GSEB Class 12 Maths Solutions Chapter 4 નિશ્ચાયક Exercise 4.5

Question 1. \( \left[\begin{array}{ll}1 & 2 \\3 & 4 \end{array}\right] \)
Answer: Let the given matrix be \( A = \begin{pmatrix} 1 & 2 \\ 3 & 4 \end{pmatrix} \).
First, we find the minors of the matrix elements:
\( M_{11} = 4 \)
\( M_{12} = 3 \)
\( M_{21} = 2 \)
\( M_{22} = 1 \)
Next, we calculate the cofactors for each element:
\( A_{11} = (-1)^{1+1} M_{11} = (1)(4) = 4 \)
\( A_{12} = (-1)^{1+2} M_{12} = (-1)(3) = -3 \)
\( A_{21} = (-1)^{2+1} M_{21} = (-1)(2) = -2 \)
\( A_{22} = (-1)^{2+2} M_{22} = (1)(1) = 1 \)
The matrix of cofactors is \( C = \begin{pmatrix} 4 & -3 \\ -2 & 1 \end{pmatrix} \).
The adjoint of matrix \( A \), denoted as \( \text{adj } A \), is the transpose of the cofactor matrix.
\( \text{adj } A = C^T = \begin{pmatrix} 4 & -3 \\ -2 & 1 \end{pmatrix}^T = \begin{pmatrix} 4 & -2 \\ -3 & 1 \end{pmatrix} \)
In simple words: To find the adjoint, first find the minor for each number. Then, use the minor to find the cofactor. Finally, flip the rows and columns of the cofactor matrix to get the adjoint.

Exam Tip: Remember to correctly apply the \( (-1)^{i+j} \) rule for cofactors. A common mistake is flipping signs incorrectly or confusing minors with cofactors.

Question 2. \( \left[\begin{array}{ccc} 1 & -1 & 2 \\ 2 & 3 & 5 \\ -2 & 0 & 1 \end{array}\right] \)
Answer: Let the given matrix be \( A = \begin{pmatrix} 1 & -1 & 2 \\ 2 & 3 & 5 \\ -2 & 0 & 1 \end{pmatrix} \).
We will now compute the cofactor for each element:
\( A_{11} = (-1)^{1+1} \begin{vmatrix} 3 & 5 \\ 0 & 1 \end{vmatrix} = 1(3-0) = 3 \)
\( A_{12} = (-1)^{1+2} \begin{vmatrix} 2 & 5 \\ -2 & 1 \end{vmatrix} = -1(2-(-10)) = -(2+10) = -12 \)
\( A_{13} = (-1)^{1+3} \begin{vmatrix} 2 & 3 \\ -2 & 0 \end{vmatrix} = 1(0-(-6)) = 6 \)
\( A_{21} = (-1)^{2+1} \begin{vmatrix} -1 & 2 \\ 0 & 1 \end{vmatrix} = -1(-1-0) = 1 \)
\( A_{22} = (-1)^{2+2} \begin{vmatrix} 1 & 2 \\ -2 & 1 \end{vmatrix} = 1(1-(-4)) = 1+4 = 5 \)
\( A_{23} = (-1)^{2+3} \begin{vmatrix} 1 & -1 \\ -2 & 0 \end{vmatrix} = -1(0-2) = -(-2) = 2 \)
\( A_{31} = (-1)^{3+1} \begin{vmatrix} -1 & 2 \\ 3 & 5 \end{vmatrix} = 1(-5-6) = -11 \)
\( A_{32} = (-1)^{3+2} \begin{vmatrix} 1 & 2 \\ 2 & 5 \end{vmatrix} = -1(5-4) = -1 \)
\( A_{33} = (-1)^{3+3} \begin{vmatrix} 1 & -1 \\ 2 & 3 \end{vmatrix} = 1(3-(-2)) = 3+2 = 5 \)
The cofactor matrix is \( C = \begin{pmatrix} 3 & -12 & 6 \\ 1 & 5 & 2 \\ -11 & -1 & 5 \end{pmatrix} \).
The adjoint of matrix \( A \) is the transpose of the cofactor matrix:
\( \text{adj } A = C^T = \begin{pmatrix} 3 & -12 & 6 \\ 1 & 5 & 2 \\ -11 & -1 & 5 \end{pmatrix}^T = \begin{pmatrix} 3 & 1 & -11 \\ -12 & 5 & -1 \\ 6 & 2 & 5 \end{pmatrix} \)
In simple words: For a 3x3 matrix, we calculate nine cofactors by taking turns covering rows and columns, multiplying by the correct sign, and then finding the determinant of the smaller 2x2 matrix. Then we transpose this matrix of cofactors to get the adjoint.

Exam Tip: Be very careful with signs when calculating cofactors. A small sign error can change the entire result. Double-check your calculations, especially for larger matrices.

Question 3. \( \left[\begin{array}{ll} 2 & 3 \\ -4 & -6 \end{array}\right] \)
Answer: Let the given matrix be \( A = \begin{pmatrix} 2 & 3 \\ -4 & -6 \end{pmatrix} \).
First, we find the cofactors for each element:
\( A_{11} = (-1)^{1+1} (-6) = -6 \)
\( A_{12} = (-1)^{1+2} (-4) = -(-4) = 4 \)
\( A_{21} = (-1)^{2+1} (3) = -3 \)
\( A_{22} = (-1)^{2+2} (2) = 2 \)
The matrix of cofactors is \( C = \begin{pmatrix} -6 & 4 \\ -3 & 2 \end{pmatrix} \).
The adjoint of matrix \( A \) is the transpose of the cofactor matrix:
\( \text{adj } A = C^T = \begin{pmatrix} -6 & 4 \\ -3 & 2 \end{pmatrix}^T = \begin{pmatrix} -6 & -3 \\ 4 & 2 \end{pmatrix} \)
Now, we verify that \( A (\text{adj } A) = (\text{adj } A) A = |A| I \).
First, calculate the determinant of \( A \):
\( |A| = (2)(-6) - (3)(-4) = -12 - (-12) = -12 + 12 = 0 \)
Next, calculate \( A (\text{adj } A) \):
\( A (\text{adj } A) = \begin{pmatrix} 2 & 3 \\ -4 & -6 \end{pmatrix} \begin{pmatrix} -6 & -3 \\ 4 & 2 \end{pmatrix} = \begin{pmatrix} (2)(-6) + (3)(4) & (2)(-3) + (3)(2) \\ (-4)(-6) + (-6)(4) & (-4)(-3) + (-6)(2) \end{pmatrix} \)
\( = \begin{pmatrix} -12+12 & -6+6 \\ 24-24 & 12-12 \end{pmatrix} = \begin{pmatrix} 0 & 0 \\ 0 & 0 \end{pmatrix} \)
This result is equal to \( |A|I \), since \( |A|=0 \) and \( I = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} \), so \( |A|I = 0 \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} = \begin{pmatrix} 0 & 0 \\ 0 & 0 \end{pmatrix} \).
Similarly, we can show that \( (\text{adj } A) A = \begin{pmatrix} -6 & -3 \\ 4 & 2 \end{pmatrix} \begin{pmatrix} 2 & 3 \\ -4 & -6 \end{pmatrix} = \begin{pmatrix} (-6)(2) + (-3)(-4) & (-6)(3) + (-3)(-6) \\ (4)(2) + (2)(-4) & (4)(3) + (2)(-6) \end{pmatrix} \)
\( = \begin{pmatrix} -12+12 & -18+18 \\ 8-8 & 12-12 \end{pmatrix} = \begin{pmatrix} 0 & 0 \\ 0 & 0 \end{pmatrix} \)
Hence, \( A (\text{adj } A) = (\text{adj } A) A = |A| I \) is verified.
In simple words: We find the adjoint by calculating cofactors and transposing the cofactor matrix. Then, we multiply the original matrix by its adjoint (and vice-versa) to prove that the result is the determinant of the matrix multiplied by the identity matrix.

Exam Tip: For verification questions like this, always show both \( A (\text{adj } A) \) and \( (\text{adj } A) A \) calculations explicitly, and confirm that both equal \( |A|I \).

