How to solve Gaussian elimination method?

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

1 Answer
Apr 12, 2017

Answer:

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_3toR_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_2-2R_1toR_2#

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

#R_3-4R_1toR_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_2toR_1#

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

#R_3-2R_2toR_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_3/10to 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, and z = 1#