Using Matrices to Solve Systems of

Equations

Matrix Equations We have solved systems using

graphing, but now we learn how to do it using matrices. This will be particularly useful when we have equations with three variables.

Matrix Equation Before you start, make sure

that both of your equations are in standard form and the variables are in the same order (alphabetical usually is best).

Setting up the Matrix Equation

Given a system of equations-2x - 6y = 0

3x + 11y = 4 Since there are 2 equations, there

will be 2 rows. Since there are 2 variables, there

will be 2 columns.

There are 3 parts to a matrix equation

1)The coefficient matrix,

2)the variable matrix, and

3)the constant matrix.

Setting up the Matrix Equation

-2x - 6y = 03x + 11y = 4

The coefficients are placed into the coefficient matrix.

2 6

3 11

-2x - 6y = 03x + 11y = 4

Your variable matrix will consist of a column.

x

y

-2x - 6y = 03x + 11y = 4

The matrices are multiplied and represent the left side of our matrix equation.

x

y

2 6

3 11

-2x - 6y = 03x + 11y = 4

The right side consists of our constants. Two equations = two rows.

0

4

-2x - 6y = 03x + 11y = 4

Now put them together.

2 6

3 11

x

y

0

4

We’ll solve it later!

Create a matrix equation 3x - 2y = 7

y + 4x = 8 Put them in Standard Form. Write your equation.

3 2

4 1

x

y

7

8

3a - 5b + 2c = 9

4a + 7b + c = 3

2a - c = 123 5 2

4 7 1

2 0 1

a

b

c

9

3

12

Create a matrix equation

To solve matrix equations, get the variable matrix alone on one side.

Get rid of the coefficient matrix by multiplying by its inverse

Solving a matrix equation

2 6

3 11

x

y

0

4

When solving matrix equations we will always multiply by the inverse matrix on the left of the coefficient and constant matrix. (remember commutative property does not hold!!)

The left side of the equation simplifies to the identity times the variable matrix. Giving us just the variable matrix.

x

y

2 6

3 11

1 0

4

2 6

3 11

1 2 6

3 11

x

y

2 6

3 11

1 0

4

Using the calculator we can simplify the left side. The coefficient matrix will be A and the constant matrix will be B. We then find A-1B.

x

y

2 6

3 11

1 0

4

The right side simplifies to give us our answer.

x = -6 y = 2 You can check the systems by

graphing, substitution or elimination.

x

y

6

2

Advantages

Basically, all you have to do is put in the coefficient matrix as A and the constant matrix as B. Then find A-1B. This will always work!!!

NO SOLVING FOR Y!!!!! :)

Solve: Plug in the coeff. matrix as A Put in the const. matrix as B Calculate A-1B.

3 2

4 1

x

y

7

8

x

y

2111 411

Solve: r - s + 3t = -8 2s - t = 15 3r + 2t = -7

1 1 3

0 2 1

3 0 2

r

s

t

8

15

7

r

s

t

3

8

1

Word Problem Systems The sum of three numbers

is 12. The 1st is 5 times the 2nd. The sum of the 1st and 3rd is 9. Find the numbers.

Word Problem Systems The sum of three numbers is 12. x + y + z = 12 The 1st is 5 times the 2nd. x = 5y The sum of the 1st and 3rd is 9. x + z = 9

Word Problem Systems x + y + z = 12 x =

5y => x - 5y = 0 x + z = 9

1 1 1

1 5 0

1 0 1

x

y

z

12

0

9

Word Problem Systems x + y + z = 12 x -

5y = 0 x + z = 9

x

y

z

15

3

6