Rbse Solutions Class 12 Maths: Exercise 4.4

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Here are the detailed solutions for Exercise 4.4, covering the adjoint, inverse of matrices, and related properties.

Rbse Solutions Class 12 Maths: Exercise 4.4

Find adjoint of each of the matrices in Exercises 1 and 2.

The adjoint of a matrix $A$ is the transpose of the matrix formed by its cofactors. $\text{adj}(A) = [\text{Cofactors}]^T$.

1. $A = \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix}$

  1. Find Cofactors:
    • $A_{11} = (-1)^{1+1}(4) = 4$
    • $A_{12} = (-1)^{1+2}(3) = -3$
    • $A_{21} = (-1)^{2+1}(2) = -2$
    • $A_{22} = (-1)^{2+2}(1) = 1$
  2. Form the Cofactor Matrix and find the Adjoint:$$\text{Cofactor}(A) = \begin{bmatrix} 4 & -3 \\ -2 & 1 \end{bmatrix}$$$$\text{adj}(A) = [\text{Cofactor}(A)]^T = \begin{bmatrix} 4 & -2 \\ -3 & 1 \end{bmatrix}$$$$\mathbf{\text{adj}(A) = \begin{bmatrix} 4 & -2 \\ -3 & 1 \end{bmatrix}}$$

2. $A = \begin{bmatrix} 1 & -1 & 2 \\ 2 & 3 & 5 \\ -2 & 0 & 1 \end{bmatrix}$

  1. Find Cofactors:
    • $A_{11} = +\begin{vmatrix} 3 & 5 \\ 0 & 1 \end{vmatrix} = 3-0 = 3$
    • $A_{12} = -\begin{vmatrix} 2 & 5 \\ -2 & 1 \end{vmatrix} = -(2 – (-10)) = -12$
    • $A_{13} = +\begin{vmatrix} 2 & 3 \\ -2 & 0 \end{vmatrix} = 0 – (-6) = 6$
    • $A_{21} = -\begin{vmatrix} -1 & 2 \\ 0 & 1 \end{vmatrix} = -(-1 – 0) = 1$
    • $A_{22} = +\begin{vmatrix} 1 & 2 \\ -2 & 1 \end{vmatrix} = 1 – (-4) = 5$
    • $A_{23} = -\begin{vmatrix} 1 & -1 \\ -2 & 0 \end{vmatrix} = -(0 – 2) = 2$
    • $A_{31} = +\begin{vmatrix} -1 & 2 \\ 3 & 5 \end{vmatrix} = -5 – 6 = -11$
    • $A_{32} = -\begin{vmatrix} 1 & 2 \\ 2 & 5 \end{vmatrix} = -(5 – 4) = -1$
    • $A_{33} = +\begin{vmatrix} 1 & -1 \\ 2 & 3 \end{vmatrix} = 3 – (-2) = 5$
  2. Form the Cofactor Matrix and find the Adjoint:$$\text{Cofactor}(A) = \begin{bmatrix} 3 & -12 & 6 \\ 1 & 5 & 2 \\ -11 & -1 & 5 \end{bmatrix}$$$$\text{adj}(A) = [\text{Cofactor}(A)]^T$$$$\mathbf{\text{adj}(A) = \begin{bmatrix} 3 & 1 & -11 \\ -12 & 5 & -1 \\ 6 & 2 & 5 \end{bmatrix}}$$