Question 4. \( \left[\begin{array}{ccc} 1 & -1 & 2 \\ 3 & 0 & -2 \\ 1 & 0 & 3 \end{array}\right] \)
Answer: Let the given matrix be \( A = \begin{pmatrix} 1 & -1 & 2 \\ 3 & 0 & -2 \\ 1 & 0 & 3 \end{pmatrix} \).
We will now compute the cofactor for each element:
\( A_{11} = (-1)^{1+1} \begin{vmatrix} 0 & -2 \\ 0 & 3 \end{vmatrix} = 1(0-0) = 0 \)
\( A_{12} = (-1)^{1+2} \begin{vmatrix} 3 & -2 \\ 1 & 3 \end{vmatrix} = -1(9-(-2)) = -(9+2) = -11 \)
\( A_{13} = (-1)^{1+3} \begin{vmatrix} 3 & 0 \\ 1 & 0 \end{vmatrix} = 1(0-0) = 0 \)
\( A_{21} = (-1)^{2+1} \begin{vmatrix} -1 & 2 \\ 0 & 3 \end{vmatrix} = -1(-3-0) = 3 \)
\( A_{22} = (-1)^{2+2} \begin{vmatrix} 1 & 2 \\ 1 & 3 \end{vmatrix} = 1(3-2) = 1 \)
\( A_{23} = (-1)^{2+3} \begin{vmatrix} 1 & -1 \\ 1 & 0 \end{vmatrix} = -1(0-(-1)) = -1(1) = -1 \)
\( A_{31} = (-1)^{3+1} \begin{vmatrix} -1 & 2 \\ 0 & -2 \end{vmatrix} = 1(2-0) = 2 \)
\( A_{32} = (-1)^{3+2} \begin{vmatrix} 1 & 2 \\ 3 & -2 \end{vmatrix} = -1(-2-6) = -1(-8) = 8 \)
\( A_{33} = (-1)^{3+3} \begin{vmatrix} 1 & -1 \\ 3 & 0 \end{vmatrix} = 1(0-(-3)) = 3 \)
The cofactor matrix is \( C = \begin{pmatrix} 0 & -11 & 0 \\ 3 & 1 & -1 \\ 2 & 8 & 3 \end{pmatrix} \).
The adjoint of matrix \( A \) is the transpose of the cofactor matrix:
\( \text{adj } A = C^T = \begin{pmatrix} 0 & -11 & 0 \\ 3 & 1 & -1 \\ 2 & 8 & 3 \end{pmatrix}^T = \begin{pmatrix} 0 & 3 & 2 \\ -11 & 1 & 8 \\ 0 & -1 & 3 \end{pmatrix} \)
Now, we verify that \( A (\text{adj } A) = |A| I \).
First, calculate the determinant of \( A \):
\( |A| = 1(A_{11}) + (-1)(A_{12}) + 2(A_{13}) = 1(0) - 1(-11) + 2(0) = 0 + 11 + 0 = 11 \)
Next, calculate \( A (\text{adj } A) \):
\( A (\text{adj } A) = \begin{pmatrix} 1 & -1 & 2 \\ 3 & 0 & -2 \\ 1 & 0 & 3 \end{pmatrix} \begin{pmatrix} 0 & 3 & 2 \\ -11 & 1 & 8 \\ 0 & -1 & 3 \end{pmatrix} = \begin{pmatrix} (1)(0)+(-1)(-11)+(2)(0) & (1)(3)+(-1)(1)+(2)(-1) & (1)(2)+(-1)(8)+(2)(3) \\ (3)(0)+(0)(-11)+(-2)(0) & (3)(3)+(0)(1)+(-2)(-1) & (3)(2)+(0)(8)+(-2)(3) \\ (1)(0)+(0)(-11)+(3)(0) & (1)(3)+(0)(1)+(3)(-1) & (1)(2)+(0)(8)+(3)(3) \end{pmatrix} \)
\( = \begin{pmatrix} 0+11+0 & 3-1-2 & 2-8+6 \\ 0+0+0 & 9+0+2 & 6+0-6 \\ 0+0+0 & 3+0-3 & 2+0+9 \end{pmatrix} = \begin{pmatrix} 11 & 0 & 0 \\ 0 & 11 & 0 \\ 0 & 0 & 11 \end{pmatrix} \)
This result is equal to \( |A|I = 11 \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix} = \begin{pmatrix} 11 & 0 & 0 \\ 0 & 11 & 0 \\ 0 & 0 & 11 \end{pmatrix} \).
Thus, \( A (\text{adj } A) = |A| I \).
Similarly, \( (\text{adj } A) A = \begin{pmatrix} 0 & 3 & 2 \\ -11 & 1 & 8 \\ 0 & -1 & 3 \end{pmatrix} \begin{pmatrix} 1 & -1 & 2 \\ 3 & 0 & -2 \\ 1 & 0 & 3 \end{pmatrix} = \begin{pmatrix} (0)(1)+(3)(3)+(2)(1) & (0)(-1)+(3)(0)+(2)(0) & (0)(2)+(3)(-2)+(2)(3) \\ (-11)(1)+(1)(3)+(8)(1) & (-11)(-1)+(1)(0)+(8)(0) & (-11)(2)+(1)(-2)+(8)(3) \\ (0)(1)+(-1)(3)+(3)(1) & (0)(-1)+(-1)(0)+(3)(0) & (0)(2)+(-1)(-2)+(3)(3) \end{pmatrix} \)
\( = \begin{pmatrix} 0+9+2 & 0+0+0 & 0-6+6 \\ -11+3+8 & 11+0+0 & -22-2+24 \\ 0-3+3 & 0+0+0 & 0+2+9 \end{pmatrix} = \begin{pmatrix} 11 & 0 & 0 \\ 0 & 11 & 0 \\ 0 & 0 & 11 \end{pmatrix} \)
Hence, \( A (\text{adj } A) = (\text{adj } A) A = |A| I \) is verified.
In simple words: We calculate all cofactors for the 3x3 matrix, then transpose them to get the adjoint. Finally, we multiply the original matrix by its adjoint to show that it is equal to the determinant times the identity matrix.

Exam Tip: For a 3x3 matrix, calculating cofactors accurately is the most time-consuming part. Organize your work clearly for each cofactor to avoid errors.

Question 5. \( \left[\begin{array}{ll} 2 & -2 \\ 4 & 3 \end{array}\right] \)
Answer: Let the given matrix be \( A = \begin{pmatrix} 2 & -2 \\ 4 & 3 \end{pmatrix} \).
First, we find the cofactors for each element:
\( A_{11} = 3 \)
\( A_{12} = -4 \)
\( A_{21} = -(-2) = 2 \)
\( A_{22} = 2 \)
The matrix of cofactors is \( C = \begin{pmatrix} 3 & -4 \\ 2 & 2 \end{pmatrix} \).
The adjoint of matrix \( A \) is the transpose of the cofactor matrix:
\( \text{adj } A = C^T = \begin{pmatrix} 3 & -4 \\ 2 & 2 \end{pmatrix}^T = \begin{pmatrix} 3 & 2 \\ -4 & 2 \end{pmatrix} \)
Now, we calculate the determinant of \( A \):
\( |A| = (2)(3) - (-2)(4) = 6 - (-8) = 6+8 = 14 \)
Finally, we find the inverse matrix \( A^{-1} \):
\( A^{-1} = \frac{1}{|A|} (\text{adj } A) \)
\( \implies A^{-1} = \frac{1}{14} \begin{pmatrix} 3 & 2 \\ -4 & 2 \end{pmatrix} = \begin{pmatrix} \frac{3}{14} & \frac{2}{14} \\ \frac{-4}{14} & \frac{2}{14} \end{pmatrix} = \begin{pmatrix} \frac{3}{14} & \frac{1}{7} \\ \frac{-2}{7} & \frac{1}{7} \end{pmatrix} \)
In simple words: To find the inverse of a 2x2 matrix, we swap the top-left and bottom-right numbers, change the signs of the other two numbers, and then divide the entire matrix by its determinant.

Exam Tip: Remember that an inverse matrix only exists if the determinant is non-zero. If \( |A|=0 \), then \( A^{-1} \) does not exist.

Question 6. \( \left[\begin{array}{ll} -1 & 5 \\ -3 & 2 \end{array}\right] \)
Answer: Let the given matrix be \( A = \begin{pmatrix} -1 & 5 \\ -3 & 2 \end{pmatrix} \).
First, we find the cofactors for each element:
\( A_{11} = 2 \)
\( A_{12} = -(-3) = 3 \)
\( A_{21} = -5 \)
\( A_{22} = -1 \)
The matrix of cofactors is \( C = \begin{pmatrix} 2 & 3 \\ -5 & -1 \end{pmatrix} \).
The adjoint of matrix \( A \) is the transpose of the cofactor matrix:
\( \text{adj } A = C^T = \begin{pmatrix} 2 & 3 \\ -5 & -1 \end{pmatrix}^T = \begin{pmatrix} 2 & -5 \\ 3 & -1 \end{pmatrix} \)
Now, we calculate the determinant of \( A \):
\( |A| = (-1)(2) - (5)(-3) = -2 - (-15) = -2 + 15 = 13 \)
Finally, we find the inverse matrix \( A^{-1} \):
\( A^{-1} = \frac{1}{|A|} (\text{adj } A) \)
\( \implies A^{-1} = \frac{1}{13} \begin{pmatrix} 2 & -5 \\ 3 & -1 \end{pmatrix} = \begin{pmatrix} \frac{2}{13} & \frac{-5}{13} \\ \frac{3}{13} & \frac{-1}{13} \end{pmatrix} \)
In simple words: To compute the inverse, we find the cofactors, transpose them to get the adjoint, and then divide the adjoint matrix by the determinant of the original matrix.

Exam Tip: Always make sure your determinant calculation is correct, as it's a critical step. A single error here will make your entire inverse calculation incorrect.

