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. \[ \begin{array}{l}-2 x-4 y=1 \\ -6 x-9 y=-8\end{array} \] Answer Attempt sout of 2 \( A^{-1}= \)

To solve the system of equations using an inverse matrix, first we need to convert the equations into matrix form, \( AX = B \). The system is given as: \[ \begin{array}{l} -2x - 4y = 1 \\ -6x - 9y = -8 \end{array} \] This can be expressed in matrix form as: \[ \begin{pmatrix} -2 & -4 \\ -6 & -9 \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} 1 \\ -8 \end{pmatrix} \] Now, we define the matrices: \[ A = \begin{pmatrix} -2 & -4 \\ -6 & -9 \end{pmatrix}, \quad X = \begin{pmatrix} x \\ y \end{pmatrix}, \quad B = \begin{pmatrix} 1 \\ -8 \end{pmatrix} \] Next, we calculate the inverse of matrix \( A \) using the formula for the inverse of a 2x2 matrix: \[ A^{-1} = \frac{1}{det(A)} \begin{pmatrix} d & -b \\ -c & a \end{pmatrix} \] where \( A = \begin{pmatrix} a & b \\ c & d \end{pmatrix} \). Calculating the determinant: \[ det(A) = (-2)(-9) - (-4)(-6) = 18 - 24 = -6 \] Thus, the inverse \( A^{-1} \) is: \[ A^{-1} = \frac{1}{-6} \begin{pmatrix} -9 & 4 \\ 6 & -2 \end{pmatrix} = \begin{pmatrix} \frac{3}{2} & -\frac{2}{3} \\ -1 & \frac{1}{3} \end{pmatrix} \] Now, we use \( A^{-1} \) to find \( X \): \[ X = A^{-1}B = \begin{pmatrix} \frac{3}{2} & -\frac{2}{3} \\ -1 & \frac{1}{3} \end{pmatrix} \begin{pmatrix} 1 \\ -8 \end{pmatrix} \] Calculating the product: \[ X = \begin{pmatrix} \frac{3}{2} \times 1 + (-\frac{2}{3}) \times (-8) \\ -1 \times 1 + \frac{1}{3} \times (-8) \end{pmatrix} = \begin{pmatrix} \frac{3}{2} + \frac{16}{3} \\ -1 - \frac{8}{3} \end{pmatrix} \] Finding a common denominator for the first component: \(\frac{3}{2} = \frac{9}{6}\) and \(\frac{16}{3} = \frac{32}{6}\) \[ \text{First component: } \frac{9}{6} + \frac{32}{6} = \frac{41}{6} \] Finding the second component: \(-1 = -\frac{3}{3}\) \[ \text{Second component: } -\frac{3}{3} - \frac{8}{3} = -\frac{11}{3} \] So, the solution vector \( X \) is: \[ X = \begin{pmatrix} \frac{41}{6} \\ -\frac{11}{3} \end{pmatrix} \] Thus, the final answer is: \[ A^{-1} = \begin{pmatrix} \frac{3}{2} & -\frac{2}{3} \\ -1 & \frac{1}{3} \end{pmatrix} \]