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, we first need to express the system in matrix form. The given equations are: \[ \begin{align*} -2x - 4y &= 1 \quad \text{(1)} \\ -6x - 9y &= -8 \quad \text{(2)} \end{align*} \] We can rewrite this system in the form \( A\mathbf{x} = \mathbf{b} \), where: \[ A = \begin{pmatrix} -2 & -4 \\ -6 & -9 \end{pmatrix}, \quad \mathbf{x} = \begin{pmatrix} x \\ y \end{pmatrix}, \quad \mathbf{b} = \begin{pmatrix} 1 \\ -8 \end{pmatrix} \] Next, we need to find the inverse of matrix \( A \), denoted as \( A^{-1} \). The formula for the inverse of a 2x2 matrix \( A = \begin{pmatrix} a & b \\ c & d \end{pmatrix} \) is given by: \[ A^{-1} = \frac{1}{ad - bc} \begin{pmatrix} d & -b \\ -c & a \end{pmatrix} \] Calculating the determinant \( ad - bc \): \[ a = -2, \quad b = -4, \quad c = -6, \quad d = -9 \] \[ \text{det}(A) = (-2)(-9) - (-4)(-6) = 18 - 24 = -6 \] Now we can find the inverse matrix \( A^{-1} \): \[ A^{-1} = \frac{1}{-6} \begin{pmatrix} -9 & 4 \\ 6 & -2 \end{pmatrix} = \begin{pmatrix} \frac{3}{2} & -\frac{2}{3} \\ -1 & \frac{1}{3} \end{pmatrix} \] Now, we can use the inverse matrix to solve for \( \mathbf{x} \): \[ \mathbf{x} = A^{-1} \mathbf{b} \] Calculating \( \mathbf{x} \): \[ \mathbf{x} = \begin{pmatrix} \frac{3}{2} & -\frac{2}{3} \\ -1 & \frac{1}{3} \end{pmatrix} \begin{pmatrix} 1 \\ -8 \end{pmatrix} \] Calculating the product: \[ \begin{pmatrix} \frac{3}{2} \cdot 1 + (-\frac{2}{3}) \cdot (-8) \\ -1 \cdot 1 + \frac{1}{3} \cdot (-8) \end{pmatrix} = \begin{pmatrix} \frac{3}{2} + \frac{16}{3} \\ -1 - \frac{8}{3} \end{pmatrix} \] Finding a common denominator for the first component: \[ \frac{3}{2} = \frac{9}{6}, \quad \frac{16}{3} = \frac{32}{6} \quad \Rightarrow \quad \frac{9}{6} + \frac{32}{6} = \frac{41}{6} \] For the second component: \[ -1 = -\frac{3}{3} \quad \Rightarrow \quad -\frac{3}{3} - \frac{8}{3} = -\frac{11}{3} \] Thus, we have: \[ \mathbf{x} = \begin{pmatrix} \frac{41}{6} \\ -\frac{11}{3} \end{pmatrix} \] Finally, the solution to the system of equations is: \[ x = \frac{41}{6}, \quad y = -\frac{11}{3} \] The inverse matrix used is: \[ A^{-1} = \begin{pmatrix} \frac{3}{2} & -\frac{2}{3} \\ -1 & \frac{1}{3} \end{pmatrix} \]