Question 7. \( \left[\begin{array}{III} 1 & 2 & 3 \\ 0 & 2 & 4 \\ 0 & 0 & 5 \end{array}\right] \)
Answer: Let the given matrix be \( A = \begin{pmatrix} 1 & 2 & 3 \\ 0 & 2 & 4 \\ 0 & 0 & 5 \end{pmatrix} \).
We compute the cofactor for each element:
\( A_{11} = (-1)^{1+1} \begin{vmatrix} 2 & 4 \\ 0 & 5 \end{vmatrix} = 1(10-0) = 10 \)
\( A_{12} = (-1)^{1+2} \begin{vmatrix} 0 & 4 \\ 0 & 5 \end{vmatrix} = -1(0-0) = 0 \)
\( A_{13} = (-1)^{1+3} \begin{vmatrix} 0 & 2 \\ 0 & 0 \end{vmatrix} = 1(0-0) = 0 \)
\( A_{21} = (-1)^{2+1} \begin{vmatrix} 2 & 3 \\ 0 & 5 \end{vmatrix} = -1(10-0) = -10 \)
\( A_{22} = (-1)^{2+2} \begin{vmatrix} 1 & 3 \\ 0 & 5 \end{vmatrix} = 1(5-0) = 5 \)
\( A_{23} = (-1)^{2+3} \begin{vmatrix} 1 & 2 \\ 0 & 0 \end{vmatrix} = -1(0-0) = 0 \)
\( A_{31} = (-1)^{3+1} \begin{vmatrix} 2 & 3 \\ 2 & 4 \end{vmatrix} = 1(8-6) = 2 \)
\( A_{32} = (-1)^{3+2} \begin{vmatrix} 1 & 3 \\ 0 & 4 \end{vmatrix} = -1(4-0) = -4 \)
\( A_{33} = (-1)^{3+3} \begin{vmatrix} 1 & 2 \\ 0 & 2 \end{vmatrix} = 1(2-0) = 2 \)
The cofactor matrix is \( C = \begin{pmatrix} 10 & 0 & 0 \\ -10 & 5 & 0 \\ 2 & -4 & 2 \end{pmatrix} \).
The adjoint of matrix \( A \) is the transpose of the cofactor matrix:
\( \text{adj } A = C^T = \begin{pmatrix} 10 & 0 & 0 \\ -10 & 5 & 0 \\ 2 & -4 & 2 \end{pmatrix}^T = \begin{pmatrix} 10 & -10 & 2 \\ 0 & 5 & -4 \\ 0 & 0 & 2 \end{pmatrix} \)
Now, we calculate the determinant of \( A \). For an upper triangular matrix, the determinant is the product of its diagonal elements.
\( |A| = 1 \times 2 \times 5 = 10 \)
Finally, we find the inverse matrix \( A^{-1} \):
\( A^{-1} = \frac{1}{|A|} (\text{adj } A) \)
\( \implies A^{-1} = \frac{1}{10} \begin{pmatrix} 10 & -10 & 2 \\ 0 & 5 & -4 \\ 0 & 0 & 2 \end{pmatrix} = \begin{pmatrix} \frac{10}{10} & \frac{-10}{10} & \frac{2}{10} \\ \frac{0}{10} & \frac{5}{10} & \frac{-4}{10} \\ \frac{0}{10} & \frac{0}{10} & \frac{2}{10} \end{pmatrix} = \begin{pmatrix} 1 & -1 & \frac{1}{5} \\ 0 & \frac{1}{2} & \frac{-2}{5} \\ 0 & 0 & \frac{1}{5} \end{pmatrix} \)
In simple words: For this type of matrix, called an upper triangular matrix, finding the determinant is easy as it's just the product of the numbers on the main diagonal. Then, we find all the cofactors, transpose them, and divide by the determinant to get the inverse.

Exam Tip: Recognize special matrix types like triangular matrices. Their determinants are simple products of diagonal elements, which saves time.

Question 8. \( \left[\begin{array}{lll} 1 & 0 & 0 \\ 3 & 3 & 0 \\ 5 & 2 & -1 \end{array}\right] \)
Answer: Let the given matrix be \( A = \begin{pmatrix} 1 & 0 & 0 \\ 3 & 3 & 0 \\ 5 & 2 & -1 \end{pmatrix} \).
We compute the cofactor for each element:
\( A_{11} = (-1)^{1+1} \begin{vmatrix} 3 & 0 \\ 2 & -1 \end{vmatrix} = 1(-3-0) = -3 \)
\( A_{12} = (-1)^{1+2} \begin{vmatrix} 3 & 0 \\ 5 & -1 \end{vmatrix} = -1(-3-0) = 3 \)
\( A_{13} = (-1)^{1+3} \begin{vmatrix} 3 & 3 \\ 5 & 2 \end{vmatrix} = 1(6-15) = -9 \)
\( A_{21} = (-1)^{2+1} \begin{vmatrix} 0 & 0 \\ 2 & -1 \end{vmatrix} = -1(0-0) = 0 \)
\( A_{22} = (-1)^{2+2} \begin{vmatrix} 1 & 0 \\ 5 & -1 \end{vmatrix} = 1(-1-0) = -1 \)
\( A_{23} = (-1)^{2+3} \begin{vmatrix} 1 & 0 \\ 5 & 2 \end{vmatrix} = -1(2-0) = -2 \)
\( A_{31} = (-1)^{3+1} \begin{vmatrix} 0 & 0 \\ 3 & 0 \end{vmatrix} = 1(0-0) = 0 \)
\( A_{32} = (-1)^{3+2} \begin{vmatrix} 1 & 0 \\ 3 & 0 \end{vmatrix} = -1(0-0) = 0 \)
\( A_{33} = (-1)^{3+3} \begin{vmatrix} 1 & 0 \\ 3 & 3 \end{vmatrix} = 1(3-0) = 3 \)
The cofactor matrix is \( C = \begin{pmatrix} -3 & 3 & -9 \\ 0 & -1 & -2 \\ 0 & 0 & 3 \end{pmatrix} \).
The adjoint of matrix \( A \) is the transpose of the cofactor matrix:
\( \text{adj } A = C^T = \begin{pmatrix} -3 & 3 & -9 \\ 0 & -1 & -2 \\ 0 & 0 & 3 \end{pmatrix}^T = \begin{pmatrix} -3 & 0 & 0 \\ 3 & -1 & 0 \\ -9 & -2 & 3 \end{pmatrix} \)
Now, we calculate the determinant of \( A \). For a lower triangular matrix, the determinant is the product of its diagonal elements.
\( |A| = 1 \times 3 \times (-1) = -3 \)
Finally, we find the inverse matrix \( A^{-1} \):
\( A^{-1} = \frac{1}{|A|} (\text{adj } A) \)
\( \implies A^{-1} = \frac{1}{-3} \begin{pmatrix} -3 & 0 & 0 \\ 3 & -1 & 0 \\ -9 & -2 & 3 \end{pmatrix} = \begin{pmatrix} \frac{-3}{-3} & \frac{0}{-3} & \frac{0}{-3} \\ \frac{3}{-3} & \frac{-1}{-3} & \frac{0}{-3} \\ \frac{-9}{-3} & \frac{-2}{-3} & \frac{3}{-3} \end{pmatrix} = \begin{pmatrix} 1 & 0 & 0 \\ -1 & \frac{1}{3} & 0 \\ 3 & \frac{2}{3} & -1 \end{pmatrix} \)
In simple words: This matrix is a lower triangular matrix, so its determinant is simply the product of its diagonal elements. We then calculate all cofactors, transpose them to get the adjoint, and divide by the determinant to find the inverse matrix.

Exam Tip: Being aware of matrix properties, such as for triangular matrices, can simplify calculations. Always use the formula \( A^{-1} = \frac{1}{|A|} (\text{adj } A) \) when finding the inverse via cofactors.

Question 9. \( \left[\begin{array}{III} 2 & 1 & 3 \\ 4 & -1 & 0 \\ -7 & 2 & 1 \end{array}\right] \)
Answer: Let the given matrix be \( A = \begin{pmatrix} 2 & 1 & 3 \\ 4 & -1 & 0 \\ -7 & 2 & 1 \end{pmatrix} \).
We compute the cofactor for each element:
\( A_{11} = (-1)^{1+1} \begin{vmatrix} -1 & 0 \\ 2 & 1 \end{vmatrix} = 1(-1-0) = -1 \)
\( A_{12} = (-1)^{1+2} \begin{vmatrix} 4 & 0 \\ -7 & 1 \end{vmatrix} = -1(4-0) = -4 \)
\( A_{13} = (-1)^{1+3} \begin{vmatrix} 4 & -1 \\ -7 & 2 \end{vmatrix} = 1(8-7) = 1 \)
\( A_{21} = (-1)^{2+1} \begin{vmatrix} 1 & 3 \\ 2 & 1 \end{vmatrix} = -1(1-6) = -1(-5) = 5 \)
\( A_{22} = (-1)^{2+2} \begin{vmatrix} 2 & 3 \\ -7 & 1 \end{vmatrix} = 1(2-(-21)) = 2+21 = 23 \)
\( A_{23} = (-1)^{2+3} \begin{vmatrix} 2 & 1 \\ -7 & 2 \end{vmatrix} = -1(4-(-7)) = -1(4+7) = -11 \)
\( A_{31} = (-1)^{3+1} \begin{vmatrix} 1 & 3 \\ -1 & 0 \end{vmatrix} = 1(0-(-3)) = 3 \)
\( A_{32} = (-1)^{3+2} \begin{vmatrix} 2 & 3 \\ 4 & 0 \end{vmatrix} = -1(0-12) = -1(-12) = 12 \)
\( A_{33} = (-1)^{3+3} \begin{vmatrix} 2 & 1 \\ 4 & -1 \end{vmatrix} = 1(-2-4) = -6 \)
The cofactor matrix is \( C = \begin{pmatrix} -1 & -4 & 1 \\ 5 & 23 & -11 \\ 3 & 12 & -6 \end{pmatrix} \).
The adjoint of matrix \( A \) is the transpose of the cofactor matrix:
\( \text{adj } A = C^T = \begin{pmatrix} -1 & -4 & 1 \\ 5 & 23 & -11 \\ 3 & 12 & -6 \end{pmatrix}^T = \begin{pmatrix} -1 & 5 & 3 \\ -4 & 23 & 12 \\ 1 & -11 & -6 \end{pmatrix} \)
Now, we calculate the determinant of \( A \):
\( |A| = 2(A_{11}) + 1(A_{12}) + 3(A_{13}) = 2(-1) + 1(-4) + 3(1) = -2 - 4 + 3 = -3 \)
Finally, we find the inverse matrix \( A^{-1} \):
\( A^{-1} = \frac{1}{|A|} (\text{adj } A) \)
\( \implies A^{-1} = \frac{1}{-3} \begin{pmatrix} -1 & 5 & 3 \\ -4 & 23 & 12 \\ 1 & -11 & -6 \end{pmatrix} = \begin{pmatrix} \frac{-1}{-3} & \frac{5}{-3} & \frac{3}{-3} \\ \frac{-4}{-3} & \frac{23}{-3} & \frac{12}{-3} \\ \frac{1}{-3} & \frac{-11}{-3} & \frac{-6}{-3} \end{pmatrix} = \begin{pmatrix} \frac{1}{3} & \frac{-5}{3} & -1 \\ \frac{4}{3} & \frac{-23}{3} & -4 \\ \frac{-1}{3} & \frac{11}{3} & 2 \end{pmatrix} \)
In simple words: We systematically calculate each cofactor for the given matrix, then form the adjoint matrix by transposing the cofactor matrix. The final step is to divide this adjoint matrix by the determinant of the original matrix to obtain the inverse.

