# How do you solve the system x+y-z=2, x-2z=1, and 2x-3y-z=8?

Dec 20, 2016

Write the coefficients of the equations into an augmented matrix and then perform Gauss-Jordan Elimination, using row operations.

#### Explanation:

Write the coefficients of the equations into an augmented matrix:

$x - 2 z = 1 \to \left[\left(1 , 0 , - 2 , | , 1\right)\right]$

$x + y - z = 2 \to \left[\begin{matrix}1 & 0 & - 2 & | & 1 \\ 1 & 1 & - 1 & | & 2\end{matrix}\right]$

$2 x - 3 y - z = 8 \to \left[\begin{matrix}1 & 0 & - 2 & | & 1 \\ 1 & 1 & - 1 & | & 2 \\ 2 & - 3 & - 1 & | & 8\end{matrix}\right]$

Perform Guass-Jordan Elimination using row operations on the augmented matrix:

[ (1, 0, -2,|,1), (1, 1, -1,|,2), (2, -3, -1,|,8) ]

${R}_{2} - {R}_{1} \to {R}_{2}$

[ (1, 0, -2,|,1), (0, 1, 1,|,1), (2, -3, -1,|,8) ]

${R}_{3} - 2 {R}_{1} \to {R}_{3}$

[ (1, 0, -2,|,1), (0, 1, 1,|,1), (0, -3, 3,|,6) ]

$3 {R}_{2} + {R}_{3} \to {R}_{3}$

[ (1, 0, -2,|,1), (0, 1, 1,|,1), (0, 0, 6,|,9) ]

${R}_{3} / 6$

[ (1, 0, -2,|,1), (0, 1, 1,|,1), (0, 0, 1,|,3/2) ]

${R}_{2} - {R}_{3} \to {R}_{2}$

[ (1, 0, -2,|,1), (0, 1, 0,|,-1/2), (0, 0, 1,|,3/2) ]

$2 {R}_{3} + {R}_{1} \to {R}_{1}$

[ (1, 0, 0,|,4), (0, 1, 0,|,-1/2), (0, 0, 1,|,3/2) ]

$x = 4 , y = - \frac{1}{2} , \mathmr{and} z = \frac{3}{2}$

check:

$4 - 2 \left(\frac{3}{2}\right) = 1$
$4 + \left(- \frac{1}{2}\right) - \frac{3}{2} = 2$
$2 \left(4\right) - 3 \left(- \frac{1}{2}\right) - \frac{3}{2} = 8$

$1 = 1$
$2 = 2$
$8 = 8$

This checks.