Write the matrix equation you would use to solve the following system. Then solve. \( \begin{array}{l}x-3 y+2 z=10 \\ \qquad \begin{aligned} 2 x+4 z=8\end{aligned} \\ -x+y+z=-6\end{array} \)

To write the matrix equation for the given system of equations, we can express the system in the form \(Ax = b\), where \(A\) is the coefficients matrix, \(x\) is the variables matrix, and \(b\) is the constants matrix. 1. The equations can be rearranged to fit the standard form: \[ \begin{aligned} 1x - 3y + 2z &= 10 \\ 2x + 0y + 4z &= 8 \\ -x + 1y + 1z &= -6 \end{aligned} \] 2. From the equations, we can write the coefficient matrix \(A\), variable matrix \(x\), and constant matrix \(b\): \[ A = \begin{bmatrix} 1 & -3 & 2 \\ 2 & 0 & 4 \\ -1 & 1 & 1 \end{bmatrix}, \quad x = \begin{bmatrix} x \\ y \\ z \end{bmatrix}, \quad b = \begin{bmatrix} 10 \\ 8 \\ -6 \end{bmatrix} \] Thus, the matrix equation is: \[ \begin{bmatrix} 1 & -3 & 2 \\ 2 & 0 & 4 \\ -1 & 1 & 1 \end{bmatrix} \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} 10 \\ 8 \\ -6 \end{bmatrix} \] To solve the system, we can use methods such as Gaussian elimination or matrix inversion. Here, I'll choose to use substitution or elimination, as it may be simpler for this case. Starting with the second equation: \[ 2x + 4z = 8 \implies x + 2z = 4 \implies x = 4 - 2z \] Substituting \(x\) in the first equation: \[ (4 - 2z) - 3y + 2z = 10 \implies 4 - 3y = 10 \implies -3y = 6 \implies y = -2 \] Next, substitute \(y\) into the expression for \(x\): \[ x = 4 - 2z \] Now, substituting \(x\) and \(y\) into the third equation: \[ -(4 - 2z) + (-2) + z = -6 \implies -4 + 2z - 2 + z = -6 \implies 3z = 0 \implies z = 0 \] Now, substituting \(z = 0\) back to find \(x\): \[ x = 4 - 2(0) = 4 \] Thus, we have: \[ x = 4, \quad y = -2, \quad z = 0 \] So the solution to the system of equations is: \[ \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} 4 \\ -2 \\ 0 \end{bmatrix} \]