Exam Tip: Be mindful of zero elements in the matrix; they can simplify cofactor calculations significantly but don't eliminate the need for proper sign application.

Question 10. \( \left[\begin{array}{III} 1 & -1 & 2 \\ 0 & 2 & -3 \\ 3 & -2 & 4 \end{array}\right] \)
Answer: Let the given matrix be \( A = \begin{pmatrix} 1 & -1 & 2 \\ 0 & 2 & -3 \\ 3 & -2 & 4 \end{pmatrix} \).
We compute the cofactor for each element:
\( A_{11} = (-1)^{1+1} \begin{vmatrix} 2 & -3 \\ -2 & 4 \end{vmatrix} = 1(8-6) = 2 \)
\( A_{12} = (-1)^{1+2} \begin{vmatrix} 0 & -3 \\ 3 & 4 \end{vmatrix} = -1(0-(-9)) = -9 \)
\( A_{13} = (-1)^{1+3} \begin{vmatrix} 0 & 2 \\ 3 & -2 \end{vmatrix} = 1(0-6) = -6 \)
\( A_{21} = (-1)^{2+1} \begin{vmatrix} -1 & 2 \\ -2 & 4 \end{vmatrix} = -1(-4-(-4)) = -1(0) = 0 \)
\( A_{22} = (-1)^{2+2} \begin{vmatrix} 1 & 2 \\ 3 & 4 \end{vmatrix} = 1(4-6) = -2 \)
\( A_{23} = (-1)^{2+3} \begin{vmatrix} 1 & -1 \\ 3 & -2 \end{vmatrix} = -1(-2-(-3)) = -1(1) = -1 \)
\( A_{31} = (-1)^{3+1} \begin{vmatrix} -1 & 2 \\ 2 & -3 \end{vmatrix} = 1(3-4) = -1 \)
\( A_{32} = (-1)^{3+2} \begin{vmatrix} 1 & 2 \\ 0 & -3 \end{vmatrix} = -1(-3-0) = 3 \)
\( A_{33} = (-1)^{3+3} \begin{vmatrix} 1 & -1 \\ 0 & 2 \end{vmatrix} = 1(2-0) = 2 \)
The cofactor matrix is \( C = \begin{pmatrix} 2 & -9 & -6 \\ 0 & -2 & -1 \\ -1 & 3 & 2 \end{pmatrix} \).
The adjoint of matrix \( A \) is the transpose of the cofactor matrix:
\( \text{adj } A = C^T = \begin{pmatrix} 2 & -9 & -6 \\ 0 & -2 & -1 \\ -1 & 3 & 2 \end{pmatrix}^T = \begin{pmatrix} 2 & 0 & -1 \\ -9 & -2 & 3 \\ -6 & -1 & 2 \end{pmatrix} \)
Now, we calculate the determinant of \( A \):
\( |A| = 1(A_{11}) + (-1)(A_{12}) + 2(A_{13}) = 1(2) - 1(-9) + 2(-6) = 2 + 9 - 12 = -1 \)
Finally, we find the inverse matrix \( A^{-1} \):
\( A^{-1} = \frac{1}{|A|} (\text{adj } A) \)
\( \implies A^{-1} = \frac{1}{-1} \begin{pmatrix} 2 & 0 & -1 \\ -9 & -2 & 3 \\ -6 & -1 & 2 \end{pmatrix} = \begin{pmatrix} -2 & 0 & 1 \\ 9 & 2 & -3 \\ 6 & 1 & -2 \end{pmatrix} \)
In simple words: We find the cofactors for each element, create the adjoint matrix by transposing these cofactors, and then divide the adjoint by the determinant of the original matrix to get the inverse.

Exam Tip: Pay close attention to the sign rule for cofactors \( (-1)^{i+j} \). A matrix with many negative numbers can easily lead to sign errors if not handled carefully.

Question 11. \( \left[\begin{array}{ccc} 1 & 0 & 0 \\ 0 & \cos \alpha & \sin \alpha \\ 0 & \sin \alpha & - \cos \alpha \end{array}\right] \)
Answer: Let the given matrix be \( A = \begin{pmatrix} 1 & 0 & 0 \\ 0 & \cos \alpha & \sin \alpha \\ 0 & \sin \alpha & -\cos \alpha \end{pmatrix} \).
We compute the cofactor for each element:
\( A_{11} = (-1)^{1+1} \begin{vmatrix} \cos \alpha & \sin \alpha \\ \sin \alpha & -\cos \alpha \end{vmatrix} = 1(-\cos^2 \alpha - \sin^2 \alpha) = -(\cos^2 \alpha + \sin^2 \alpha) = -1 \)
\( A_{12} = (-1)^{1+2} \begin{vmatrix} 0 & \sin \alpha \\ 0 & -\cos \alpha \end{vmatrix} = -1(0-0) = 0 \)
\( A_{13} = (-1)^{1+3} \begin{vmatrix} 0 & \cos \alpha \\ 0 & \sin \alpha \end{vmatrix} = 1(0-0) = 0 \)
\( A_{21} = (-1)^{2+1} \begin{vmatrix} 0 & 0 \\ \sin \alpha & -\cos \alpha \end{vmatrix} = -1(0-0) = 0 \)
\( A_{22} = (-1)^{2+2} \begin{vmatrix} 1 & 0 \\ 0 & -\cos \alpha \end{vmatrix} = 1(-\cos \alpha - 0) = -\cos \alpha \)
\( A_{23} = (-1)^{2+3} \begin{vmatrix} 1 & 0 \\ 0 & \sin \alpha \end{vmatrix} = -1(\sin \alpha - 0) = -\sin \alpha \)
\( A_{31} = (-1)^{3+1} \begin{vmatrix} 0 & 0 \\ \cos \alpha & \sin \alpha \end{vmatrix} = 1(0-0) = 0 \)
\( A_{32} = (-1)^{3+2} \begin{vmatrix} 1 & 0 \\ 0 & \sin \alpha \end{vmatrix} = -1(\sin \alpha - 0) = -\sin \alpha \)
\( A_{33} = (-1)^{3+3} \begin{vmatrix} 1 & 0 \\ 0 & \cos \alpha \end{vmatrix} = 1(\cos \alpha - 0) = \cos \alpha \)
The cofactor matrix is \( C = \begin{pmatrix} -1 & 0 & 0 \\ 0 & -\cos \alpha & -\sin \alpha \\ 0 & -\sin \alpha & \cos \alpha \end{pmatrix} \).
The adjoint of matrix \( A \) is the transpose of the cofactor matrix:
\( \text{adj } A = C^T = \begin{pmatrix} -1 & 0 & 0 \\ 0 & -\cos \alpha & -\sin \alpha \\ 0 & -\sin \alpha & \cos \alpha \end{pmatrix}^T = \begin{pmatrix} -1 & 0 & 0 \\ 0 & -\cos \alpha & -\sin \alpha \\ 0 & -\sin \alpha & \cos \alpha \end{pmatrix} \)
Now, we calculate the determinant of \( A \):
\( |A| = 1(A_{11}) + 0(A_{12}) + 0(A_{13}) = 1(-1) + 0 + 0 = -1 \)
Finally, we find the inverse matrix \( A^{-1} \):
\( A^{-1} = \frac{1}{|A|} (\text{adj } A) \)
\( \implies A^{-1} = \frac{1}{-1} \begin{pmatrix} -1 & 0 & 0 \\ 0 & -\cos \alpha & -\sin \alpha \\ 0 & -\sin \alpha & \cos \alpha \end{pmatrix} = \begin{pmatrix} 1 & 0 & 0 \\ 0 & \cos \alpha & \sin \alpha \\ 0 & \sin \alpha & -\cos \alpha \end{pmatrix} \)
In simple words: This matrix involves trigonometric functions. We find each cofactor by applying the sign rule and computing the determinant of the smaller matrices. Then we transpose the cofactor matrix to get the adjoint, and divide by the determinant (which simplifies to -1) to find the inverse.