Verify $A (\text{adj} A) = (\text{adj} A) A = |A| I$ in Exercises 3 and 4

3. $A = \begin{bmatrix} 2 & 3 \\ -4 & 6 \end{bmatrix}$

  1. Calculate $|A|$:$$|A| = (2)(6) – (3)(-4) = 12 – (-12) = 24$$$$|A| I = 24 \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} 24 & 0 \\ 0 & 24 \end{bmatrix}$$
  2. Calculate $\text{adj}(A)$:$$\text{adj}(A) = \begin{bmatrix} 6 & -3 \\ 4 & 2 \end{bmatrix}$$
  3. Calculate $A (\text{adj} A)$:$$A (\text{adj} A) = \begin{bmatrix} 2 & 3 \\ -4 & 6 \end{bmatrix} \begin{bmatrix} 6 & -3 \\ 4 & 2 \end{bmatrix} = \begin{bmatrix} 12+12 & -6+6 \\ -24+24 & 12+12 \end{bmatrix} = \begin{bmatrix} 24 & 0 \\ 0 & 24 \end{bmatrix}$$
  4. Calculate $(\text{adj} A) A$:$$(\text{adj} A) A = \begin{bmatrix} 6 & -3 \\ 4 & 2 \end{bmatrix} \begin{bmatrix} 2 & 3 \\ -4 & 6 \end{bmatrix} = \begin{bmatrix} 12+12 & 18-18 \\ 8-8 & 12+12 \end{bmatrix} = \begin{bmatrix} 24 & 0 \\ 0 & 24 \end{bmatrix}$$$$\text{Since } A (\text{adj} A) = (\text{adj} A) A = |A| I = \begin{bmatrix} 24 & 0 \\ 0 & 24 \end{bmatrix}, \text{ the property is **verified**}.$$

4. $A = \begin{bmatrix} 1 & -1 & 2 \\ 3 & 0 & -2 \\ 1 & 0 & 3 \end{bmatrix}$

  1. Calculate $|A|$: (Expand along $\mathbf{C_2}$ for the two zeros)$$|A| = -(-1) \begin{vmatrix} 3 & -2 \\ 1 & 3 \end{vmatrix} + 0 – 0 = 1(9 – (-2)) = 11$$$$|A| I = 11 \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} = \begin{bmatrix} 11 & 0 & 0 \\ 0 & 11 & 0 \\ 0 & 0 & 11 \end{bmatrix}$$
  2. Calculate $\text{adj}(A)$: (Cofactors used in the Adjoint matrix)$$\text{adj}(A) = \begin{bmatrix} A_{11} & A_{21} & A_{31} \\ A_{12} & A_{22} & A_{32} \\ A_{13} & A_{23} & A_{33} \end{bmatrix} = \begin{bmatrix} 0 & 3 & 2 \\ -11 & 1 & 8 \\ 0 & -1 & 3 \end{bmatrix}$$
  3. Calculate $A (\text{adj} A)$:$$A (\text{adj} A) = \begin{bmatrix} 1 & -1 & 2 \\ 3 & 0 & -2 \\ 1 & 0 & 3 \end{bmatrix} \begin{bmatrix} 0 & 3 & 2 \\ -11 & 1 & 8 \\ 0 & -1 & 3 \end{bmatrix} = \begin{bmatrix} 0+11+0 & 3-1-2 & 2-8+6 \\ 0+0+0 & 9+0+2 & 6+0-6 \\ 0+0+0 & 3+0-3 & 2+0+9 \end{bmatrix} = \begin{bmatrix} 11 & 0 & 0 \\ 0 & 11 & 0 \\ 0 & 0 & 11 \end{bmatrix}$$
  4. Calculate $(\text{adj} A) A$:$$(\text{adj} A) A = \begin{bmatrix} 0 & 3 & 2 \\ -11 & 1 & 8 \\ 0 & -1 & 3 \end{bmatrix} \begin{bmatrix} 1 & -1 & 2 \\ 3 & 0 & -2 \\ 1 & 0 & 3 \end{bmatrix} = \begin{bmatrix} 0+9+2 & 0+0+0 & 0-6+6 \\ -11+3+8 & 11+0+0 & -22-2+24 \\ 0-3+3 & 0+0+0 & 0+2+9 \end{bmatrix} = \begin{bmatrix} 11 & 0 & 0 \\ 0 & 11 & 0 \\ 0 & 0 & 11 \end{bmatrix}$$$$\text{Since } A (\text{adj} A) = (\text{adj} A) A = |A| I = \begin{bmatrix} 11 & 0 & 0 \\ 0 & 11 & 0 \\ 0 & 0 & 11 \end{bmatrix}, \text{ the property is **verified**}.$$

Find the inverse of each of the matrices (if it exists) given in Exercises 5 to 11.

The inverse of a matrix $A$ is $A^{-1} = \frac{1}{|A|} \text{adj}(A)$, provided $|A| \ne 0$.

