Use Cramer's Rule to solve the following system of equations. Write each answer in simplest form. \[ \begin{array}{r}-4 x+9 y-z=-8 \\ 2 x-3 y-6 z=4 \\ 3 x-5 y-9 z=6\end{array} \] Write the four relevant determinants in any order:

To use Cramer's Rule, we need to calculate the determinants of the coefficient matrix and the matrices formed by replacing each column with the constants from the right-hand side of the equations. The coefficient matrix \( A \) is: \[ A = \begin{bmatrix} -4 & 9 & -1 \\ 2 & -3 & -6 \\ 3 & -5 & -9 \end{bmatrix} \] ### 1. Calculate the determinant of \( A \) (denoted as \( D \)): \[ D = \begin{vmatrix} -4 & 9 & -1 \\ 2 & -3 & -6 \\ 3 & -5 & -9 \end{vmatrix} = -4 \begin{vmatrix} -3 & -6 \\ -5 & -9 \end{vmatrix} - 9 \begin{vmatrix} 2 & -6 \\ 3 & -9 \end{vmatrix} - 1 \begin{vmatrix} 2 & -3 \\ 3 & -5 \end{vmatrix} \] Calculating these smaller 2x2 determinants: - First: \((-3)(-9) - (-6)(-5) = 27 - 30 = -3\) - Second: \((2)(-9) - (-6)(3) = -18 + 18 = 0\) - Third: \((2)(-5) - (-3)(3) = -10 + 9 = -1\) Now substituting back to find \( D \): \[ D = -4(-3) - 9(0) - 1(-1) = 12 + 0 + 1 = 13 \] ### 2. Calculate the determinant \( D_x \): Replace the first column of \( A \) with the constants from the equations: \[ D_x = \begin{vmatrix} -8 & 9 & -1 \\ 4 & -3 & -6 \\ 6 & -5 & -9 \end{vmatrix} \] Calculating it: \[ D_x = -8 \begin{vmatrix} -3 & -6 \\ -5 & -9 \end{vmatrix} - 9 \begin{vmatrix} 4 & -6 \\ 6 & -9 \end{vmatrix} - 1 \begin{vmatrix} 4 & -3 \\ 6 & -5 \end{vmatrix} \] Calculating the smaller 2x2 determinants: - First: \((-3)(-9) - (-6)(-5) = 27 - 30 = -3\) - Second: \((4)(-9) - (-6)(6) = -36 + 36 = 0\) - Third: \((4)(-5) - (-3)(6) = -20 + 18 = -2\) Now substituting back to find \( D_x \): \[ D_x = -8(-3) - 9(0) - 1(-2) = 24 + 0 + 2 = 26 \] ### 3. Calculate the determinant \( D_y \): Replace the second column of \( A \): \[ D_y = \begin{vmatrix} -4 & -8 & -1 \\ 2 & 4 & -6 \\ 3 & 6 & -9 \end{vmatrix} \] Calculating it: \[ D_y = -4 \begin{vmatrix} 4 & -6 \\ 6 & -9 \end{vmatrix} - (-8) \begin{vmatrix} 2 & -6 \\ 3 & -9 \end{vmatrix} - 1 \begin{vmatrix} 2 & 4 \\ 3 & 6 \end{vmatrix} \] Calculating the smaller 2x2 determinants: - First: \((4)(-9) - (-6)(6) = -36 + 36 = 0\) - Second: \((2)(-9) - (-6)(3) = -18 + 18 = 0\) - Third: \((2)(6) - (4)(3) = 12 - 12 = 0\) Now substituting back to find \( D_y \): \[ D_y = -4(0) + 8(0) - 0 = 0 \] ### 4. Calculate the determinant \( D_z \): Replace the third column of \( A \): \[ D_z = \begin{vmatrix} -4 & 9 & -8 \\ 2