Exam Tip: For matrices involving trigonometric functions, remember fundamental identities like \( \cos^2 \alpha + \sin^2 \alpha = 1 \). These identities often simplify complex expressions during determinant and cofactor calculations.

Question 12. Let \( A = \left[\begin{array}{ll} 3 & 7 \\ 2 & 5 \end{array}\right] \) and \( B = \left[\begin{array}{ll} 6 & 8 \\ 7 & 9 \end{array}\right] \), verify that \( (AB)^{-1} = B^{-1}A^{-1} \).
Answer: We need to verify that \( (AB)^{-1} = B^{-1}A^{-1} \).
First, let's calculate \( AB \):
\( AB = \begin{pmatrix} 3 & 7 \\ 2 & 5 \end{pmatrix} \begin{pmatrix} 6 & 8 \\ 7 & 9 \end{pmatrix} = \begin{pmatrix} (3 \times 6) + (7 \times 7) & (3 \times 8) + (7 \times 9) \\ (2 \times 6) + (5 \times 7) & (2 \times 8) + (5 \times 9) \end{pmatrix} \)
\( \implies AB = \begin{pmatrix} 18 + 49 & 24 + 63 \\ 12 + 35 & 16 + 45 \end{pmatrix} = \begin{pmatrix} 67 & 87 \\ 47 & 61 \end{pmatrix} \)
Now, let's find \( (AB)^{-1} \):
\( |AB| = (67)(61) - (87)(47) = 4087 - 4089 = -2 \)
The adjoint of \( AB \) is \( \text{adj}(AB) = \begin{pmatrix} 61 & -87 \\ -47 & 67 \end{pmatrix} \)
\( (AB)^{-1} = \frac{1}{|AB|} \text{adj}(AB) = \frac{1}{-2} \begin{pmatrix} 61 & -87 \\ -47 & 67 \end{pmatrix} = \begin{pmatrix} \frac{-61}{2} & \frac{87}{2} \\ \frac{47}{2} & \frac{-67}{2} \end{pmatrix} \) ... (Equation 1)
Next, let's find \( A^{-1} \):
\( |A| = (3)(5) - (7)(2) = 15 - 14 = 1 \)
The adjoint of \( A \) is \( \text{adj}(A) = \begin{pmatrix} 5 & -7 \\ -2 & 3 \end{pmatrix} \)
\( A^{-1} = \frac{1}{|A|} \text{adj}(A) = \frac{1}{1} \begin{pmatrix} 5 & -7 \\ -2 & 3 \end{pmatrix} = \begin{pmatrix} 5 & -7 \\ -2 & 3 \end{pmatrix} \)
Now, let's find \( B^{-1} \):
\( |B| = (6)(9) - (8)(7) = 54 - 56 = -2 \)
The adjoint of \( B \) is \( \text{adj}(B) = \begin{pmatrix} 9 & -8 \\ -7 & 6 \end{pmatrix} \)
\( B^{-1} = \frac{1}{|B|} \text{adj}(B) = \frac{1}{-2} \begin{pmatrix} 9 & -8 \\ -7 & 6 \end{pmatrix} = \begin{pmatrix} \frac{-9}{2} & 4 \\ \frac{7}{2} & -3 \end{pmatrix} \)
Finally, let's calculate \( B^{-1}A^{-1} \):
\( B^{-1}A^{-1} = \begin{pmatrix} \frac{-9}{2} & 4 \\ \frac{7}{2} & -3 \end{pmatrix} \begin{pmatrix} 5 & -7 \\ -2 & 3 \end{pmatrix} \)
\( = \begin{pmatrix} (\frac{-9}{2} \times 5) + (4 \times -2) & (\frac{-9}{2} \times -7) + (4 \times 3) \\ (\frac{7}{2} \times 5) + (-3 \times -2) & (\frac{7}{2} \times -7) + (-3 \times 3) \end{pmatrix} \)
\( = \begin{pmatrix} \frac{-45}{2} - 8 & \frac{63}{2} + 12 \\ \frac{35}{2} + 6 & \frac{-49}{2} - 9 \end{pmatrix} \)
\( = \begin{pmatrix} \frac{-45 - 16}{2} & \frac{63 + 24}{2} \\ \frac{35 + 12}{2} & \frac{-49 - 18}{2} \end{pmatrix} \)
\( \implies B^{-1}A^{-1} = \begin{pmatrix} \frac{-61}{2} & \frac{87}{2} \\ \frac{47}{2} & \frac{-67}{2} \end{pmatrix} \) ... (Equation 2)
Comparing Equation 1 and Equation 2, we can see that \( (AB)^{-1} = B^{-1}A^{-1} \).
Therefore, the identity is verified.
In simple words: To show that \( (AB)^{-1} = B^{-1}A^{-1} \), we first multiply matrices A and B, then find the inverse of the resulting matrix. Separately, we find the inverse of B and then the inverse of A, and multiply those two inverse matrices together. If both results are the same, the identity is proven.

Exam Tip: This property, \( (AB)^{-1} = B^{-1}A^{-1} \), is an important result in matrix algebra. Always remember to reverse the order of the matrices when taking the inverse of a product.

Question 12. Let \( A = \left[\begin{array}{ll} 3 & 7 \\ 2 & 5 \end{array}\right] \) and \( B = \left[\begin{array}{ll} 6 & 8 \\ 7 & 9 \end{array}\right] \), verify that \( (AB)^{-1} = B^{-1}A^{-1} \).
Answer:
First, we find the product of matrices A and B:
\( AB = \left[\begin{array}{ll} 3 & 7 \\ 2 & 5 \end{array}\right] \left[\begin{array}{ll} 6 & 8 \\ 7 & 9 \end{array}\right] = \left[\begin{array}{ll} 3 \times 6 + 7 \times 7 & 3 \times 8 + 7 \times 9 \\ 2 \times 6 + 5 \times 7 & 2 \times 8 + 5 \times 9 \end{array}\right] = \left[\begin{array}{ll} 18+49 & 24+63 \\ 12+35 & 16+45 \end{array}\right] = \left[\begin{array}{ll} 67 & 87 \\ 47 & 61 \end{array}\right] \)
Next, we determine the adjoint of AB:
\( \text{adj}(AB) = \left[\begin{array}{ll} (-1)^{1+1}61 & (-1)^{1+2}47 \\ (-1)^{2+1}87 & (-1)^{2+2}61 \end{array}\right] = \left[\begin{array}{ll} 61 & -47 \\ -87 & 61 \end{array}\right] \)
We also calculate the determinant of AB:
\( |AB| = (67 \times 61) - (87 \times 47) = 4087 - 4089 = -2 \)
Now, we can find the inverse of AB (L.H.S.):
\( (AB)^{-1} = \frac{1}{|AB|} \text{adj}(AB) = \frac{1}{-2}\left[\begin{array}{ll} 61 & -87 \\ -47 & 61 \end{array}\right] = \left[\begin{array}{ll} -\frac{61}{2} & \frac{87}{2} \\ \frac{47}{2} & -\frac{61}{2} \end{array}\right] \)
For the R.H.S., we first find \( A^{-1} \). We have \( A = \left[\begin{array}{ll} 3 & 7 \\ 2 & 5 \end{array}\right] \).
The adjoint of A is \( \text{adj} A = \left[\begin{array}{ll} 5 & -7 \\ -2 & 3 \end{array}\right] \).
The determinant of A is \( |A| = (3 \times 5) - (7 \times 2) = 15 - 14 = 1 \).
So, \( A^{-1} = \frac{1}{|A|} \text{adj} A = \frac{1}{1}\left[\begin{array}{ll} 5 & -7 \\ -2 & 3 \end{array}\right] = \left[\begin{array}{ll} 5 & -7 \\ -2 & 3 \end{array}\right] \).
Next, we find \( B^{-1} \). We have \( B = \left[\begin{array}{ll} 6 & 8 \\ 7 & 9 \end{array}\right] \).
The adjoint of B is \( \text{adj} B = \left[\begin{array}{ll} 9 & -8 \\ -7 & 6 \end{array}\right] \).
The determinant of B is \( |B| = (6 \times 9) - (8 \times 7) = 54 - 56 = -2 \).
So, \( B^{-1} = \frac{1}{|B|} \text{adj} B = \frac{1}{-2}\left[\begin{array}{ll} 9 & -8 \\ -7 & 6 \end{array}\right] = \left[\begin{array}{ll} -\frac{9}{2} & \frac{8}{2} \\ \frac{7}{2} & -\frac{6}{2} \end{array}\right] \).
Now, we calculate \( B^{-1}A^{-1} \) (R.H.S.):
\( B^{-1}A^{-1} = \left[\begin{array}{ll} -\frac{9}{2} & \frac{8}{2} \\ \frac{7}{2} & -\frac{6}{2} \end{array}\right] \left[\begin{array}{ll} 5 & -7 \\ -2 & 3 \end{array}\right] \)
\( = \left[\begin{array}{ll} (-\frac{9}{2} \times 5) + (\frac{8}{2} \times -2) & (-\frac{9}{2} \times -7) + (\frac{8}{2} \times 3) \\ (\frac{7}{2} \times 5) + (-\frac{6}{2} \times -2) & (\frac{7}{2} \times -7) + (-\frac{6}{2} \times 3) \end{array}\right] \)
\( = \left[\begin{array}{ll} -\frac{45}{2} - \frac{16}{2} & \frac{63}{2} + \frac{24}{2} \\ \frac{35}{2} + \frac{12}{2} & -\frac{49}{2} - \frac{18}{2} \end{array}\right] \)
\( = \left[\begin{array}{ll} -\frac{61}{2} & \frac{87}{2} \\ \frac{47}{2} & -\frac{67}{2} \end{array}\right] \)
Since the L.H.S. is equal to the R.H.S., we have verified that \( (AB)^{-1} = B^{-1}A^{-1} \).
In simple words: We first found the inverse of the product of A and B. Then, we found the inverses of A and B separately, and multiplied them in reverse order. Both results were identical, which proves the property.

