Solve the system of equations by converting to a matrix equation and using the inverse of the coefficient matrix. \[ \begin{array}{l}9 x-5 y=-134 \\ -8 x+9 y=192\end{array} \] \[ \begin{array}{l}\text { Question Help: Video } \\ \text { Submit Question }\end{array} \]

To solve the system of equations using a matrix equation, we start by rewriting the equations in matrix form \(AX = B\), where: \[ A = \begin{pmatrix} 9 & -5 \\ -8 & 9 \end{pmatrix}, \quad X = \begin{pmatrix} x \\ y \end{pmatrix}, \quad B = \begin{pmatrix} -134 \\ 192 \end{pmatrix} \] Next, we calculate the inverse of the coefficient matrix \(A\). The inverse of a 2x2 matrix \( \begin{pmatrix} a & b \\ c & d \end{pmatrix} \) is given by \( \frac{1}{ad - bc} \begin{pmatrix} d & -b \\ -c & a \end{pmatrix} \). Here, \(a=9\), \(b=-5\), \(c=-8\), and \(d=9\). First, calculate the determinant: \[ \text{det}(A) = (9)(9) - (-5)(-8) = 81 - 40 = 41 \] Now we find the inverse: \[ A^{-1} = \frac{1}{41} \begin{pmatrix} 9 & 5 \\ 8 & 9 \end{pmatrix} = \begin{pmatrix} \frac{9}{41} & \frac{5}{41} \\ \frac{8}{41} & \frac{9}{41} \end{pmatrix} \] Now we can find \(X\) by multiplying \(A^{-1}\) by \(B\): \[ X = A^{-1}B = \begin{pmatrix} \frac{9}{41} & \frac{5}{41} \\ \frac{8}{41} & \frac{9}{41} \end{pmatrix} \begin{pmatrix} -134 \\ 192 \end{pmatrix} \] Calculating this gives: \[ x = \frac{9 \cdot -134 + 5 \cdot 192}{41} = \frac{-1206 + 960}{41} = \frac{-246}{41} = -6 \] \[ y = \frac{8 \cdot -134 + 9 \cdot 192}{41} = \frac{-1072 + 1728}{41} = \frac{656}{41} = 16 \] Thus, the solution to the system of equations is: \[ x = -6, \quad y = 16 \]