# How do you solve equations with matrices?

Ex:2x+7y=9

5x-6y=-1. Ik the answer is obvious. But how do you solve it? Please show step by step. Remember, matrices!!!!! Thank you so much to all that answer!

### 5 Answers

- Anonymous8 years agoFavorite Answer
Too long to explain. Look up Cramer's Rule. It is also in your textbook.

- 8 years ago
first turn it into the following matrix:

[2 7 | 9]

[5 -6 | -1]

now you do row operations until the left side is the identity matrix, which is

[1 0]

[0 1]

row operations are defined as:

multiplying a row by a non-zero scalar

adding some multiple of one row to another

once you have gotten the left side to be the identity matrix, the right part of the matrix gives you the values for x and y

R1 is the first row, R2 is the second row.

[2 7 | 9]

[5 -6 | -1]

first multiply R1 by 1/2

[1 7/2 | 9/2]

[5 -6 | -1]

now add -5*R1 to R2

[1 7/2 | 9/2]

[0 -6-35/2 | -1-45/2]

simplify

[1 7/2 | 9/2]

[0 -47/2 | -47/2]

multiply R2 by -2/47

[1 7/2 | 9/2]

[0 1 | 1]

add -7/2 * R2 to R1

[1 0 | 1]

[0 1 | 1]

now the left is the identity matrix, so we have:

[x] = [1]

[y] = [1]

We can check by plugging it into the sytem we started with:

2x+7y=9

5x-6y=-1

2 + 7 = 9

5 - 6 = -1

it checks out.

- 8 years ago
Make the augmented matrix:

[2 7 | 9]

[5 -6 | -1]

You want the part before the pipes (|s) to be the identity matrix:

[1 0]

[0 1]

Divide the top row by 2 to get the first 1

[1 3.5 | 4.5]

[5 -6 | -1 ]

Multiply the first row by -5, and add to the second row, and store this result in the second row. This gets the bottom left 0

[1 3.5__| __4.5]

[0 -23.5 | -23.5]

Divide the bottom row by -23.5 to get the other 1

[1 3.5 | 4.5]

[0 1__|__1]

Multiply the second row by -3.5, add to the first row, and store the result in the first row. This completes the identity matrix.

[1 0 | 1]

[0 1 | 1]

The answer is:

x=1

y=1

This is one way to solve any system of equations, no matter how many terms it has. The numbers don't always work out so well, but it always works.

- Divide By ZeroLv 78 years ago
For that system I know of three ways. So far everyone has mentioned using row-reduced echelon form. One person mentioned Cramer's rule.

The third, unmentioned way is to form two matrices:

[2, 7]

[5 -6]

and:

[9

-1]

Then multiply the inverse of the first matrix by the second matrix. (This works when the left side of the system forms an invertible matrix with the correct dimensions to be multiplied by the right-hand column. So a square matrix.)

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- 8 years ago
Answering this question would take half of a semester of linear algebra, but I'll give you a start. First, you set up the augmented matrix that will appear as follows:

[2 7 | 9]

[5 -6 | -1]

From there, you row reduce until you get your answer in the following form:

[1 0 | 1]

[0 1 | 1]

Look up row reduction and matrix manipulation and you should get the idea.