# Solve the system of equations (Use matrices.):x-2y+z = 16, 2x-y-z = 14, 3x+5y-4z =-10

Solve the system of equations (Use matrices.):
$x-2y+z=16$,
$2x-y-z=14$,
$3x+5y-4z=-10$

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Step 1
Given,
$x-2y+z=16,$
$2x-y-z=14,$
$3x+5y-4z=-10$
Step 2
Consider the Augmented matrix is of the form $AX=B$
$\left[\begin{array}{ccc}1& -2& 1\\ 2& -1& -1\\ 3& 5& -4\end{array}\right]\left[\begin{array}{c}x\\ y\\ z\end{array}\right]=\left[\begin{array}{c}16\\ 14\\ -10\end{array}\right]$
Here,

Use Gauss Elimination method,
Consider,
$\left[A/B\right]=\left[\begin{array}{cccc}1& -2& 1& 16\\ 2& -1& -1& 14\\ 3& 5& -4& -10\end{array}\right]$
${R}_{2}\to {R}_{2}-2{R}_{1}$
$\sim \left[\begin{array}{cccc}1& -2& 1& 16\\ 0& 3& -3& -18\\ 3& 5& -4& -10\end{array}\right]$
${R}_{3}\to {R}_{3}-3{R}_{1}$
$\sim \left[\begin{array}{cccc}1& -2& 1& 16\\ 0& 3& -3& -18\\ 0& 11& -7& -58\end{array}\right]$
${R}_{2}\to \frac{1}{3}{R}_{2}$
$\sim \left[\begin{array}{cccc}1& -2& 1& 16\\ 0& 1& -1& -6\\ 0& 11& -7& -58\end{array}\right]$
${R}_{3}\to {R}_{3}-11{R}_{2}$
$\sim \left[\begin{array}{cccc}1& -2& 1& 16\\ 0& 1& -1& -6\\ 0& 0& 4& 8\end{array}\right]$
${R}_{3}\to \frac{1}{4}{R}_{3}$
$\sim \left[\begin{array}{cccc}1& -2& 1& 16\\ 0& 1& -1& -6\\ 0& 0& 1& 2\end{array}\right]$
Step 3
The above matrix is in the row echelon form
By back substitution we get,
$z=2\dots \left(i\right)$
$y-z=-6\dots \left(ii\right)$
$x-2y+z=16\dots \left(iii\right)$
Substitute the value of z in (ii) we get,
$y-2=-6$
$y=-4$
Substitute the value of y and z in (iii) we get,
$x-2\left(-4\right)+2=16$
$x+8+2=16$
$x+10=16$
$x=16-10$
$x=6$
Therefore the solution set is $\left(x,y,z\right)=\left(6,-4,2\right)$

Jeffrey Jordon