# How do you solve using gaussian elimination or gauss-jordan elimination, x +2y +3z = 1, 2x +5y +7z = 2, 3x +5y +7z = 4?

Apr 15, 2017

Write an Augmented Matrix .
Perform Elementary Row Operations, until an identity matrix is obtained.
The solutions will be on the right.
Check.

#### Explanation:

Write $x + 2 y + 3 z = 1$ as a row in an Augmented Matrix:

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

Add row for the equation $2 x + 5 y + 7 z = 2$:

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

Add row for the equation $3 x + 5 y + 7 z = 4$:

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

The augmented matrix is complete. Perform Elementary Row Operations.

We want the coefficient in position $\left(1 , 1\right)$ to be one and it is, therefore, no operation is required.

We want the other two coefficients is column 1 to be 0, therefore, we perform the following two row operations:

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

 [ (1, 2, 3,|,1), (0,1,1,|, 0), (3,5, 7,|, 4) ]

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

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

We want the coefficient in position $\left(2 , 2\right)$ to be 1 and it is, therefore, no operation is required.

We want the other two coefficients in column 2 to be 0, therefore, we perform the following two row operations:

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

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

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

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

We want the coefficient in position $\left(3 , 3\right)$ to be 1 and it is, therefore, we multiply the row by -1:

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

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

We want the other two coefficients in column 3 to be 0, therefore, we perform the following two row operations:

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

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

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

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

We have an identity matrix on the left, therefore, the solutions are on the right:

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

Check:

$2 + 2 \left(1\right) + 3 \left(- 1\right) = 1$
$2 \left(2\right) + 5 \left(1\right) + 7 \left(- 1\right) = 2$
$3 \left(2\right) + 5 \left(1\right) + 7 \left(- 1\right) = 4$

$1 = 1$
$2 = 2$
$4 = 4$

This checks.