# How to solve Gaussian elimination method?

## [( 5, -3, 2,|, 13), (2, -1, -3,|,1 ), (4, -2, 4,|, 12) ]

Apr 12, 2017

Perform elementary row operations, until the matrix on the left is an identity matrix, then the column vector on the right will contain the solution set.

#### Explanation:

Given:

[( 5, -3, 2,|, 13), (2, -1, -3,|,1 ), (4, -2, 4,|, 12) ]

Perform elementary row operations.

We want the coefficient ${A}_{1 , 1}$ to be 1 so we perform the following operation:

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

[( 1, -1, -2,|, 1), (2, -1, -3,|,1 ), (4, -2, 4,|, 12) ]

We want the remainder of the first column to be zeros so we perform the next 2 row operations:

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

[( 1, -1, -2,|, 1), (0, 1, 1,|,-1 ), (4, -2, 4,|, 12) ]

${R}_{3} - 4 {R}_{1} \to {R}_{3}$

[( 1, -1, -2,|, 1), (0, 1, 1,|,-1 ), (0, 2, 12,|, 8) ]

We want the coefficient ${A}_{2 , 2}$ to be 1 and it is, therefore, no operation is done. We want the other coefficients in column 2 to be 0 so we do the following 2 operations:

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

[( 1, 0, -1,|, 0), (0, 1, 1,|,-1 ), (0, 2, 12,|, 8) ]

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

[( 1, 0, -1,|, 0), (0, 1, 1,|,-1 ), (0, 0, 10,|,10) ]

We want the coefficient ${A}_{3 , 3}$ to be 1, therefore, we merely divide that row by 10:

${R}_{3} / 10 \to {R}_{3}$

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

We want the rest of the column to be zeros so we do the following 2 row operations:

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

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

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

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

There is an identity matrix on the left, therefore, the solution set is on the right, $x = 1 , y = - 2 , \mathmr{and} z = 1$