Calculate: x=sqrt((-9)^2)

Expression: $\left\{\begin{array} { l } 2x+y-z+2w=-6 \\ 3x+4y+w=1 \\ x+5y+2z+6w=-3 \\ 5x+2y-z-w=3\end{array} \right.$

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

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

Evaluate the determinant

$\begin{array} { l }D=-71,\\D_1=\begin{vmatrix}\\-6 & 1 & -1 & 2\\1 & 4 & 0 & 1\\-3 & 5 & 2 & 6\\3 & 2 & -1 & -1\\\end{vmatrix},\\D_2=\begin{vmatrix}\\2 & -6 & -1 & 2\\3 & 1 & 0 & 1\\1 & -3 & 2 & 6\\5 & 3 & -1 & -1\\\end{vmatrix},\\D_3=\begin{vmatrix}\\2 & 1 & -6 & 2\\3 & 4 & 1 & 1\\1 & 5 & -3 & 6\\5 & 2 & 3 & -1\\\end{vmatrix},\\D_4=\begin{vmatrix}\\2 & 1 & -1 & -6\\3 & 4 & 0 & 1\\1 & 5 & 2 & -3\\5 & 2 & -1 & 3\\\end{vmatrix}\end{array}$

Evaluate the determinant

$\begin{array} { l }D=-71,\\D_1=-71,\\D_2=\begin{vmatrix}\\2 & -6 & -1 & 2\\3 & 1 & 0 & 1\\1 & -3 & 2 & 6\\5 & 3 & -1 & -1\\\end{vmatrix},\\D_3=\begin{vmatrix}\\2 & 1 & -6 & 2\\3 & 4 & 1 & 1\\1 & 5 & -3 & 6\\5 & 2 & 3 & -1\\\end{vmatrix},\\D_4=\begin{vmatrix}\\2 & 1 & -1 & -6\\3 & 4 & 0 & 1\\1 & 5 & 2 & -3\\5 & 2 & -1 & 3\\\end{vmatrix}\end{array}$

Evaluate the determinant

$\begin{array} { l }D=-71,\\D_1=-71,\\D_2=0,\\D_3=\begin{vmatrix}\\2 & 1 & -6 & 2\\3 & 4 & 1 & 1\\1 & 5 & -3 & 6\\5 & 2 & 3 & -1\\\end{vmatrix},\\D_4=\begin{vmatrix}\\2 & 1 & -1 & -6\\3 & 4 & 0 & 1\\1 & 5 & 2 & -3\\5 & 2 & -1 & 3\\\end{vmatrix}\end{array}$

Evaluate the determinant

$\begin{array} { l }D=-71,\\D_1=-71,\\D_2=0,\\D_3=-284,\\D_4=\begin{vmatrix}\\2 & 1 & -1 & -6\\3 & 4 & 0 & 1\\1 & 5 & 2 & -3\\5 & 2 & -1 & 3\\\end{vmatrix}\end{array}$

Evaluate the determinant

$\begin{array} { l }D=-71,\\D_1=-71,\\D_2=0,\\D_3=-284,\\D_4=142\end{array}$

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

$\begin{array} { l }x=1,\\y=0,\\z=4,\\w=-2\end{array}$

The possible solution of the system is the ordered quadruple $\left( x, y, z, w\right)$

$\left( x, y, z, w\right)=\left( 1, 0, 4, -2\right)$

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

$\left\{\begin{array} { l } 2 \times 1+0-4+2 \times \left( -2 \right)=-6 \\ 3 \times 1+4 \times 0+\left( -2 \right)=1 \\ 1+5 \times 0+2 \times 4+6 \times \left( -2 \right)=-3 \\ 5 \times 1+2 \times 0-4-\left( -2 \right)=3\end{array} \right.$

Simplify the equalities

$\left\{\begin{array} { l } -6=-6 \\ 1=1 \\ -3=-3 \\ 3=3\end{array} \right.$

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

$\left( x, y, z, w\right)=\left( 1, 0, 4, -2\right)$