Step 1: Use the inverse property.
For matrices, \[ (B^{-1}A^{-1})^{-1} = A B \]
Step 2: Multiply matrices \( A \) and \( B \).
\[ AB = \begin{bmatrix}2&3\\1&2\end{bmatrix} \begin{bmatrix}2&-3\\-1&2\end{bmatrix} \] \[ = \begin{bmatrix} 4-3 & -6+6 \\ 2-2 & -3+4 \end{bmatrix} = \begin{bmatrix}1&0\\0&1\end{bmatrix} \]
Step 3: Conclusion.
\[ \boxed{\begin{bmatrix}1&0\\0&1\end{bmatrix}} \]
If matrix A =\(\begin{bmatrix} 1 & 2 \\ 4 & 3 \end{bmatrix}\) is such that AX = I, where I is 2 x 2 unit matrix, then X =
If \(\begin{bmatrix} 2 & 1 \\ 3 & 2\end{bmatrix}\) A \(\begin{bmatrix} -3 & 2 \\ 5 & -3\end{bmatrix}\) =\(\begin{bmatrix} 1 & 0 \\ 0 & 1\end{bmatrix}\), then A =?