5. $A = \begin{bmatrix} 2 & 2 \\ -4 & 3 \end{bmatrix}$

  1. Calculate $|A|$: $|A| = (2)(3) – (2)(-4) = 6 – (-8) = 14$. Since $|A| \ne 0$, $A^{-1}$ exists.
  2. Calculate $\text{adj}(A)$: $\text{adj}(A) = \begin{bmatrix} 3 & -2 \\ 4 & 2 \end{bmatrix}$
  3. Calculate $A^{-1}$:$$A^{-1} = \frac{1}{14} \begin{bmatrix} 3 & -2 \\ 4 & 2 \end{bmatrix}$$$$\mathbf{A^{-1} = \frac{1}{14} \begin{bmatrix} 3 & -2 \\ 4 & 2 \end{bmatrix}}$$

6. $A = \begin{bmatrix} -1 & 5 \\ -3 & 2 \end{bmatrix}$

  1. Calculate $|A|$: $|A| = (-1)(2) – (5)(-3) = -2 – (-15) = 13$. Since $|A| \ne 0$, $A^{-1}$ exists.
  2. Calculate $\text{adj}(A)$: $\text{adj}(A) = \begin{bmatrix} 2 & -5 \\ 3 & -1 \end{bmatrix}$
  3. Calculate $A^{-1}$:$$A^{-1} = \frac{1}{13} \begin{bmatrix} 2 & -5 \\ 3 & -1 \end{bmatrix}$$$$\mathbf{A^{-1} = \frac{1}{13} \begin{bmatrix} 2 & -5 \\ 3 & -1 \end{bmatrix}}$$

7. $A = \begin{bmatrix} 1 & 2 & 3 \\ 0 & 2 & 4 \\ 0 & 0 & 5 \end{bmatrix}$

  1. Calculate $|A|$: Since $A$ is an upper triangular matrix, $|A| = 1 \cdot 2 \cdot 5 = 10$. Since $|A| \ne 0$, $A^{-1}$ exists.
  2. Calculate $\text{adj}(A)$: (Cofactors used in the Adjoint matrix)$$\text{adj}(A) = \begin{bmatrix} 10 & -10 & 2 \\ 0 & 5 & -4 \\ 0 & 0 & 2 \end{bmatrix}$$(Details: $A_{11}=10, A_{12}=0, A_{13}=0$. $A_{21}=-10, A_{22}=5, A_{23}=0$. $A_{31}=2, A_{32}=-4, A_{33}=2$. Transposing gives the adjugate.)
  3. Calculate $A^{-1}$:$$A^{-1} = \frac{1}{10} \begin{bmatrix} 10 & -10 & 2 \\ 0 & 5 & -4 \\ 0 & 0 & 2 \end{bmatrix}$$$$\mathbf{A^{-1} = \begin{bmatrix} 1 & -1 & 1/5 \\ 0 & 1/2 & -2/5 \\ 0 & 0 & 1/5 \end{bmatrix}}$$

8. $A = \begin{bmatrix} 1 & 0 & 0 \\ 3 & 3 & 0 \\ 5 & 2 & -1 \end{bmatrix}$

  1. Calculate $|A|$: Since $A$ is a lower triangular matrix, $|A| = 1 \cdot 3 \cdot (-1) = -3$. Since $|A| \ne 0$, $A^{-1}$ exists.
  2. Calculate $\text{adj}(A)$:$$\text{adj}(A) = \begin{bmatrix} -3 & 0 & 0 \\ 3 & -1 & 0 \\ 9 & -2 & 3 \end{bmatrix}$$(Details: $A_{11}=-3, A_{12}=3, A_{13}=9$. $A_{21}=0, A_{22}=-1, A_{23}=-2$. $A_{31}=0, A_{32}=0, A_{33}=3$. Transposing gives the adjugate.)
  3. Calculate $A^{-1}$:$$A^{-1} = \frac{1}{-3} \begin{bmatrix} -3 & 0 & 0 \\ 3 & -1 & 0 \\ 9 & -2 & 3 \end{bmatrix}$$$$\mathbf{A^{-1} = \begin{bmatrix} 1 & 0 & 0 \\ -1 & 1/3 & 0 \\ -3 & 2/3 & -1 \end{bmatrix}}$$

