What is the inverse of A? \( \left[\begin{array}{ccc}1 & 3 & 1 \\ 2 & 7 & 4 \\ 1 & -1 & -5\end{array}\right] \)
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To find the inverse of matrix \( A \), we first need to check if it’s invertible by calculating its determinant. If the determinant is non-zero, we can proceed to find the inverse. For the matrix: \[ A = \begin{bmatrix} 1 & 3 & 1 \\ 2 & 7 & 4 \\ 1 & -1 & -5 \end{bmatrix} \] Calculating the determinant using the rule for \(3 \times 3\) matrices, we find that it equals \(A_{11}(A_{22}A_{33}-A_{23}A_{32}) - A_{12}(A_{21}A_{33}-A_{23}A_{31}) + A_{13}(A_{21}A_{32}-A_{22}A_{31})\). After carrying out those calculations, you should find that the determinant is indeed \( -8 \) (non-zero), meaning the inverse exists. The inverse can then be calculated using the adjugate method or through row reduction. Calculating the adjugate and dividing each entry by the determinant gives us: \[ A^{-1} = \frac{1}{-8} \begin{bmatrix} -29 & -5 & 8 \\ -8 & -1 & 6 \\ 7 & 1 & -3 \end{bmatrix} = \begin{bmatrix} \frac{29}{8} & \frac{5}{8} & -1 \\ 1 & \frac{1}{8} & -\frac{3}{4} \\ -\frac{7}{8} & -\frac{1}{8} & \frac{3}{8} \end{bmatrix} \] Now you have the inverse of matrix \( A \)!