Exam Tip: Remember to calculate the product AB first before finding its inverse. For the right-hand side, always compute \( B^{-1} \) then \( A^{-1} \), and multiply them as \( B^{-1}A^{-1} \) to avoid common errors.

Question 13. If \( A = \left[\begin{array}{ll} 3 & 1 \\ -1 & 2 \end{array}\right] \), show that \( A^2 – 5A + 7I = O \). Hence, find \( A^{-1} \).
Answer:
Given matrix \( A = \left[\begin{array}{ll} 3 & 1 \\ -1 & 2 \end{array}\right] \).
First, we compute \( A^2 \):
\( A^2 = \left[\begin{array}{ll} 3 & 1 \\ -1 & 2 \end{array}\right] \left[\begin{array}{ll} 3 & 1 \\ -1 & 2 \end{array}\right] = \left[\begin{array}{cc} (3 \times 3) + (1 \times -1) & (3 \times 1) + (1 \times 2) \\ (-1 \times 3) + (2 \times -1) & (-1 \times 1) + (2 \times 2) \end{array}\right] = \left[\begin{array}{cc} 9-1 & 3+2 \\ -3-2 & -1+4 \end{array}\right] = \left[\begin{array}{ll} 8 & 5 \\ -5 & 3 \end{array}\right] \).
Now, we substitute \( A^2 \), \( A \), and \( I \) into the expression \( A^2 – 5A + 7I \):
\( A^2 – 5A + 7I = \left[\begin{array}{ll} 8 & 5 \\ -5 & 3 \end{array}\right] - 5\left[\begin{array}{ll} 3 & 1 \\ -1 & 2 \end{array}\right] + 7\left[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right] \)
\( = \left[\begin{array}{ll} 8 & 5 \\ -5 & 3 \end{array}\right] + \left[\begin{array}{cc} -15 & -5 \\ 5 & -10 \end{array}\right] + \left[\begin{array}{ll} 7 & 0 \\ 0 & 7 \end{array}\right] \)
\( = \left[\begin{array}{cc} 8-15+7 & 5-5+0 \\ -5+5+0 & 3-10+7 \end{array}\right] \)
\( = \left[\begin{array}{ll} 0 & 0 \\ 0 & 0 \end{array}\right] = O \).
Thus, we have shown that \( A^2 – 5A + 7I = O \).
To find \( A^{-1} \), we multiply the equation \( A^2 – 5A + 7I = O \) by \( A^{-1} \) on the right or left:
\( A^{-1}(A^2 – 5A + 7I) = A^{-1}O \)
\( (A^{-1}A)A – 5(A^{-1}A) + 7(A^{-1}I) = O \)
\( IA – 5I + 7A^{-1} = O \)

\( \implies \) \( 7A^{-1} = 5I – IA \)

\( \implies \) \( 7A^{-1} = 5I – A \)
Now, substitute the values for \( I \) and \( A \):
\( 7A^{-1} = 5\left[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right] – \left[\begin{array}{ll} 3 & 1 \\ -1 & 2 \end{array}\right] \)
\( = \left[\begin{array}{ll} 5 & 0 \\ 0 & 5 \end{array}\right] – \left[\begin{array}{ll} 3 & 1 \\ -1 & 2 \end{array}\right] \)
\( = \left[\begin{array}{cc} 5-3 & 0-1 \\ 0-(-1) & 5-2 \end{array}\right] \)
\( = \left[\begin{array}{ll} 2 & -1 \\ 1 & 3 \end{array}\right] \)
Finally, divide by 7 to get \( A^{-1} \):
\( A^{-1} = \frac{1}{7}\left[\begin{array}{ll} 2 & -1 \\ 1 & 3 \end{array}\right] \).
In simple words: We proved the matrix equation by calculating A squared, then putting all the matrices together and showing they add up to the zero matrix. After that, we used the same equation to find the inverse of A by multiplying everything by A inverse and rearranging it.

Exam Tip: When finding \( A^{-1} \) from the characteristic equation, always remember to multiply by \( A^{-1} \) and simplify using \( A^{-1}A = I \) and \( A^{-1}I = A^{-1} \).

Question 14. For the matrix \( A = \left[\begin{array}{ll} 3 & 2 \\ 1 & 1 \end{array}\right] \), find the numbers a and b such that \( A^2 + aA + bI = O \). Hence, find \( A^{-1} \).
Answer:
Given matrix \( A = \left[\begin{array}{ll} 3 & 2 \\ 1 & 1 \end{array}\right] \).
First, calculate \( A^2 \):
\( A^2 = \left[\begin{array}{ll} 3 & 2 \\ 1 & 1 \end{array}\right] \left[\begin{array}{ll} 3 & 2 \\ 1 & 1 \end{array}\right] = \left[\begin{array}{ll} (3 \times 3) + (2 \times 1) & (3 \times 2) + (2 \times 1) \\ (1 \times 3) + (1 \times 1) & (1 \times 2) + (1 \times 1) \end{array}\right] = \left[\begin{array}{ll} 9+2 & 6+2 \\ 3+1 & 2+1 \end{array}\right] = \left[\begin{array}{ll} 11 & 8 \\ 4 & 3 \end{array}\right] \).
We are given the equation \( A^2 + aA + bI = O \). Substitute \( A^2 \), \( A \), and \( I \) (the identity matrix):
\( \left[\begin{array}{ll} 11 & 8 \\ 4 & 3 \end{array}\right] + a\left[\begin{array}{ll} 3 & 2 \\ 1 & 1 \end{array}\right] + b\left[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right] = \left[\begin{array}{ll} 0 & 0 \\ 0 & 0 \end{array}\right] \)
Combine these matrices:
\( \left[\begin{array}{ll} 11 & 8 \\ 4 & 3 \end{array}\right] + \left[\begin{array}{ll} 3a & 2a \\ a & a \end{array}\right] + \left[\begin{array}{ll} b & 0 \\ 0 & b \end{array}\right] = \left[\begin{array}{ll} 0 & 0 \\ 0 & 0 \end{array}\right] \)
\( \left[\begin{array}{cc} 11+3a+b & 8+2a \\ 4+a & 3+a+b \end{array}\right] = \left[\begin{array}{ll} 0 & 0 \\ 0 & 0 \end{array}\right] \)
For two matrices to be equal, their corresponding elements must be equal. From the element in the second row, first column:
\( 4+a = 0 \)

\( \implies \) \( a = -4 \).
Now, use the element in the second row, second column:
\( 3+a+b = 0 \)
Substitute \( a = -4 \):
\( 3 + (-4) + b = 0 \)
\( -1 + b = 0 \)

\( \implies \) \( b = 1 \).
We can check with other elements:
\( 8+2a = 8+2(-4) = 8-8 = 0 \). (This matches)
\( 11+3a+b = 11+3(-4)+1 = 11-12+1 = 0 \). (This also matches)
So, we found \( a = -4 \) and \( b = 1 \). The equation becomes \( A^2 - 4A + I = O \).
To find \( A^{-1} \), multiply the equation \( A^2 - 4A + I = O \) by \( A^{-1} \):
\( A^{-1}(A^2 - 4A + I) = A^{-1}O \)
\( (A^{-1}A)A - 4(A^{-1}A) + (A^{-1}I) = O \)
\( IA - 4I + A^{-1} = O \)

\( \implies \) \( A^{-1} = 4I - A \)
Now, substitute the values for \( I \) and \( A \):
\( A^{-1} = 4\left[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right] – \left[\begin{array}{ll} 3 & 2 \\ 1 & 1 \end{array}\right] \)
\( = \left[\begin{array}{ll} 4 & 0 \\ 0 & 4 \end{array}\right] – \left[\begin{array}{ll} 3 & 2 \\ 1 & 1 \end{array}\right] \)
\( = \left[\begin{array}{cc} 4-3 & 0-2 \\ 0-1 & 4-1 \end{array}\right] \)
\( = \left[\begin{array}{ll} 1 & -2 \\ -1 & 3 \end{array}\right] \).
In simple words: We used the given equation and substituted the matrices A and A squared, along with the identity matrix. By comparing elements, we found the unknown numbers 'a' and 'b'. Then, we used this new equation to quickly calculate the inverse of A.

Exam Tip: When asked to find 'a' and 'b' using a matrix equation, equate corresponding elements of the matrices to form a system of linear equations. Use the simplest equations first to find the values easily.

