Solve the following system of equations using an inverse matrix. You must also indicate the inverse matrix, \( A^{-1} \), that was used to solve the system. You may optionally write the inverse matrix with a scalar coefficient. \[ \begin{array}{r}-2 x-5 y=-6 \\ -x+y=8\end{array} \] Answer Attempt out of 2

First, let's represent the system of equations in matrix form. We can express it as \( AX = B \), where: \[ A = \begin{bmatrix} -2 & -5 \\ -1 & 1 \end{bmatrix}, \quad X = \begin{bmatrix} x \\ y \end{bmatrix}, \quad B = \begin{bmatrix} -6 \\ 8 \end{bmatrix} \] Next, we need to find the inverse of matrix \( A \). The formula for the inverse of a 2x2 matrix \( \begin{bmatrix} a & b \\ c & d \end{bmatrix} \) is given by: \[ A^{-1} = \frac{1}{ad - bc} \begin{bmatrix} d & -b \\ -c & a \end{bmatrix} \] For our matrix \( A = \begin{bmatrix} -2 & -5 \\ -1 & 1 \end{bmatrix} \), we can calculate the determinant \( ad - bc \): \[ \text{det}(A) = (-2)(1) - (-5)(-1) = -2 - 5 = -7 \] Now we can calculate \( A^{-1} \): \[ A^{-1} = \frac{1}{-7} \begin{bmatrix} 1 & 5 \\ 1 & -2 \end{bmatrix} = \begin{bmatrix} -\frac{1}{7} & -\frac{5}{7} \\ -\frac{1}{7} & \frac{2}{7} \end{bmatrix} \] Now we can solve for \( X \): \[ X = A^{-1}B = \begin{bmatrix} -\frac{1}{7} & -\frac{5}{7} \\ -\frac{1}{7} & \frac{2}{7} \end{bmatrix} \begin{bmatrix} -6 \\ 8 \end{bmatrix} \] Calculating this product, we find: \[ X = \begin{bmatrix} -\frac{1}{7}(-6) + -\frac{5}{7}(8) \\ -\frac{1}{7}(-6) + \frac{2}{7}(8) \end{bmatrix} = \begin{bmatrix} \frac{6}{7} - \frac{40}{7} \\ \frac{6}{7} + \frac{16}{7} \end{bmatrix} = \begin{bmatrix} -\frac{34}{7} \\ \frac{22}{7} \end{bmatrix} \] Thus, the solution is: \[ x = -\frac{34}{7}, \quad y = \frac{22}{7} \] The inverse matrix used is: \[ A^{-1} = \begin{bmatrix} -\frac{1}{7} & -\frac{5}{7} \\ -\frac{1}{7} & \frac{2}{7} \end{bmatrix} \]