# How do you solve the system -3a-b-3c=-8, -5a+3b+6c=-4, and -6a-4b+c=-20?

Jan 5, 2017

Verify that the determinant of the matrix of coefficients is not zero. If so, then append the constants to the matrix to form an augmented matrix and perform elementary row operations to solve it.

#### Explanation:

The matrix of coefficients is:

[ (-3,-1,-3), (-5,3,6), (-6,-4,1) ]

The determinant is:

| (-3,-1,-3), (-5,3,6), (-6,-4,1) | = -164

The determinant is not zero, therefore, the system has a unique solution.

Append the constants to the matrix of coefficients, to create an augmented matrix :

[ (-3,-1,-3,|,-8), (-5,3,6,|,-4), (-6,-4,1,|,-20) ]

Peform elementary row operations on the augmented matrix, until you obtain an identity matrix on the left:

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

[ (-3,-1,-3,|,-8), (1,7,7,|,16), (-6,-4,1,|,-20) ]

${R}_{1} \leftrightarrow {R}_{2}$

[ (1,7,7,|,16), (-3,-1,-3,|,-8), (-6,-4,1,|,-20) ]

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

[ (1,7,7,|,16), (-3,-1,-3,|,-8), (0,-2,7,|,-4) ]

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

[ (1,7,7,|,16), (0,20,18,|,40), (0,-2,7,|,-4) ]

${R}_{2} + 10 {R}_{3} \to {R}_{3}$

[ (1,7,7,|,16), (0,20,18,|,40), (0,0,88,|,0) ]

$\left(\frac{1}{88}\right) {R}_{3} \to {R}_{3}$

[ (1,7,7,|,16), (0,20,18,|,40), (0,0,1,|,0) ]

${R}_{2} - 18 {R}_{2} \to {R}_{2}$

[ (1,7,7,|,16), (0,20,0,|,40), (0,0,1,|,0) ]

${R}_{1} - 7 {R}_{3} \to {R}_{1}$

[ (1,7,0,|,16), (0,20,0,|,40), (0,0,1,|,0) ]

$\left(\frac{1}{20}\right) {R}_{2} \to {R}_{2}$

[ (1,7,0,|,16), (0,1,0,|,2), (0,0,1,|,0) ]

${R}_{1} - 7 {R}_{2} \to {R}_{1}$

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

We have an identity matrix on the left, therefore, we can read the values of the variables on the right:

$a = 2 , b = 2 , \mathmr{and} c = 0$

Check:

$- 3 a - 1 b - 3 c = - 8$
$- 5 a + 3 b + 6 c = - 4$
$- 6 a - 4 b + 1 c = - 20$

$- 3 \left(2\right) - 1 \left(2\right) - 3 \left(0\right) = - 8$
$- 5 \left(2\right) + 3 \left(2\right) + 6 \left(0\right) = - 4$
$- 6 \left(2\right) - 4 \left(2\right) + 1 \left(0\right) = - 20$

$- 8 = - 8$
$- 4 = - 4$
$- 20 = - 20$

This checks.

The solution is: $a = 2 , b = 2 , \mathmr{and} c = 0$