Question

If \[ A = \begin{bmatrix} 2 & 3 \\ 1 & 2 \end{bmatrix}, \quad B = \begin{bmatrix} 1 & 0 \\ 3 & 1 \end{bmatrix}, \] then find \((AB)^{-1}\).

• A. \( \begin{bmatrix} 2 & -3 \\ -7 & 11 \end{bmatrix} \)
• B. \( \begin{bmatrix} 2 & -3 \\ 7 & 11 \end{bmatrix} \)
• C. \( \begin{bmatrix} 2 & -3 \\ -7 & -11 \end{bmatrix} \)
• D. \( \begin{bmatrix} -2 & -3 \\ -7 & 11 \end{bmatrix} \)

Hint:

For matrices, \((AB)^{-1} = B^{-1}A^{-1}\), but direct multiplication is often faster in MCQs.

Answer 1

The Correct Option is A

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 \(AB\).
\[ (AB)^{-1} = \begin{bmatrix} 2 & -3 \\ -7 & 11 \end{bmatrix} \]