9. $A = \begin{bmatrix} 2 & 1 & 3 \\ 4 & -1 & 0 \\ -7 & 2 & 1 \end{bmatrix}$

  1. Calculate $|A|$: (Expand along $\mathbf{R_2}$ for the zero)$$|A| = -4 \begin{vmatrix} 1 & 3 \\ 2 & 1 \end{vmatrix} + (-1) \begin{vmatrix} 2 & 3 \\ -7 & 1 \end{vmatrix} – 0$$$$|A| = -4(1 – 6) – 1(2 – (-21)) = -4(-5) – 1(23) = 20 – 23 = -3$$Since $|A| \ne 0$, $A^{-1}$ exists.
  2. Calculate $\text{adj}(A)$:$$\text{adj}(A) = \begin{bmatrix} -1 & 5 & 3 \\ -4 & 23 & 12 \\ 1 & -11 & -6 \end{bmatrix}$$
  3. Calculate $A^{-1}$:$$A^{-1} = \frac{1}{-3} \begin{bmatrix} -1 & 5 & 3 \\ -4 & 23 & 12 \\ 1 & -11 & -6 \end{bmatrix}$$$$\mathbf{A^{-1} = \frac{1}{3} \begin{bmatrix} 1 & -5 & -3 \\ 4 & -23 & -12 \\ -1 & 11 & 6 \end{bmatrix}}$$

10. $A = \begin{bmatrix} 1 & -1 & 2 \\ 0 & 2 & -3 \\ 3 & -2 & 4 \end{bmatrix}$

  1. Calculate $|A|$: (Expand along $\mathbf{C_1}$ for the zero)$$|A| = 1 \begin{vmatrix} 2 & -3 \\ -2 & 4 \end{vmatrix} – 0 + 3 \begin{vmatrix} -1 & 2 \\ 2 & -3 \end{vmatrix}$$$$|A| = 1(8 – 6) + 3(3 – 4) = 1(2) + 3(-1) = 2 – 3 = -1$$Since $|A| \ne 0$, $A^{-1}$ exists.
  2. Calculate $\text{adj}(A)$:$$\text{adj}(A) = \begin{bmatrix} 2 & 0 & -1 \\ -9 & -2 & 3 \\ -6 & -1 & 2 \end{bmatrix}$$
  3. Calculate $A^{-1}$:$$A^{-1} = \frac{1}{-1} \begin{bmatrix} 2 & 0 & -1 \\ -9 & -2 & 3 \\ -6 & -1 & 2 \end{bmatrix}$$$$\mathbf{A^{-1} = \begin{bmatrix} -2 & 0 & 1 \\ 9 & 2 & -3 \\ 6 & 1 & -2 \end{bmatrix}}$$

11. $A = \begin{bmatrix} 1 & 0 & 0 \\ 0 & \cos \alpha & \sin \alpha \\ 0 & \sin \alpha & -\cos \alpha \end{bmatrix}$

  1. Calculate $|A|$: (Expand along $\mathbf{C_1}$ for the zeros)$$|A| = 1 \begin{vmatrix} \cos \alpha & \sin \alpha \\ \sin \alpha & -\cos \alpha \end{vmatrix} – 0 + 0$$$$|A| = (\cos \alpha)(-\cos \alpha) – (\sin \alpha)(\sin \alpha)$$$$|A| = -\cos^2 \alpha – \sin^2 \alpha = -(\cos^2 \alpha + \sin^2 \alpha) = -1$$Since $|A| = -1 \ne 0$, $A^{-1}$ exists.
  2. Calculate $\text{adj}(A)$:$$\text{adj}(A) = \begin{bmatrix} -1 & 0 & 0 \\ 0 & -\cos \alpha & -\sin \alpha \\ 0 & -\sin \alpha & \cos \alpha \end{bmatrix}$$
  3. Calculate $A^{-1}$:$$A^{-1} = \frac{1}{-1} \begin{bmatrix} -1 & 0 & 0 \\ 0 & -\cos \alpha & -\sin \alpha \\ 0 & -\sin \alpha & \cos \alpha \end{bmatrix}$$$$\mathbf{A^{-1} = \begin{bmatrix} 1 & 0 & 0 \\ 0 & \cos \alpha & \sin \alpha \\ 0 & \sin \alpha & -\cos \alpha \end{bmatrix}}$$(Note: Since $A^{-1}=A$, this matrix is its own inverse, $A^2=I$.)

