Question:
If
\[
A = \begin{bmatrix} 2 & 3 \\ 1 & 2 \end{bmatrix}, \quad
B = \begin{bmatrix} 1 & 0 \\ 3 & 1 \end{bmatrix},
\]
then find \((AB)^{-1}\).
If
\[
A = \begin{bmatrix} 2 & 3 \\ 1 & 2 \end{bmatrix}, \quad
B = \begin{bmatrix} 1 & 0 \\ 3 & 1 \end{bmatrix},
\]
then find \((AB)^{-1}\).
Show Hint
For matrices, \((AB)^{-1} = B^{-1}A^{-1}\), but direct multiplication is often faster in MCQs.
Updated On: Apr 27, 2026
- \( \begin{bmatrix} 2 & -3 \\ -7 & 11 \end{bmatrix} \)
- \( \begin{bmatrix} 2 & -3 \\ 7 & 11 \end{bmatrix} \)
- \( \begin{bmatrix} 2 & -3 \\ -7 & -11 \end{bmatrix} \)
- \( \begin{bmatrix} -2 & -3 \\ -7 & 11 \end{bmatrix} \)
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The Correct Option is A
Solution and Explanation
Step 1: Find the product \(AB\).
\[ AB = \begin{bmatrix} 2 & 3 \\ 1 & 2 \end{bmatrix} \begin{bmatrix} 1 & 0 \\ 3 & 1 \end{bmatrix} = \begin{bmatrix} 11 & 3 \\ 7 & 2 \end{bmatrix} \]
Step 2: Find the determinant of \(AB\).
\[ |AB| = (11)(2) - (3)(7) = 22 - 21 = 1 \]
Step 3: Find the inverse of \(AB\).
\[ (AB)^{-1} = \begin{bmatrix} 2 & -3 \\ -7 & 11 \end{bmatrix} \]
\[ AB = \begin{bmatrix} 2 & 3 \\ 1 & 2 \end{bmatrix} \begin{bmatrix} 1 & 0 \\ 3 & 1 \end{bmatrix} = \begin{bmatrix} 11 & 3 \\ 7 & 2 \end{bmatrix} \]
Step 2: Find the determinant of \(AB\).
\[ |AB| = (11)(2) - (3)(7) = 22 - 21 = 1 \]
Step 3: Find the inverse of \(AB\).
\[ (AB)^{-1} = \begin{bmatrix} 2 & -3 \\ -7 & 11 \end{bmatrix} \]
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