# How do you solve the system of equations: \begin{array}{ l }{ 3x + 2y + 4z = 11} \\ { 2x - y + 3z = 4} \\ { 5x - 3y + 5z = - 1} \end{array}?

Jan 2, 2017

Verify that the determinant is not zero; if so, then write an augmented matrix and perform elementary row operations, until you obtain an identity matrix on the left.

#### Explanation:

Verify that the Determinant is not zero:

$| \left(3 , 2 , 4\right) , \left(2 , - 1 , 3\right) , \left(5 , - 3 , 5\right) | = 18$

The system has a unique solution.

Write the Augmented Matrix

[ (3,2,4,|,11), (2,-1,3,|,4), (5,-3,5,|,-1) ]

Perform Elementary Row Operations until you obtain an Identity Matrix on the left:

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

[ (1,3,1,|,7), (2,-1,3,|,4), (5,-3,5,|,-1) ]

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

[ (1,3,1,|,7), (0,-7,1,|,-10), (5,-3,5,|,-1) ]

$R 3 - 5 {R}_{1} \to {R}_{3}$

[ (1,3,1,|,7), (0,-7,1,|,-10), (0,-18,0,|,-36) ]

${R}_{2} \leftrightarrow {R}_{3}$

[ (1,3,1,|,7), (0,-18,0,|,-36), (0,-7,1,|,-10) ]

$\left(- \frac{1}{18}\right) {R}_{2} \to {R}_{2}$

[ (1,3,1,|,7), (0,1,0,|,2), (0,-7,1,|,-10) ]

${R}_{3} + 7 {R}_{2} \to {R}_{3}$

[ (1,3,1,|,7), (0,1,0,|,2), (0,0,1,|,4) ]

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

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

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

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

We have an identity matrix on the left, therefore, we can read the solution set on the right:

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

Check:

$3 \left(- 3\right) + 2 \left(2\right) + 4 \left(4\right) = 11$
2(-3)−(2)+3(4)=4
5(-3)−3(2)+5(4)=−1

$11 = 11$
$4 = 4$
−1=−1

This checks

The ordered triple is $\left(- 3 , 2 , 4\right)$