Solve 3x + y = 13, x + y = 7 with a matrix Step by Step

Step by step

The question

Question: solve 3x + y = 13 and x + y = 7 with a matrix.

3x + y = 13①

We want the x and y that make both true.

x + y = 7②

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 and y, then the number on the right.

The bar stands for the = signs.

This is called an augmented matrix.

= [3, 1, 13; 1, 1, 7]

We call the rows R 1 and R 2.

2 Get a 1 in row 1

→ [1, 1, 7; 3, 1, 13]

3 Get 0s in column 1

→ [1, 1, 7; 0, −2, −8]

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, 7; 0, 1, 4]

5 Get 0s in column 2

→ [1, 0, 3; 0, 1, 4]

6 Read the answer

Each row is an equation again.

Row 1 is 1x + 0y = 3.

So x = 3.

x = 3

Row 2: y = 4.

y = 4

We write it as (3, 4).

(x, y) = (3, 4)

7 Check

Let's check in every equation.

We put x = 3, y = 4 into ①.

① 3(3) + 4 = 9 + 4 = 13

We get 13, the same as the right side.

Now ②.

② 3 + 4 = 7

We get 7, the same as the right side.

Both 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 in column 1.

Then: get a 1 in row 2.

Then: get 0s in column 2.

Then: read the answer.

Last: check.

Answer