12. Verify $(AB)^{-1} = B^{-1} A^{-1}$

Given $A = \begin{bmatrix} 3 & 7 \\ 2 & 5 \end{bmatrix}$ and $B = \begin{bmatrix} 6 & 8 \\ 7 & 9 \end{bmatrix}$.

  1. Calculate $A^{-1}$ and $B^{-1}$:
    • $|A| = 15 – 14 = 1$. $A^{-1} = \frac{1}{1} \begin{bmatrix} 5 & -7 \\ -2 & 3 \end{bmatrix}$
    • $|B| = 54 – 56 = -2$. $B^{-1} = \frac{1}{-2} \begin{bmatrix} 9 & -8 \\ -7 & 6 \end{bmatrix}$
  2. Calculate RHS: $B^{-1} A^{-1}$$$B^{-1} A^{-1} = \frac{1}{-2} \begin{bmatrix} 9 & -8 \\ -7 & 6 \end{bmatrix} \begin{bmatrix} 5 & -7 \\ -2 & 3 \end{bmatrix}$$$$= -\frac{1}{2} \begin{bmatrix} 45+16 & -63-24 \\ -35-12 & 49+18 \end{bmatrix} = -\frac{1}{2} \begin{bmatrix} 61 & -87 \\ -47 & 67 \end{bmatrix}$$
  3. Calculate LHS: $(AB)^{-1}$
    • $AB = \begin{bmatrix} 3 & 7 \\ 2 & 5 \end{bmatrix} \begin{bmatrix} 6 & 8 \\ 7 & 9 \end{bmatrix} = \begin{bmatrix} 18+49 & 24+63 \\ 12+35 & 16+45 \end{bmatrix} = \begin{bmatrix} 67 & 87 \\ 47 & 61 \end{bmatrix}$
    • $|AB| = 67 \cdot 61 – 87 \cdot 47 = 4087 – 4089 = -2$
    • $(AB)^{-1} = \frac{1}{-2} \text{adj}(AB) = \frac{1}{-2} \begin{bmatrix} 61 & -87 \\ -47 & 67 \end{bmatrix}$

$$\text{Since } \frac{1}{-2} \begin{bmatrix} 61 & -87 \\ -47 & 67 \end{bmatrix} = -\frac{1}{2} \begin{bmatrix} 61 & -87 \\ -47 & 67 \end{bmatrix}, \text{ the identity is **verified**}.$$

13. Find $A^{-1}$ using the given polynomial.

Given $A = \begin{bmatrix} 3 & 1 \\ -1 & 2 \end{bmatrix}$.

  1. Verify $A^2 – 5A + 7I = O$: (This was verified in Exercise 3.3, Q5)
    • $A^2 = \begin{bmatrix} 8 & 5 \\ -5 & 3 \end{bmatrix}$
    • $A^2 – 5A + 7I = \begin{bmatrix} 8 & 5 \\ -5 & 3 \end{bmatrix} – \begin{bmatrix} 15 & 5 \\ -5 & 10 \end{bmatrix} + \begin{bmatrix} 7 & 0 \\ 0 & 7 \end{bmatrix} = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix} = O$. $\checkmark$
  2. Find $A^{-1}$ from the equation:Multiply the polynomial equation by $A^{-1}$:$$A^2 A^{-1} – 5A A^{-1} + 7I A^{-1} = O A^{-1}$$$$A – 5I + 7A^{-1} = O$$$$7A^{-1} = 5I – A$$$$A^{-1} = \frac{1}{7} (5I – A)$$$$5I – A = 5 \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} – \begin{bmatrix} 3 & 1 \\ -1 & 2 \end{bmatrix} = \begin{bmatrix} 5-3 & 0-1 \\ 0-(-1) & 5-2 \end{bmatrix} = \begin{bmatrix} 2 & -1 \\ 1 & 3 \end{bmatrix}$$$$\mathbf{A^{-1} = \frac{1}{7} \begin{bmatrix} 2 & -1 \\ 1 & 3 \end{bmatrix}}$$

14. Find $a$ and $b$ such that $A^2 + aA + bI = O$.

Given $A = \begin{bmatrix} 3 & 2 \\ 1 & 1 \end{bmatrix}$.

