If A=≤ft[{array}{cc}3 & 1 \\ -1 & 2{array}], C=≤ft[{array}{ll}7 & 3 \\ 0 & 6{array}] and A B=C, then the inverse of matrix B is

Correct Answer is : 16[30−12]1/6≤ft[{array}{cc}3 & 0 \\ -1 & 2{array}]61[3−102]

1/42≤ft[{array}{cc}3 & 0 \\ -1 & 2{array}]

1/6≤ft[{array}{cc}3 & 0 \\ -1 & 2{array}]

1/42≤ft[{array}{cc}6 & 3 \\ -1 & 2{array}]

1/6≤ft[{array}{cc}7 & 3 \\ -1 & 3{array}]

Correct Answer is : 16[30−12]1/6≤ft[{array}{cc}3 & 0 \\ -1 & 2{array}]61[3−102]

Test Index

MHTCET 15 Apr 2026 Shift 2 Paper

Question : 117 of 150

Marks: +1, -0

A=\left[\begin{array}{cc}3 & 1 \\ -1 & 2\end{array}\right], C=\left[\begin{array}{ll}7 & 3 \\ 0 & 6\end{array}\right]

A B=C

, then the inverse of matrix B is

Solution:

Concept:

Use the inverse of a product to express

B^{-1}

in terms of

A

and

C

Explanation:

Given

AB = C

Taking inverse on both sides:

(AB)^{-1} = C^{-1}

Using

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

B^{-1}A^{-1} = C^{-1}

Multiply both sides on the right by

A

B^{-1} = C^{-1}A

First compute

C^{-1}

C = \begin{bmatrix} 7 & 3 \\ 0 & 6 \end{bmatrix}, \quad |C| = 7 \times 6 - 0 \times 3 = 42

So,

C^{-1} = \frac{1}{42} \begin{bmatrix} 6 & -3 \\ 0 & 7 \end{bmatrix}

Now multiply by

A

B^{-1} = \frac{1}{42} \begin{bmatrix} 6 & -3 \\ 0 & 7 \end{bmatrix} \begin{bmatrix} 3 & 1 \\ -1 & 2 \end{bmatrix}

B^{-1} = \frac{1}{42} \begin{bmatrix} 18+3 & 6-6 \\ 0-7 & 0+14 \end{bmatrix}

B^{-1} = \frac{1}{42} \begin{bmatrix} 21 & 0 \\ -7 & 14 \end{bmatrix}

Taking

7

common:

B^{-1} = \frac{7}{42} \begin{bmatrix} 3 & 0 \\ -1 & 2 \end{bmatrix} = \frac{1}{6} \begin{bmatrix} 3 & 0 \\ -1 & 2 \end{bmatrix}

Answer:

Option B.

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