Solve x - y - z = -2, 2x + y + z = 8, x + y + 2z = 9 by gaussian… Step by Step

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The question

Question: solve x − y − z = −2, 2x + y + z = 8 and x + y + 2z = 9 by Gaussian elimination.

x − y − z = −2①

2x + y + z = 8②

We want the x, y and z that make all three true.

x + y + 2z = 9③

We will use three moves on the rows.

Swap two rows.

Times a row by a number that is not 0.

Add a multiple of one row to another.

Each move keeps the equations true, like in elimination.

swap two rows

× a row by a number (not 0)

add a multiple of one row to another

1 Write the augmented matrix

Each row is one equation: the numbers in front of x, y and z, then the number on the right.

The bar stands for the = signs.

This is called an augmented matrix.

= [1, −1, −1, −2; 2, 1, 1, 8; 1, 1, 2, 9]

We call the rows R 1, R 2 and R 3.

Our goal: 1s down the diagonal, and 0s under them.

This shape is called row-echelon form.

Then we go back up, one letter at a time.

goal: 1s down the diagonal, 0s under them

2 Get 0s under the 1 in column 1

→ [1, −1, −1, −2; 0, 3, 3, 12; 0, 2, 3, 11]

Careful!

A row move works on the whole row, the number after the bar too.

That number is the right side of the equation.

Careful: the whole row, after the bar too

3 Get a 1 in row 2

→ [1, −1, −1, −2; 0, 1, 1, 4; 0, 2, 3, 11]

4 Get 0s under the 1 in column 2

→ [1, −1, −1, −2; 0, 1, 1, 4; 0, 0, 1, 3]

5 Back substitution

Start at the bottom.

Each row is an equation again.

Row 3 is 1z = 3, with 0 for the other letters.

So z = 3.

z = 3

Row 2 is the equation y + z = 4.

y + z = 4

We know z = 3.

We put it in.

y + 3 = 4

Work it out: y + 3 = 4.

y + 3 = 4

To remove the + 3, we take away 3 on both sides.

y + 3 − 3 = 4 − 3

So y = 1.

y = 1

Row 1 is the equation x − y − z = −2.

x − y − z = −2

We know y = 1 and z = 3.

We put them in.

x − 1 − 3 = −2

Work it out: x − 4 = −2.

x − 4 = −2

To remove the − 4, we add 4 on both sides.

x − 4 + 4 = −2 + 4

So x = 2.

x = 2

We write it as (2, 1, 3).

(x, y, z) = (2, 1, 3)

6 Check

Let's check in every equation.

We put x = 2, y = 1, z = 3 into ①.

① 2 − 1 − 3 = −2

We get −2, the same as the right side.

Now ②.

② 2(2) + 1 + 3 = 4 + 1 + 3 = 8

We get 8, the same as the right side.

Now ③.

③ 2 + 1 + 2(3) = 2 + 1 + 6 = 9

We get 9, the same as the right side.

All of them work.

So the answer is right!

Let's look back.

First: write the augmented matrix.

Then: get 0s under the 1 in column 1.

Then: get a 1 in row 2.

Then: get 0s under the 1 in column 2.

Then: back substitution.

Last: check.

Answer