Question 24

BONUS
Solve the system using Cramer's Rule.
\[
\begin{array}{l}
\left\{\begin{array}{l}
2 x-4 y-5 z=-37 \\
4 x-2 y-4 z=-38 \\
6 x-4 y+2 z=-36
\end{array}\right. \\
D=\square \\
D_{x}=\square \\
D_{y}=\square \\
\dot{D}_{z}=\square
\end{array}
\]

Solution:
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Step 1 ：Define the coefficient matrix A and the constant vector b as follows: \(A = \begin{bmatrix} 2 & -4 & -5 \ 4 & -2 & -4 \ 6 & -4 & 2 \end{bmatrix}\) and \(b = \begin{bmatrix} -37 \ -38 \ -36 \end{bmatrix}\)

Step 2 ：Calculate the determinant of the coefficient matrix A, denoted as D. The result is \(D = \boxed{108}\)

Step 3 ：Replace the first column of A with b to get a new matrix A_x, then calculate its determinant, denoted as D_x. The result is \(D_x = \boxed{-540}\)

Step 4 ：Replace the second column of A with b to get a new matrix A_y, then calculate its determinant, denoted as D_y. The result is \(D_y = \boxed{324}\)

Step 5 ：Replace the third column of A with b to get a new matrix A_z, then calculate its determinant, denoted as D_z. The result is \(D_z = \boxed{324}\)

Step 6 ：Calculate the solutions x, y, and z by dividing D_x, D_y, and D_z by D respectively. The results are \(x = \boxed{-5}\), \(y = \boxed{3}\), and \(z = \boxed{3}\)

Step 7 ：The solutions to the system of equations are \(x = -5\), \(y = 3\), and \(z = 3\). The determinants are \(D = 108\), \(D_x = -540\), \(D_y = 324\), and \(D_z = 324\)