Question 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. \[ \begin{array}{r}-x-6 y=8 \\ -3 x-10 y=6\end{array} \]

To solve the system of equations using an inverse matrix, we first need to express the system in matrix form. The given equations are: 1. \(-x - 6y = 8\) 2. \(-3x - 10y = 6\) We can rewrite this system in the form \(AX = B\), where: \[ A = \begin{pmatrix} -1 & -6 \\ -3 & -10 \end{pmatrix}, \quad X = \begin{pmatrix} x \\ y \end{pmatrix}, \quad B = \begin{pmatrix} 8 \\ 6 \end{pmatrix} \] Next, we need to find the inverse of matrix \(A\), denoted as \(A^{-1}\). The formula for the inverse of a 2x2 matrix \[ \begin{pmatrix} a & b \\ c & d \end{pmatrix} \] is given by \[ \frac{1}{ad - bc} \begin{pmatrix} d & -b \\ -c & a \end{pmatrix} \] Now, let's calculate the determinant \(ad - bc\) for our matrix \(A\): - \(a = -1\), \(b = -6\), \(c = -3\), \(d = -10\) - \(ad = (-1)(-10) = 10\) - \(bc = (-6)(-3) = 18\) - Therefore, \(ad - bc = 10 - 18 = -8\) Now we can find the inverse matrix \(A^{-1}\): \[ A^{-1} = \frac{1}{-8} \begin{pmatrix} -10 & 6 \\ 3 & -1 \end{pmatrix} = \begin{pmatrix} \frac{10}{8} & -\frac{6}{8} \\ -\frac{3}{8} & \frac{1}{8} \end{pmatrix} = \begin{pmatrix} \frac{5}{4} & -\frac{3}{4} \\ -\frac{3}{8} & \frac{1}{8} \end{pmatrix} \] Now, we can use the inverse matrix to solve for \(X\): \[ X = A^{-1}B \] Let's calculate \(X\): \[ X = \begin{pmatrix} \frac{5}{4} & -\frac{3}{4} \\ -\frac{3}{8} & \frac{1}{8} \end{pmatrix} \begin{pmatrix} 8 \\ 6 \end{pmatrix} \] Now, we will perform the matrix multiplication: 1. For \(x\): \[ x = \frac{5}{4} \cdot 8 + \left(-\frac{3}{4}\right) \cdot 6 = 10 - 4.5 = 5.5 \] 2. For \(y\): \[ y = -\frac{3}{8} \cdot 8 + \frac{1}{8} \cdot 6 = -3 + 0.75 = -2.25 \] Thus, the solution to the system of equations is: \[ x = 5.5, \quad y = -2.25 \] In summary, the inverse matrix used is: \[ A^{-1} = \begin{pmatrix} \frac{5}{4} & -\frac{3}{4} \\ -\frac{3}{8} & \frac{1}{8} \end{pmatrix} \] And the solution to the system is: \[ \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} 5.5 \\ -2.25 \end{pmatrix} \]