Solve 2x + y - z = 2, x + y + z = 6, x - y + z = 4 by gaussian… Step by Step

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The question

Question: solve 2x + y − z = 2, x + y + z = 6 and x − y + z = 4 by Gaussian elimination.

2x + y − z = 2①

x + y + z = 6②

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

x − y + z = 4③

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.

= [2, 1, −1, 2; 1, 1, 1, 6; 1, −1, 1, 4]

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 a 1 in row 1

→ [1, 1, 1, 6; 2, 1, −1, 2; 1, −1, 1, 4]

3 Get 0s under the 1 in column 1

→ [1, 1, 1, 6; 0, −1, −3, −10; 0, −2, 0, −2]

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

4 Get a 1 in row 2

→ [1, 1, 1, 6; 0, 1, 3, 10; 0, −2, 0, −2]

5 Get 0s under the 1 in column 2

→ [1, 1, 1, 6; 0, 1, 3, 10; 0, 0, 6, 18]

6 Get a 1 in row 3

→ [1, 1, 1, 6; 0, 1, 3, 10; 0, 0, 1, 3]

7 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 + 3z = 10.

y + 3z = 10

We know z = 3.

We put it in.

y + 3 × 3 = 10

Work it out: y + 9 = 10.

y + 9 = 10

To remove the + 9, we take away 9 on both sides.

y + 9 − 9 = 10 − 9

So y = 1.

y = 1

Row 1 is the equation x + y + z = 6.

x + y + z = 6

We know y = 1 and z = 3.

We put them in.

x + 1 + 3 = 6

Work it out: x + 4 = 6.

x + 4 = 6

To remove the + 4, we take away 4 on both sides.

x + 4 − 4 = 6 − 4

So x = 2.

x = 2

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

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

8 Check

Let's check in every equation.

We put x = 2, y = 1, z = 3 into ①.

① 2(2) + 1 − 3 = 4 + 1 − 3 = 2

We get 2, the same as the right side.

Now ②.

② 2 + 1 + 3 = 6

We get 6, the same as the right side.

Now ③.

③ 2 − 1 + 3 = 4

We get 4, 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 a 1 in row 1.

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: get a 1 in row 3.

Then: back substitution.

Last: check.

Answer