Find the inverse of \( A B \) if \( A^{-1}=\left[\begin{array}{ll}1 & 3 \\ 3 & -1\end{array}\right] \) and \( B^{-1}=\left[\begin{array}{ll}-4 & 5 \\ -2 & -2\end{array}\right] \) \( (A B)^{-1}=\left[\begin{array}{l}\square\end{array}\right] \)

To find the inverse of the product \( A B \), we can use the property that \( (AB)^{-1} = B^{-1} A^{-1} \). Given the inverses of matrices \( A \) and \( B \): \[ A^{-1} = \left[\begin{array}{ll} 1 & 3 \\ 3 & -1 \end{array}\right], \quad B^{-1} = \left[\begin{array}{ll} -4 & 5 \\ -2 & -2 \end{array}\right] \] We can now compute \( (AB)^{-1} \): \[ (AB)^{-1} = B^{-1} A^{-1} = \left[\begin{array}{ll} -4 & 5 \\ -2 & -2 \end{array}\right] \left[\begin{array}{ll} 1 & 3 \\ 3 & -1 \end{array}\right] \] Calculating this product: \[ (AB)^{-1} = \left[\begin{array}{ll} (-4)(1) + (5)(3) & (-4)(3) + (5)(-1) \\ (-2)(1) + (-2)(3) & (-2)(3) + (-2)(-1) \end{array}\right] \] \[ = \left[\begin{array}{ll} -4 + 15 & -12 - 5 \\ -2 - 6 & -6 + 2 \end{array}\right] \] This simplifies to: \[ = \left[\begin{array}{ll} 11 & -17 \\ -8 & -4 \end{array}\right] \] Thus, the inverse of \( A B \) is: \[ (AB)^{-1} = \left[\begin{array}{ll} 11 & -17 \\ -8 & -4 \end{array}\right] \] So, the final answer is: \[ (AB)^{-1}=\left[\begin{array}{ll} 11 & -17 \\ -8 & -4 \end{array}\right] \]