If A=≤ft[{array}{cc}2 & 3 \\ 5 & -2{array}], B^{-1}=≤ft[{array}{cc}1/5 & 2/5 \\ 2/5 & -1/5{array}], then (A B)^{-1}=

Correct Answer is : 195[12−1−18]1/95≤ft[{array}{cc}12 & -1 \\ -1 & 8{array}]951[12−1−18]

1/95≤ft[{array}{cc}8 & -1 \\ -1 & 12{array}]

1/95≤ft[{array}{cc}12 & -1 \\ -1 & 8{array}]

1/95≤ft[{array}{cc}-12 & 1 \\ 1 & -8{array}]

1/95≤ft[{array}{cc}-8 & 1 \\ 1 & -12{array}]

Correct Answer is : 195[12−1−18]1/95≤ft[{array}{cc}12 & -1 \\ -1 & 8{array}]951[12−1−18]

Test Index

MHTCET 16 Apr 2026 Shift 2 Paper

Question : 113 of 150

Marks: +1, -0

A=\left[\begin{array}{cc}2 & 3 \\ 5 & -2\end{array}\right], B^{-1}=\left[\begin{array}{cc}\frac{1}{5} & \frac{2}{5} \\ \frac{2}{5} & -\frac{1}{5}\end{array}\right]

(A B)^{-1}=

Solution:

Concept:

Use the property

(AB)^{-1} = B^{-1} A^{-1}

for invertible matrices.

Explanation:

Given

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

and

B^{-1} = \begin{bmatrix} \frac{1}{5} & \frac{2}{5} \\ \frac{2}{5} & -\frac{1}{5} \end{bmatrix}

First, compute

A^{-1}

using the formula:

\begin{bmatrix} a & b \\ c & d \end{bmatrix}^{-1} = \frac{1}{ad-bc} \begin{bmatrix} d & -b \\ -c & a \end{bmatrix}

Here,

ad-bc = (2)(-2) - (3)(5) = -4 - 15 = -19

So,

A^{-1} = \frac{1}{-19} \begin{bmatrix} -2 & -3 \\ -5 & 2 \end{bmatrix} = \frac{1}{19} \begin{bmatrix} 2 & 3 \\ 5 & -2 \end{bmatrix}

Now apply the inverse product property:

(AB)^{-1} = B^{-1} A^{-1}

= \frac{1}{5}\begin{bmatrix} 1 & 2 \\ 2 & -1 \end{bmatrix} \cdot \frac{1}{19}\begin{bmatrix} 2 & 3 \\ 5 & -2 \end{bmatrix}

= \frac{1}{95} \begin{bmatrix} 1(2)+2(5) & 1(3)+2(-2) \\ 2(2)+(-1)(5) & 2(3)+(-1)(-2) \end{bmatrix}

= \frac{1}{95} \begin{bmatrix} 12 & -1 \\ -1 & 8 \end{bmatrix}

This matches Option B.

Answer:

Option B:

\frac{1}{95}\left[\begin{array}{cc}12 & -1 \\ -1 & 8\end{array}\right]

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