# How do you solve 3x + 2y + z = 8, 2x -3y + 2z = -16, and x + 4y -z = 20 using matrices?

$x = 1$ and $y = 4$ and $z = - 3$

#### Explanation:

from the given
$3 x + 2 y + z = 8$
$2 x - 3 y + 2 z = - 16$
$x + 4 y - z = 20$

For the matrix

$D = \left(\begin{matrix}{a}_{11} & {a}_{12} & {a}_{13} \\ {a}_{21} & {a}_{22} & {a}_{23} \\ {a}_{31} & {a}_{32} & {a}_{33}\end{matrix}\right) = \left(\begin{matrix}3 & 2 & 1 \\ 2 & - 3 & 2 \\ 1 & 4 & - 1\end{matrix}\right)$

$C = \left(\begin{matrix}{c}_{1} \\ {c}_{2} \\ {c}_{3}\end{matrix}\right) = \left(\begin{matrix}8 \\ - 16 \\ 20\end{matrix}\right)$

We will use the formula

D=(a_11* a_22*a_33+a_12* a_23*a_31+a_13* a_21*a_32
-a_31* a_22*a_13-a_32* a_23*a_11-a_33* a_21*a_12)

For x:

$x = \frac{\left(\begin{matrix}{c}_{1} & {a}_{12} & {a}_{13} \\ {c}_{2} & {a}_{22} & {a}_{23} \\ {c}_{3} & {a}_{32} & {a}_{33}\end{matrix}\right)}{D}$

For y:

$y = \frac{\left(\begin{matrix}{a}_{11} & {c}_{1} & {a}_{13} \\ {a}_{21} & {c}_{2} & {a}_{23} \\ {a}_{31} & {c}_{3} & {a}_{33}\end{matrix}\right)}{D}$

For z:

$z = \frac{\left(\begin{matrix}{a}_{11} & {a}_{12} & {c}_{1} \\ {a}_{21} & {a}_{22} & {c}_{2} \\ {a}_{31} & {a}_{32} & {c}_{3}\end{matrix}\right)}{D}$

We form the matrix

$D = \left(\begin{matrix}3 & 2 & 1 \\ 2 & - 3 & 2 \\ 1 & 4 & - 1\end{matrix}\right) = 4$

For x:

$x = \frac{\left(\begin{matrix}{c}_{1} & {a}_{12} & {a}_{13} \\ {c}_{2} & {a}_{22} & {a}_{23} \\ {c}_{3} & {a}_{32} & {a}_{33}\end{matrix}\right)}{D} = \frac{\left(\begin{matrix}8 & 2 & 1 \\ - 16 & - 3 & 2 \\ 20 & 4 & - 1\end{matrix}\right)}{4} = 1$

For y:

$y = \frac{\left(\begin{matrix}{a}_{11} & {c}_{1} & {a}_{13} \\ {a}_{21} & {c}_{2} & {a}_{23} \\ {a}_{31} & {c}_{3} & {a}_{33}\end{matrix}\right)}{D} = \frac{\left(\begin{matrix}3 & 8 & 1 \\ 2 & - 16 & 2 \\ 1 & 20 & - 1\end{matrix}\right)}{4} = 4$

For z:

$z = \frac{\left(\begin{matrix}{a}_{11} & {a}_{12} & {c}_{1} \\ {a}_{21} & {a}_{22} & {c}_{2} \\ {a}_{31} & {a}_{32} & {c}_{3}\end{matrix}\right)}{D} = \frac{\left(\begin{matrix}3 & 2 & 8 \\ 2 & - 3 & - 16 \\ 1 & 4 & 20\end{matrix}\right)}{4} = - 3$

solution set $x = 1$ and $y = 4$ and $z = - 3$

God bless...I hope the explanation is useful.