \end{bmatrix} \).
Now, substitute these into the characteristic equation \( A^2 - 6A + 32I \):
\( A^2 - 6A + 32I = \begin{bmatrix} -20 & -48 \\ 18 & -8 \end{bmatrix} - \begin{bmatrix} 12 & -48 \\ 18 & 24 \end{bmatrix} + \begin{bmatrix} 32 & 0 \\ 0 & 32 \end{bmatrix} \)
\( = \begin{bmatrix} -20-12+32 & -48-(-48)+0 \\ 18-18+0 & -8-24+32 \end{bmatrix} \)
\( = \begin{bmatrix} -32+32 & -48+48 \\ 0 & -32+32 \end{bmatrix} = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix} = O \).
Thus, the matrix A satisfies the equation \( A^2 - 6A + 32I = O \).
To find \( A^{-1} \), we use the satisfied equation. Multiply the equation by \( A^{-1} \):
\( A^{-1} (A^2 - 6A + 32I) = A^{-1} O \)
\( A^{-1} A^2 - 6A^{-1} A + 32A^{-1} I = O \)
\( A - 6I + 32A^{-1} = O \)
\( 32A^{-1} = 6I - A \)
\( A^{-1} = \frac{1}{32} (6I - A) \).
Now, calculate \( 6I - A \):
\( 6I - A = \begin{bmatrix} 6 & 0 \\ 0 & 6 \end{bmatrix} - \begin{bmatrix} 2 & -8 \\ 3 & 4 \end{bmatrix} = \begin{bmatrix} 6-2 & 0-(-8) \\ 0-3 & 6-4 \end{bmatrix} = \begin{bmatrix} 4 & 8 \\ -3 & 2 \end{bmatrix} \).
Therefore, \( A^{-1} = \frac{1}{32} \begin{bmatrix} 4 & 8 \\ -3 & 2 \end{bmatrix} \).
In simple words: We first found A squared, 6 times A, and 32 times the identity matrix. When we combined them, the result was a zero matrix, which shows the equation is true. Then, we used this proven equation to calculate the inverse of A.

🎯 Exam Tip: Every square matrix satisfies its own characteristic equation (Cayley-Hamilton theorem). If an equation is given to be satisfied by a matrix, verify it by direct substitution. Use the characteristic equation to find the inverse when asked, as it's a common method.