Q64 / 150 · MHT CET 2026 (16 April Shift 2) · Mathematics
If $A = \begin{bmatrix} 2 & 3 \\ 5 & -2 \end{bmatrix}$, $B^{-1} = \begin{bmatrix} \dfrac{1}{5} & \dfrac{2}{5} \\ \dfrac{2}{5} & -\dfrac{1}{5} \end{bmatrix}$, then $(AB)^{-1} = $
- A$\dfrac{1}{95}\begin{bmatrix} 8 & -1 \\ -1 & 12 \end{bmatrix}$
- Correct answerB$\dfrac{1}{95}\begin{bmatrix} 12 & -1 \\ -1 & 8 \end{bmatrix}$
- C$\dfrac{1}{95}\begin{bmatrix} -12 & 1 \\ 1 & -8 \end{bmatrix}$
- D$\dfrac{1}{95}\begin{bmatrix} -8 & 1 \\ 1 & -12 \end{bmatrix}$
Solution
Using the reversal law for the inverse of a product of matrices, we have:
$(AB)^{-1} = B^{-1}A^{-1}$
First, we find the inverse of matrix $A$. The determinant of $A$ is:
$|A| = (2)(-2) - (3)(5) = -4 - 15 = -19$
The adjugate of $A$ is obtained by swapping the diagonal elements and changing the signs of the off-diagonal elements:
$\text{adj}(A) = \begin{bmatrix} -2 & -3 \\ -5 & 2 \end{bmatrix}$
Thus, $A^{-1}$ is given by:
$A^{-1} = \dfrac{1}{|A|} \text{adj}(A) = \dfrac{1}{-19} \begin{bmatrix} -2 & -3 \\ -5 & 2 \end{bmatrix} = \dfrac{1}{19} \begin{bmatrix} 2 & 3 \\ 5 & -2 \end{bmatrix}$
We are given $B^{-1} = \begin{bmatrix} \dfrac{1}{5} & \dfrac{2}{5} \\ \dfrac{2}{5} & -\dfrac{1}{5} \end{bmatrix} = \dfrac{1}{5} \begin{bmatrix} 1 & 2 \\ 2 & -1 \end{bmatrix}$.
Now, we compute $(AB)^{-1}$ by multiplying $B^{-1}$ and $A^{-1}$:
$(AB)^{-1} = B^{-1}A^{-1} = \left( \dfrac{1}{5} \begin{bmatrix} 1 & 2 \\ 2 & -1 \end{bmatrix} \right) \left( \dfrac{1}{19} \begin{bmatrix} 2 & 3 \\ 5 & -2 \end{bmatrix} \right)$
$(AB)^{-1} = \dfrac{1}{95} \begin{bmatrix} 1 & 2 \\ 2 & -1 \end{bmatrix} \begin{bmatrix} 2 & 3 \\ 5 & -2 \end{bmatrix}$
Performing the matrix multiplication:
$(AB)^{-1} = \dfrac{1}{95} \begin{bmatrix} (1)(2) + (2)(5) & (1)(3) + (2)(-2) \\ (2)(2) + (-1)(5) & (2)(3) + (-1)(-2) \end{bmatrix}$
$(AB)^{-1} = \dfrac{1}{95} \begin{bmatrix} 2 + 10 & 3 - 4 \\ 4 - 5 & 6 + 2 \end{bmatrix} = \dfrac{1}{95} \begin{bmatrix} 12 & -1 \\ -1 & 8 \end{bmatrix}$
Answer: $\dfrac{1}{95}\begin{bmatrix} 12 & -1 \\ -1 & 8 \end{bmatrix}$
$(AB)^{-1} = B^{-1}A^{-1}$
First, we find the inverse of matrix $A$. The determinant of $A$ is:
$|A| = (2)(-2) - (3)(5) = -4 - 15 = -19$
The adjugate of $A$ is obtained by swapping the diagonal elements and changing the signs of the off-diagonal elements:
$\text{adj}(A) = \begin{bmatrix} -2 & -3 \\ -5 & 2 \end{bmatrix}$
Thus, $A^{-1}$ is given by:
$A^{-1} = \dfrac{1}{|A|} \text{adj}(A) = \dfrac{1}{-19} \begin{bmatrix} -2 & -3 \\ -5 & 2 \end{bmatrix} = \dfrac{1}{19} \begin{bmatrix} 2 & 3 \\ 5 & -2 \end{bmatrix}$
We are given $B^{-1} = \begin{bmatrix} \dfrac{1}{5} & \dfrac{2}{5} \\ \dfrac{2}{5} & -\dfrac{1}{5} \end{bmatrix} = \dfrac{1}{5} \begin{bmatrix} 1 & 2 \\ 2 & -1 \end{bmatrix}$.
Now, we compute $(AB)^{-1}$ by multiplying $B^{-1}$ and $A^{-1}$:
$(AB)^{-1} = B^{-1}A^{-1} = \left( \dfrac{1}{5} \begin{bmatrix} 1 & 2 \\ 2 & -1 \end{bmatrix} \right) \left( \dfrac{1}{19} \begin{bmatrix} 2 & 3 \\ 5 & -2 \end{bmatrix} \right)$
$(AB)^{-1} = \dfrac{1}{95} \begin{bmatrix} 1 & 2 \\ 2 & -1 \end{bmatrix} \begin{bmatrix} 2 & 3 \\ 5 & -2 \end{bmatrix}$
Performing the matrix multiplication:
$(AB)^{-1} = \dfrac{1}{95} \begin{bmatrix} (1)(2) + (2)(5) & (1)(3) + (2)(-2) \\ (2)(2) + (-1)(5) & (2)(3) + (-1)(-2) \end{bmatrix}$
$(AB)^{-1} = \dfrac{1}{95} \begin{bmatrix} 2 + 10 & 3 - 4 \\ 4 - 5 & 6 + 2 \end{bmatrix} = \dfrac{1}{95} \begin{bmatrix} 12 & -1 \\ -1 & 8 \end{bmatrix}$
Answer: $\dfrac{1}{95}\begin{bmatrix} 12 & -1 \\ -1 & 8 \end{bmatrix}$
Hence, the correct answer is option: 'B'.