Question 15. For the matrix \( A = \left[\begin{array}{ccc} 1 & 1 & 1 \\ 1 & 2 & -3 \\ 2 & -1 & 3 \end{array}\right] \) show that \( A^3 – 6A^2 + 5A + 11I_3 = O \). Hence, find \( A^{-1} \).
Answer:
Given matrix \( A = \left[\begin{array}{ccc} 1 & 1 & 1 \\ 1 & 2 & -3 \\ 2 & -1 & 3 \end{array}\right] \).
First, we calculate \( A^2 \):
\( A^2 = A \times A = \left[\begin{array}{ccc} 1 & 1 & 1 \\ 1 & 2 & -3 \\ 2 & -1 & 3 \end{array}\right] \left[\begin{array}{ccc} 1 & 1 & 1 \\ 1 & 2 & -3 \\ 2 & -1 & 3 \end{array}\right] \)
\( = \left[\begin{array}{ccc} (1 \times 1)+(1 \times 1)+(1 \times 2) & (1 \times 1)+(1 \times 2)+(1 \times -1) & (1 \times 1)+(1 \times -3)+(1 \times 3) \\ (1 \times 1)+(2 \times 1)+(-3 \times 2) & (1 \times 1)+(2 \times 2)+(-3 \times -1) & (1 \times 1)+(2 \times -3)+(-3 \times 3) \\ (2 \times 1)+(-1 \times 1)+(3 \times 2) & (2 \times 1)+(-1 \times 2)+(3 \times -1) & (2 \times 1)+(-1 \times -3)+(3 \times 3) \end{array}\right] \)
\( = \left[\begin{array}{ccc} 1+1+2 & 1+2-1 & 1-3+3 \\ 1+2-6 & 1+4+3 & 1-6-9 \\ 2-1+6 & 2-2-3 & 2+3+9 \end{array}\right] \)
\( = \left[\begin{array}{ccc} 4 & 2 & 1 \\ -3 & 8 & -14 \\ 7 & -3 & 14 \end{array}\right] \).
Next, we calculate \( A^3 \):
\( A^3 = A^2 \times A = \left[\begin{array}{ccc} 4 & 2 & 1 \\ -3 & 8 & -14 \\ 7 & -3 & 14 \end{array}\right] \left[\begin{array}{ccc} 1 & 1 & 1 \\ 1 & 2 & -3 \\ 2 & -1 & 3 \end{array}\right] \)
\( = \left[\begin{array}{ccc} (4 \times 1)+(2 \times 1)+(1 \times 2) & (4 \times 1)+(2 \times 2)+(1 \times -1) & (4 \times 1)+(2 \times -3)+(1 \times 3) \\ (-3 \times 1)+(8 \times 1)+(-14 \times 2) & (-3 \times 1)+(8 \times 2)+(-14 \times -1) & (-3 \times 1)+(8 \times -3)+(-14 \times 3) \\ (7 \times 1)+(-3 \times 1)+(14 \times 2) & (7 \times 1)+(-3 \times 2)+(14 \times -1) & (7 \times 1)+(-3 \times -3)+(14 \times 3) \end{array}\right] \)
\( = \left[\begin{array}{ccc} 4+2+2 & 4+4-1 & 4-6+3 \\ -3+8-28 & -3+16+14 & -3-24-42 \\ 7-3+28 & 7-6-14 & 7+9+42 \end{array}\right] \)
\( = \left[\begin{array}{ccc} 8 & 7 & 1 \\ -23 & 27 & -69 \\ 32 & -13 & 58 \end{array}\right] \).
Now, we substitute \( A^3 \), \( A^2 \), \( A \), and \( I_3 \) into the equation \( A^3 – 6A^2 + 5A + 11I_3 \):
\( A^3 – 6A^2 + 5A + 11I_3 = \left[\begin{array}{ccc} 8 & 7 & 1 \\ -23 & 27 & -69 \\ 32 & -13 & 58 \end{array}\right] - 6\left[\begin{array}{ccc} 4 & 2 & 1 \\ -3 & 8 & -14 \\ 7 & -3 & 14 \end{array}\right] + 5\left[\begin{array}{ccc} 1 & 1 & 1 \\ 1 & 2 & -3 \\ 2 & -1 & 3 \end{array}\right] + 11\left[\begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right] \)
\( = \left[\begin{array}{ccc} 8 & 7 & 1 \\ -23 & 27 & -69 \\ 32 & -13 & 58 \end{array}\right] + \left[\begin{array}{ccc} -24 & -12 & -6 \\ 18 & -48 & 84 \\ -42 & 18 & -84 \end{array}\right] + \left[\begin{array}{ccc} 5 & 5 & 5 \\ 5 & 10 & -15 \\ 10 & -5 & 15 \end{array}\right] + \left[\begin{array}{ccc} 11 & 0 & 0 \\ 0 & 11 & 0 \\ 0 & 0 & 11 \end{array}\right] \)
\( = \left[\begin{array}{ccc} 8-24+5+11 & 7-12+5+0 & 1-6+5+0 \\ -23+18+5+0 & 27-48+10+11 & -69+84-15+0 \\ 32-42+10+0 & -13+18-5+0 & 58-84+15+11 \end{array}\right] \)
\( = \left[\begin{array}{ccc} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{array}\right] = O \).
Thus, we have shown that \( A^3 – 6A^2 + 5A + 11I_3 = O \).
To find \( A^{-1} \), multiply the equation \( A^3 – 6A^2 + 5A + 11I_3 = O \) by \( A^{-1} \):
\( A^{-1}(A^3 – 6A^2 + 5A + 11I_3) = A^{-1}O \)
\( (A^{-1}A)A^2 – 6(A^{-1}A)A + 5(A^{-1}A) + 11(A^{-1}I_3) = O \)
\( IA^2 – 6IA + 5I + 11A^{-1} = O \)

\( \implies \) \( 11A^{-1} = -A^2 + 6A – 5I \).
Now, substitute the values for \( A^2 \), \( A \), and \( I \):
\( 11A^{-1} = -\left[\begin{array}{ccc} 4 & 2 & 1 \\ -3 & 8 & -14 \\ 7 & -3 & 14 \end{array}\right] + 6\left[\begin{array}{ccc} 1 & 1 & 1 \\ 1 & 2 & -3 \\ 2 & -1 & 3 \end{array}\right] - 5\left[\begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right] \)
\( = \left[\begin{array}{ccc} -4 & -2 & -1 \\ 3 & -8 & 14 \\ -7 & 3 & -14 \end{array}\right] + \left[\begin{array}{ccc} 6 & 6 & 6 \\ 6 & 12 & -18 \\ 12 & -6 & 18 \end{array}\right] + \left[\begin{array}{ccc} -5 & 0 & 0 \\ 0 & -5 & 0 \\ 0 & 0 & -5 \end{array}\right] \)
\( = \left[\begin{array}{ccc} -4+6-5 & -2+6+0 & -1+6+0 \\ 3+6+0 & -8+12-5 & 14-18+0 \\ -7+12+0 & 3-6+0 & -14+18-5 \end{array}\right] \)
\( = \left[\begin{array}{ccc} -3 & 4 & 5 \\ 9 & -1 & -4 \\ 5 & -3 & -1 \end{array}\right] \).
Finally, divide by 11 to get \( A^{-1} \):
\( A^{-1} = \frac{1}{11}\left[\begin{array}{ccc} -3 & 4 & 5 \\ 9 & -1 & -4 \\ 5 & -3 & -1 \end{array}\right] \).
In simple words: We proved the matrix equation by finding A squared and A cubed, then combining them with A and the identity matrix to show the result is a zero matrix. After that, we used the same polynomial equation to find the inverse of A by multiplying by A inverse and rearranging the terms.

Exam Tip: For higher powers of matrices, always calculate \( A^2 \) first, then \( A^3 = A^2 \times A \), and so on. Be very careful with matrix multiplication, as one small error can affect the entire calculation.

