Let \(A = \begin{bmatrix}3 & 1\\ 2 & 1\end{bmatrix}\) and \(B = \begin{bmatrix}4 & 1\\ 3 & 1\end{bmatrix}\). The matrix X such that \(AXB = I\) is
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \(\begin{bmatrix}4 & -5\\ -11 & 14\end{bmatrix}\)
Both A and B have determinant 1, so both are invertible. Multiplying \(AXB = I\) by \(A^{-1}\) on the left and by \(B^{-1}\) on the right gives
\(X = A^{-1}\,I\,B^{-1} = A^{-1}B^{-1}\)
\(\displaystyle A^{-1} = \begin{bmatrix}1 & -1\\ -2 & 3\end{bmatrix},\quad B^{-1} = \begin{bmatrix}1 & -1\\ -3 & 4\end{bmatrix}\)
\(\displaystyle X = \begin{bmatrix}1 + 3 & -1 - 4\\ -2 - 9 & 2 + 12\end{bmatrix}\)
\(\displaystyle X = \begin{bmatrix}4 & -5\\ -11 & 14\end{bmatrix}\)
Why the other options are incorrect
Option (1) \(\begin{bmatrix}3 & -4\\ -11 & 15\end{bmatrix}\) — Incorrect. This is \(B^{-1}A^{-1}\); matrix products cannot be reordered, and A must be cancelled from the left.
Option (3) \(\begin{bmatrix}4 & 5\\ 11 & 14\end{bmatrix}\) — Incorrect. The minus signs of \(A^{-1}B^{-1}\) are lost; with this X the product AXB is not I.
Option (4) \(\begin{bmatrix}1 & 0\\ 1 & 1\end{bmatrix}\) — Incorrect. This is \(A^{-1}B\), which solves AX = B, not AXB = I.