Solving Using Matrices

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Solving Using Matrices

• By Helen Chin
• Kitty Luo
• Anita La

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How to find the Determinant
3
How to find the Inverse
-1
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Example of (2 x 2) Inverse
-1
4 -2 -3 1
1 (4x1) - (2x3) -2


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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.
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Solving Matrix Equations with Inverses
B
A
4 0 6 3 -1 5
2 4 0 1

x
B
inverse of A
1/2 -2 0 1
4 0 6 3 -1 5
x

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.

-4 2 -7 3 -1
5
x
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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 1 4
• -x 3y - z 6 -1 3 -1

x y z
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.)
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matrix A
1 -1 1 2 1 4 -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.
Replace the x column values with the answer
column values of the system.
x - y z 2 2x y 4z 4 -x 3y - z 6
2 -1 1 4 1 4 6 3 -1
matrix X

• 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.

-48/-4 x is 12
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Do the same thing to solve for y and z.
1 -1 2 2 1 4 -1 3 6
1 2 1 2 4 4 -1 6 -1
z
y
determinant of matrix Y is -16
determinant of matrix Z is 24
y determinant A / determinant Y
z determinant A / determinant Z
-16/-4
24/-4
y 4
z -6
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Word Problems

• You 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 5000 A B C 60,000 .02A .08B
.10C .06(60,000)
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• Next, type the coefficient matrix.

1 -1 1 1 1 1 .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 -1 1
• 1 1 1
• .02 .08 .10

matrix A
matrix B
A B C
5000 60,000 3600
Multiply matrix A's inverse and matrix B (the
order is important) to get the values for A, B,
and C.
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• The third method to solving systems with matrices
  is by using rref, also known as augmented
  matrices.

Surprise! It's a 3x4 matrix now. The fourth
column is the same as the answer column.
1 -1 1 5,000 1 1 1 60,000 .02 .08 .10 3,600
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.
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You should've gotten this

• 1 0 0 21,875
• 0 1 0 32,500
• 0 0 1 5,625

x is 21,875 y is 32,500 z is 5,625
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Solving Matrices using addition - Tip-

• Given
• A ,B
• AB

5 -3 4 3

• 1 3
• 2 -1

6 0 6 2
1 5 3 (-3) 2 4 (-1)
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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
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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

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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
• XA-1 B (3.) Definition of identity
  

note A-1 is never written as 1/A