Solve x + y + z = 3, x - y + 2z = 5, 2x + y - z = 1 by gaussian… Step by Step

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Question: solve x + y + z = 3, x − y + 2z = 5 and 2x + y − z = 1 by Gaussian elimination.

x + y + z = 3①

x − y + 2z = 5②

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

2x + y − z = 1③

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, 3; 1, −1, 2, 5; 2, 1, −1, 1]

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, 3; 0, −2, 1, 2; 0, −1, −3, −5]

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, 3; 0, 1, −1 over 2, −1; 0, −1, −3, −5]

4 Get 0s under the 1 in column 2

→ [1, 1, 1, 3; 0, 1, −1 over 2, −1; 0, 0, −7 over 2, −6]

5 Get a 1 in row 3

→ [1, 1, 1, 3; 0, 1, −1 over 2, −1; 0, 0, 1, 12 over 7]

6 Back substitution

Start at the bottom.

Each row is an equation again.

Row 3 is 1z = 12/7, with 0 for the other letters.

So z = 12/7.

z = 127

Row 2 is the equation y − z/2 = −1.

y − z2 = −1

We know z = 12/7.

We put it in.

y − 12 × 127 = −1

Work it out: y − 6/7 = −1.

y − 67 = −1

To remove the − 6/7, we add 6/7 on both sides.

y − 67 + 67 = −1 + 67

So y = −1/7.

y = −17

Row 1 is the equation x + y + z = 3.

x + y + z = 3

We know y = −1/7 and z = 12/7.

We put them in.

x + (−17) + 127 = 3

Work it out: x + 11/7 = 3.

x + 117 = 3

To remove the + 11/7, we take away 11/7 on both sides.

x + 117 − 117 = 3 − 117

So x = 10/7.

x = 107

We write it as (10/7, −1/7, 12/7).

(x, y, z) = (107, −17, 127)

7 Check

Let's check in every equation.

We put x = 10/7, y = −1/7, z = 12/7 into ①.

① (107) + (−17) + (127) = 107 − 17 + 127

= 3

We get 3, the same as the right side.

Now ②.

② (107) − (−17) + 2(127) = 107 + 17 + 247

= 5

We get 5, the same as the right side.

Now ③.

③ 2(107) + (−17) − (127) = 207 − 17 − 127

= 1

We get 1, 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: get a 1 in row 3.

Then: back substitution.

Last: check.

Answer