\( \left.\begin{array}{l}4 \\ 6 \\ 6\end{array}\right] \begin{array}{c}\text { What is the inverse of } \\ A \text { ? } \\ 27 \\ 2\end{array} \quad A=\left[\begin{array}{ccc}1 & 3 & 1 \\ 2 & 7 & 4 \\ 1 & -1 & -5\end{array}\right] \)

To find the inverse of matrix \( A \): \[ A = \begin{bmatrix} 1 & 3 & 1 \\ 2 & 7 & 4 \\ 1 & -1 & -5 \end{bmatrix} \] we can use the formula: \[ A^{-1} = \frac{1}{\det(A)} \cdot \text{adj}(A) \] ### Step 1: Compute the Determinant of \( A \) \[ \det(A) = 1 \cdot (7 \cdot (-5) - 4 \cdot (-1)) - 3 \cdot (2 \cdot (-5) - 4 \cdot 1) + 1 \cdot (2 \cdot (-1) - 7 \cdot 1) \] \[ \det(A) = 1 \cdot (-35 + 4) - 3 \cdot (-10 - 4) + 1 \cdot (-2 - 7) \] \[ \det(A) = -31 + 42 - 9 = 2 \] ### Step 2: Compute the Cofactor Matrix Calculate the minors and apply the checkerboard of signs to get the cofactors: \[ \text{Cof}(A) = \begin{bmatrix} -31 & 14 & -9 \\ 14 & -6 & 4 \\ 5 & -2 & 1 \end{bmatrix} \] ### Step 3: Compute the Adjugate Matrix The adjugate is the transpose of the cofactor matrix: \[ \text{adj}(A) = \begin{bmatrix} -31 & 14 & 5 \\ 14 & -6 & -2 \\ -9 & 4 & 1 \end{bmatrix} \] ### Step 4: Compute the Inverse \[ A^{-1} = \frac{1}{2} \cdot \begin{bmatrix} -31 & 14 & 5 \\ 14 & -6 & -2 \\ -9 & 4 & 1 \end{bmatrix} \] Simplifying, we get: \[ A^{-1} = \begin{bmatrix} -\frac{31}{2} & 7 & \frac{5}{2} \\ 7 & -3 & -1 \\ -\frac{9}{2} & 2 & \frac{1}{2} \end{bmatrix} \] ### Final Answer \[ A^{-1} = \begin{bmatrix} -\dfrac{31}{2} & 7 & \dfrac{5}{2} \\ 7 & -3 & -1 \\ -\dfrac{9}{2} & 2 & \dfrac{1}{2} \end{bmatrix} \]