Solving Using Matrices

By Helen ChinKitty Luo

Anita La

How to find the Determinant:

How to find the Inverse:

-1

Example of (2 x 2) Inverse

-1

= 1

(4x1) - (2x3) = -2

4 -2 -3 1

=

Extra Inverse Information:

Why are inverses useful?

How would you solve 5x = 10? Normally you would simply divide both sides by 5, but matrices cannot be divided. There is another method -- multiplying each side by the inverse. 10 times 1/5 is still 2; division and multiplying by the inverse are the same thing. A number times its inverse will always equal one. You would solve matrix equations with inverses.

Solving Matrix Equations with Inverses

-4 2 -73 -1 5x =

Multiply each side by the inverse of A. The inverse of A multiplied by A would cancel each other out, leaving x = inverse of A times matrix B.

2 4

0 1

A

=4 0

63 -1

5

B

x

1/2 -20 1

4 06

3 -15

= x

Binverse of A

Cramer's Rule...is a way of solving for one variable

instead of solving the entire system of equations.

x - y + z = 2 1-1 1

2x + y + 4z = 4 2 14

-x + 3y - z = 6 -1 3-1

First, set up the coefficient matrix. Be sure to put the variables' values in the same column as the others. (So the x values would always be in the same column, etc.)

x yz

1 -11

2 14

-1 3-1

Find the determinant of matrix A. In this case, it would be -4.

Let's say you want to solve for x.

x - y + z = 22x + y + 4z = 4-x + 3y - z = 6

Replace the x column values with the answer column values of the system.

2 -11

4 14

6 3-1

Find the determinant of the new matrix. In this case, it would be -48. You would then divide the determinant of matrix X by the determinant of matrix A to get the x value.

matrix X

matrix A

-48/-4

x is 12

Do the same thing to solve for y and z.

1 21

2 44

-1 6-1

1 -12

2 14

-1 36

y z

determinant of matrix Y is -16

y = determinant A / determinant Y

-16/-4

y = 4

determinant of matrix Z is 24

z = determinant A / determinant Z

24/-4

z = -6

Word ProblemsYou inherit $60,000 from a distant relative. You decide to invest it in

three different stocks with returns of 2%, 8%, and 10% respectively. You place $5000 more in the second stock than the first and third stock combined. You receive an annual return of 6% of the original $60,000. How much did you invest in each stock?

First, write 3 equations that represent this situation.

B = A + C + 5000A + B + C = 60,000.02A + .08B + .10C = .06(60,000)

Next, type the coefficient matrix.

1 -11

1 11

.02 .08

.10

You can go three ways from here: Cramer's rule, setting up an inverse equation, or using rref.

Cramer's rule is very time consuming, however.

The inverse equation would look something like this:1 -11

1 11

.02 .08

.10

A

B

C

500060,0003600

Multiply matrix A's inverse and matrix B (the order is important) to get the values for A, B, and C.

matrix A matrix B

The third method to solving systems with matrices is by using rref, also known as augmented matrices.

1 -1 15,000

1 1 160,000

.02 .08 .10 3,600

Surprise! It's a 3x4 matrix now. The fourth column is the same as the answer column.

Go to your matrix window and hit the left arrow to go to the MATH column. Go down until you see B: rref . Hit that and then choose your augmented matrix. Hit enter and your answers should be there.

You should've gotten this:1 0 0 21,8750 1 0 32,5000 0 1 5,625

x is 21,875y is 32,500z is 5,625

Solving Matrices using addition - Tip-

Given:

A= ,B=

A+B= =

1 32 -1

5 -34 3

1 + 5 3 + (-3)2 + 4 (-1) + 3

6 0 6 2

You can only add matrices together if A and B or more is in the same order.Ex: 2 by 2 pairs up with another 2 by 2

Laws of Matrix -Tip-

(1) A + B = B+ A (2) A + (B + C) = (A + B) + C (3) 0A = 0, where 0 is the zero matrix. (4) A + 0 = A.(5) A(B + C) = AB + AC(6) A (BC) = (AB)C(7) If A^-1 and B^-1 exist then (AB)^-1

= B^-1 A^-1 Note: These are other ways of writing

matrices in which the operations will make sense either way

Solving for X for Matrix Equations -Tip-

Finding solutions for Ax= B (A^-1 A) X= A^-1B (1.) known as the associative Law)

IX =A^-1B (2.) Definition of Inverse

X=A^-1 B (3.) Definition of identity note: A^-1 is never written as 1/A