Solve 2x + y + z = 7, x + y + z = 6, x - y + 2z = 5 by gaussian… Step by Step

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The question

Question: solve 2x + y + z = 7, x + y + z = 6 and x − y + 2z = 5 by Gaussian elimination.

2x + y + z = 7①

x + y + z = 6②

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

x − y + 2z = 5③

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, 7; 1, 1, 1, 6; 1, −1, 2, 5]

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, 7; 1, −1, 2, 5]

3 Get 0s under the 1 in column 1

→ [1, 1, 1, 6; 0, −1, −1, −5; 0, −2, 1, −1]

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, 1, 5; 0, −2, 1, −1]

5 Get 0s under the 1 in column 2

→ [1, 1, 1, 6; 0, 1, 1, 5; 0, 0, 3, 9]

6 Get a 1 in row 3

→ [1, 1, 1, 6; 0, 1, 1, 5; 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 + z = 5.

y + z = 5

We know z = 3.

We put it in.

y + 3 = 5

Work it out: y + 3 = 5.

y + 3 = 5

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

y + 3 − 3 = 5 − 3

So y = 2.

y = 2

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

x + y + z = 6

We know y = 2 and z = 3.

We put them in.

x + 2 + 3 = 6

Work it out: x + 5 = 6.

x + 5 = 6

To remove the + 5, we take away 5 on both sides.

x + 5 − 5 = 6 − 5

So x = 1.

x = 1

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

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

8 Check

Let's check in every equation.

We put x = 1, y = 2, z = 3 into ①.

① 2(1) + 2 + 3 = 2 + 2 + 3 = 7

We get 7, the same as the right side.

Now ②.

② 1 + 2 + 3 = 6

We get 6, the same as the right side.

Now ③.

③ 1 − 2 + 2(3) = 1 − 2 + 6 = 5

We get 5, 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