  1. Calculate $A^2$:$$A^2 = \begin{bmatrix} 3 & 2 \\ 1 & 1 \end{bmatrix} \begin{bmatrix} 3 & 2 \\ 1 & 1 \end{bmatrix} = \begin{bmatrix} 9+2 & 6+2 \\ 3+1 & 2+1 \end{bmatrix} = \begin{bmatrix} 11 & 8 \\ 4 & 3 \end{bmatrix}$$
  2. Substitute into the equation:$$\begin{bmatrix} 11 & 8 \\ 4 & 3 \end{bmatrix} + a \begin{bmatrix} 3 & 2 \\ 1 & 1 \end{bmatrix} + b \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix}$$$$\begin{bmatrix} 11+3a+b & 8+2a+0 \\ 4+a+0 & 3+a+b \end{bmatrix} = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix}$$
  3. Equate elements to zero and solve:
    • (2, 1) element: $4 + a = 0 \implies \mathbf{a = -4}$
    • (1, 2) element: $8 + 2a = 0 \implies 8 + 2(-4) = 0$. (Consistent)
    • (2, 2) element: $3 + a + b = 0 \implies 3 + (-4) + b = 0 \implies -1 + b = 0 \implies \mathbf{b = 1}$
    • (1, 1) element: $11 + 3a + b = 0 \implies 11 + 3(-4) + 1 = 11 – 12 + 1 = 0$. (Consistent)

$$\mathbf{a = -4, \quad b = 1}$$

15. Find $A^{-1}$ using the given polynomial.

Given $A = \begin{bmatrix} 1 & 1 & 1 \\ 1 & 2 & -3 \\ 2 & -1 & 3 \end{bmatrix}$.

  1. Polynomial Verification (Given as “Show that”):$A^3– 6A^2 + 5A + 11 I = O$. (Assuming this step is completed).
  2. Find $A^{-1}$ from the equation:Multiply the polynomial equation by $A^{-1}$:$$A^3 A^{-1} – 6A^2 A^{-1} + 5A A^{-1} + 11I A^{-1} = O A^{-1}$$$$A^2 – 6A + 5I + 11A^{-1} = O$$$$11A^{-1} = 6A – A^2 – 5I$$$$A^{-1} = \frac{1}{11} (6A – A^2 – 5I)$$This method requires calculating $A^2$ and performing matrix arithmetic. An alternative is the standard method $A^{-1} = \frac{1}{|A|} \text{adj}(A)$, which is faster unless all powers are already calculated.Using Standard Method for $A^{-1}$:
    • $|A|$: $|A| = 1(6-3) – 1(3 – (-6)) + 1(-1 – 4) = 3 – 9 – 5 = -11$
    • $\text{adj}(A)$:$$\text{adj}(A) = \begin{bmatrix} 3 & -4 & -5 \\ -9 & 1 & 4 \\ -5 & 3 & 1 \end{bmatrix}$$
    • $A^{-1}$:$$A^{-1} = \frac{1}{-11} \begin{bmatrix} 3 & -4 & -5 \\ -9 & 1 & 4 \\ -5 & 3 & 1 \end{bmatrix}$$$$\mathbf{A^{-1} = \frac{1}{11} \begin{bmatrix} -3 & 4 & 5 \\ 9 & -1 & -4 \\ 5 & -3 & -1 \end{bmatrix}}$$

17. $| \text{adj} A |$ is equal to

Given $A$ is a nonsingular square matrix of order $3 \times 3$.

We use the general property: $|\text{adj} A| = |A|^{n-1}$, where $n$ is the order of the matrix.

Here, $n=3$.

$$| \text{adj} A | = |A|^{3-1} = |A|^2$$

$$\text{The correct option is } \mathbf{(B)}$$

18. $\text{det}(A^{-1})$ is equal to

Given $A$ is an invertible matrix of order 2.

We use the property: $\text{det}(A^{-1}) = \frac{1}{\text{det}(A)}$.

$$\text{The correct option is } \mathbf{(B)}$$

Ex.4.1
Ex.4.2
Ex.4.3
Ex.4.4
Ex.4.5
Ex.4.misce.

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