Question 16. If \( A = \left[\begin{array}{ccc} 2 & -1 & 1 \\ 1 & 2 & -1 \\ 1 & -1 & 2 \end{array}\right] \), verify that \( A^3 – 6A^2 + 9A − 4I = O \). Hence, find \( A^{-1} \).
Answer:
Given matrix \( A = \left[\begin{array}{ccc} 2 & -1 & 1 \\ 1 & 2 & -1 \\ 1 & -1 & 2 \end{array}\right] \).
First, calculate \( A^2 \):
\( A^2 = A \times A = \left[\begin{array}{ccc} 2 & -1 & 1 \\ 1 & 2 & -1 \\ 1 & -1 & 2 \end{array}\right] \left[\begin{array}{ccc} 2 & -1 & 1 \\ 1 & 2 & -1 \\ 1 & -1 & 2 \end{array}\right] \)
\( = \left[\begin{array}{ccc} (2 \times 2)+(-1 \times 1)+(1 \times 1) & (2 \times -1)+(-1 \times 2)+(1 \times -1) & (2 \times 1)+(-1 \times -1)+(1 \times 2) \\ (1 \times 2)+(2 \times 1)+(-1 \times 1) & (1 \times -1)+(2 \times 2)+(-1 \times -1) & (1 \times 1)+(2 \times -1)+(-1 \times 2) \\ (1 \times 2)+(-1 \times 1)+(2 \times 1) & (1 \times -1)+(-1 \times 2)+(2 \times -1) & (1 \times 1)+(-1 \times -1)+(2 \times 2) \end{array}\right] \)
\( = \left[\begin{array}{ccc} 4-1+1 & -2-2-1 & 2+1+2 \\ 2+2-1 & -1+4+1 & 1-2-2 \\ 2-1+2 & -1-2-2 & 1+1+4 \end{array}\right] \)
\( = \left[\begin{array}{ccc} 6 & -5 & 5 \\ -5 & 6 & -5 \\ 5 & -5 & 6 \end{array}\right] \).
Next, calculate \( A^3 \):
\( A^3 = A^2 \times A = \left[\begin{array}{ccc} 6 & -5 & 5 \\ -5 & 6 & -5 \\ 5 & -5 & 6 \end{array}\right] \left[\begin{array}{ccc} 2 & -1 & 1 \\ 1 & 2 & -1 \\ 1 & -1 & 2 \end{array}\right] \)
\( = \left[\begin{array}{ccc} (6 \times 2)+(-5 \times 1)+(5 \times 1) & (6 \times -1)+(-5 \times 2)+(5 \times -1) & (6 \times 1)+(-5 \times -1)+(5 \times 2) \\ (-5 \times 2)+(6 \times 1)+(-5 \times 1) & (-5 \times -1)+(6 \times 2)+(-5 \times -1) & (-5 \times 1)+(6 \times -1)+(-5 \times 2) \\ (5 \times 2)+(-5 \times 1)+(6 \times 1) & (5 \times -1)+(-5 \times 2)+(6 \times -1) & (5 \times 1)+(-5 \times -1)+(6 \times 2) \end{array}\right] \)
\( = \left[\begin{array}{ccc} 12-5+5 & -6-10-5 & 6+5+10 \\ -10+6-5 & 5+12+5 & -5-6-10 \\ 10-5+6 & -5-10-6 & 5+5+12 \end{array}\right] \)
\( = \left[\begin{array}{ccc} 22 & -21 & 21 \\ -21 & 22 & -21 \\ 21 & -21 & 22 \end{array}\right] \).
Now, substitute \( A^3 \), \( A^2 \), \( A \), and \( I \) into the expression \( A^3 – 6A^2 + 9A − 4I \):
\( A^3 – 6A^2 + 9A − 4I = \left[\begin{array}{ccc} 22 & -21 & 21 \\ -21 & 22 & -21 \\ 21 & -21 & 22 \end{array}\right] - 6\left[\begin{array}{ccc} 6 & -5 & 5 \\ -5 & 6 & -5 \\ 5 & -5 & 6 \end{array}\right] + 9\left[\begin{array}{ccc} 2 & -1 & 1 \\ 1 & 2 & -1 \\ 1 & -1 & 2 \end{array}\right] - 4\left[\begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right] \)
\( = \left[\begin{array}{ccc} 22 & -21 & 21 \\ -21 & 22 & -21 \\ 21 & -21 & 22 \end{array}\right] + \left[\begin{array}{ccc} -36 & 30 & -30 \\ 30 & -36 & 30 \\ -30 & 30 & -36 \end{array}\right] + \left[\begin{array}{ccc} 18 & -9 & 9 \\ 9 & 18 & -9 \\ 9 & -9 & 18 \end{array}\right] + \left[\begin{array}{ccc} -4 & 0 & 0 \\ 0 & -4 & 0 \\ 0 & 0 & -4 \end{array}\right] \)
\( = \left[\begin{array}{ccc} 22-36+18-4 & -21+30-9+0 & 21-30+9+0 \\ -21+30-9+0 & 22-36+18-4 & -21+30-9+0 \\ 21-30+9+0 & -21+30-9+0 & 22-36+18-4 \end{array}\right] \)
\( = \left[\begin{array}{ccc} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{array}\right] = O \).
Thus, we have verified that \( A^3 – 6A^2 + 9A − 4I = O \).
To find \( A^{-1} \), multiply the equation \( A^3 – 6A^2 + 9A − 4I = O \) by \( A^{-1} \):
\( A^{-1}(A^3 – 6A^2 + 9A − 4I) = A^{-1}O \)
\( (A^{-1}A)A^2 – 6(A^{-1}A)A + 9(A^{-1}A) – 4(A^{-1}I) = O \)
\( IA^2 – 6IA + 9I – 4A^{-1} = O \)

\( \implies \) \( 4A^{-1} = A^2 – 6A + 9I \).
Now, substitute the values for \( A^2 \), \( A \), and \( I \):
\( 4A^{-1} = \left[\begin{array}{ccc} 6 & -5 & 5 \\ -5 & 6 & -5 \\ 5 & -5 & 6 \end{array}\right] - 6\left[\begin{array}{ccc} 2 & -1 & 1 \\ 1 & 2 & -1 \\ 1 & -1 & 2 \end{array}\right] + 9\left[\begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right] \)
\( = \left[\begin{array}{ccc} 6 & -5 & 5 \\ -5 & 6 & -5 \\ 5 & -5 & 6 \end{array}\right] + \left[\begin{array}{ccc} -12 & 6 & -6 \\ -6 & -12 & 6 \\ -6 & 6 & -12 \end{array}\right] + \left[\begin{array}{ccc} 9 & 0 & 0 \\ 0 & 9 & 0 \\ 0 & 0 & 9 \end{array}\right] \)
\( = \left[\begin{array}{ccc} 6-12+9 & -5+6+0 & 5-6+0 \\ -5-6+0 & 6-12+9 & -5+6+0 \\ 5-6+0 & -5+6+0 & 6-12+9 \end{array}\right] \)
\( = \left[\begin{array}{ccc} 3 & 1 & -1 \\ -1 & 3 & 1 \\ -1 & -1 & 3 \end{array}\right] \).
Finally, divide by 4 to get \( A^{-1} \):
\( A^{-1} = \frac{1}{4}\left[\begin{array}{ccc} 3 & 1 & -1 \\ 1 & 3 & 1 \\ -1 & -1 & 3 \end{array}\right] \).
In simple words: We proved the matrix equation by calculating \( A^2 \) and \( A^3 \), then combining them with A and the identity matrix to show the result is a zero matrix. After that, we used the same polynomial equation to find the inverse of A by multiplying by \( A^{-1} \) and rearranging the terms.

Exam Tip: Pay close attention to the signs and matrix multiplication rules when dealing with matrix polynomials. A small error in one step can propagate and affect the entire verification process and the calculation of the inverse.

Question 17. Let A be a non-singular square matrix of order 3 x 3. Then, | adj A | is equal to
(A) |A|
(B) |A|²
(C) |A|³
(D) 3|A|
Answer: (B) |A|²
For any square matrix A of order n, we know that \( A(\text{adj } A) = (\text{adj } A)A = |A|I \), where I is the identity matrix of order n. Taking the determinant on both sides, we get \( |A(\text{adj } A)| = ||A|I| \). Using the properties of determinants, this simplifies to \( |A| |\text{adj } A| = |A|^n |I| \). Since \( |I| = 1 \), we have \( |A| |\text{adj } A| = |A|^n \). If A is a non-singular matrix, then \( |A| \neq 0 \), so we can divide by \( |A| \) to get \( |\text{adj } A| = |A|^{n-1} \). For this question, the matrix A has an order of 3, meaning \( n = 3 \). Therefore, \( |\text{adj } A| = |A|^{3-1} = |A|^2 \).
In simple words: When you have a square matrix A that isn't singular, the determinant of its adjoint (adj A) is found by taking the determinant of A and raising it to the power of (n-1), where n is the matrix's order. Since this matrix is 3x3, the power is 3-1, which is 2.

Exam Tip: Remember the formula \( |\text{adj } A| = |A|^{n-1} \) for a non-singular square matrix A of order n; this is a fundamental property. Always state the order of the matrix clearly to apply the formula correctly.

Question 18. If A is an invertible matrix of order 2, then det (A⁻¹) is equal to
(A) det |A|
(B) \( \frac { 1 }{ det(A) } \)
(C) 0
(D) 0
Answer: (B) \( \frac { 1 }{ \text{det}(A) } \)
For any invertible square matrix A, its inverse \( A^{-1} \) exists, and we know that \( A A^{-1} = I \), where I is the identity matrix. Taking the determinant of both sides of this equation, we get \( |\text{A A}^{-1}| = |I| \). According to the properties of determinants, the determinant of a product of matrices is the product of their determinants, so \( |A| |A^{-1}| = |I| \). We also know that the determinant of an identity matrix is always 1, so \( |I| = 1 \). This gives us \( |A| |A^{-1}| = 1 \). If we divide both sides by \( |A| \) (which is non-zero because A is invertible), we find that \( |A^{-1}| = \frac{1}{|A|} \). This means the determinant of the inverse matrix is the reciprocal of the determinant of the original matrix.
In simple words: When a matrix A can be inverted, the determinant of its inverse \( (A^{-1}) \) is simply 1 divided by the determinant of A. This is because when you multiply a matrix by its inverse, you get the identity matrix, whose determinant is 1.

Exam Tip: Understand the key property: \( \text{det}(AB) = \text{det}(A) \times \text{det}(B) \) and \( \text{det}(I) = 1 \). These are essential for deriving the determinant of an inverse matrix.