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Expression: $\left\{\begin{array} { l } x_1-x_2+2x_3+x_4=0 \\ 2x_1-3x_2+4x_3=5 \\ x_1+x_2+x_3=2 \\ 4x_1-4x_2+8x_3+6x_4=0\end{array} \right.$

To solve the system using Cramer's rule, first list all needed determinants

$\begin{array} { l }D=\begin{vmatrix}\\1 & -1 & 2 & 1\\2 & -3 & 4 & 0\\1 & 1 & 1 & 0\\4 & -4 & 8 & 6\\\end{vmatrix},\\D_1=\begin{vmatrix}\\0 & -1 & 2 & 1\\5 & -3 & 4 & 0\\2 & 1 & 1 & 0\\0 & -4 & 8 & 6\\\end{vmatrix},\\D_2=\begin{vmatrix}\\1 & 0 & 2 & 1\\2 & 5 & 4 & 0\\1 & 2 & 1 & 0\\4 & 0 & 8 & 6\\\end{vmatrix},\\D_3=\begin{vmatrix}\\1 & -1 & 0 & 1\\2 & -3 & 5 & 0\\1 & 1 & 2 & 0\\4 & -4 & 0 & 6\\\end{vmatrix},\\D_4=\begin{vmatrix}\\1 & -1 & 2 & 0\\2 & -3 & 4 & 5\\1 & 1 & 1 & 2\\4 & -4 & 8 & 0\\\end{vmatrix}\end{array}$

Evaluate the determinant

$\begin{array} { l }D=2,\\D_1=\begin{vmatrix}\\0 & -1 & 2 & 1\\5 & -3 & 4 & 0\\2 & 1 & 1 & 0\\0 & -4 & 8 & 6\\\end{vmatrix},\\D_2=\begin{vmatrix}\\1 & 0 & 2 & 1\\2 & 5 & 4 & 0\\1 & 2 & 1 & 0\\4 & 0 & 8 & 6\\\end{vmatrix},\\D_3=\begin{vmatrix}\\1 & -1 & 0 & 1\\2 & -3 & 5 & 0\\1 & 1 & 2 & 0\\4 & -4 & 0 & 6\\\end{vmatrix},\\D_4=\begin{vmatrix}\\1 & -1 & 2 & 0\\2 & -3 & 4 & 5\\1 & 1 & 1 & 2\\4 & -4 & 8 & 0\\\end{vmatrix}\end{array}$

Evaluate the determinant

$\begin{array} { l }D=2,\\D_1=38,\\D_2=\begin{vmatrix}\\1 & 0 & 2 & 1\\2 & 5 & 4 & 0\\1 & 2 & 1 & 0\\4 & 0 & 8 & 6\\\end{vmatrix},\\D_3=\begin{vmatrix}\\1 & -1 & 0 & 1\\2 & -3 & 5 & 0\\1 & 1 & 2 & 0\\4 & -4 & 0 & 6\\\end{vmatrix},\\D_4=\begin{vmatrix}\\1 & -1 & 2 & 0\\2 & -3 & 4 & 5\\1 & 1 & 1 & 2\\4 & -4 & 8 & 0\\\end{vmatrix}\end{array}$

Evaluate the determinant

$\begin{array} { l }D=2,\\D_1=38,\\D_2=-10,\\D_3=\begin{vmatrix}\\1 & -1 & 0 & 1\\2 & -3 & 5 & 0\\1 & 1 & 2 & 0\\4 & -4 & 0 & 6\\\end{vmatrix},\\D_4=\begin{vmatrix}\\1 & -1 & 2 & 0\\2 & -3 & 4 & 5\\1 & 1 & 1 & 2\\4 & -4 & 8 & 0\\\end{vmatrix}\end{array}$

Evaluate the determinant

$\begin{array} { l }D=2,\\D_1=38,\\D_2=-10,\\D_3=-24,\\D_4=\begin{vmatrix}\\1 & -1 & 2 & 0\\2 & -3 & 4 & 5\\1 & 1 & 1 & 2\\4 & -4 & 8 & 0\\\end{vmatrix}\end{array}$

Evaluate the determinant

$\begin{array} { l }D=2,\\D_1=38,\\D_2=-10,\\D_3=-24,\\D_4=0\end{array}$

Since $D≠0$, Cramer's rule can be applied, so find $\begin{array} { l }x_1,& x_2,& x_3,& x_4\end{array}$ using the formulas $\begin{array} { l }x_1=\frac{ D_1 }{ D },& x_2=\frac{ D_2 }{ D },& x_3=\frac{ D_3 }{ D },& x_4=\frac{ D_4 }{ D }\end{array}$

$\begin{array} { l }x_1=19,\\x_2=-5,\\x_3=-12,\\x_4=0\end{array}$

The possible solution of the system is the ordered quadruple $\left( x_1, x_2, x_3, x_4\right)$

$\left( x_1, x_2, x_3, x_4\right)=\left( 19, -5, -12, 0\right)$

Check if the given ordered quadruple is the solution of the system of equations

$\left\{\begin{array} { l } 19-\left( -5 \right)+2 \times \left( -12 \right)+0=0 \\ 2 \times 19-3 \times \left( -5 \right)+4 \times \left( -12 \right)=5 \\ 19+\left( -5 \right)+\left( -12 \right)=2 \\ 4 \times 19-4 \times \left( -5 \right)+8 \times \left( -12 \right)+6 \times 0=0\end{array} \right.$

Simplify the equalities

$\left\{\begin{array} { l } 0=0 \\ 5=5 \\ 2=2 \\ 0=0\end{array} \right.$

Since all of the equalities are true, the ordered quadruple is the solution of the system

$\left( x_1, x_2, x_3, x_4\right)=\left( 19, -5, -12, 0\right)$

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