Matrix
Revolutions:
Solving Matrix
Equations
Matrix 3
MathScience
Innovation
Center
Betsey Davis
1 2
3 4


0 0 1 0 1
1
1
0
1 0 1 0
0 0 00 1
1
0
0 1
10
1 1
1
1
11
1
1
1
0
1
1 0
01
0 0
0
0
0 0 00 0
1
1
0 0
00
0 0
1
0 10
1
0 1 1 1
11
1 0
0 1 0
0
11
1
1 1 1 1
01
0 0
0 0 0
1
10
0 0 0
1
1
0 1
01
011
1
100
011
0
011
100
1
000
110
01
01
1
0
1
11
00
0
1
0
1
0
1
0
0 0
1 1
1 1
0 0
1 0
0 0
1
1
1
0
1
1
0
0
1
It’s time to use matrices!
How would you solve 3 x = 6 for x?
We need to think of multiplying, not
dividing because with matrices there is
no “divide”.
Multiply 3x and 6 by the multiplicative
inverse of 3.
(1/3)3x = (1/3)6
so…. X = 2
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The Big Idea
If [A] [x] = [B] where A, B, x are
matrices,
Then [A]-1[A] [x] = [A]-1[B]
So [x] = [A]-1[B]
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(1/3)3x = (1/3)6
Multiplication by real numbers is
commutative, so order is not important.
Multiplication by matrices is NOT
commutative, so order is VERY important.
Let’s solve for x:
1 2
3 4


X=
Matrix Revolutions
7  1
4 2 


B. Davis MathScience Innovation Center
1 2
3 4


X=
7  1
4 2 


First, find the inverse of the left matrix
 4
1 2
1  4  2   2
3 4  4  6  3 1     3



 
 2
1
Matrix Revolutions
 2
 2 1 

2   3
1

 
1

 2
2
 2
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1 2
3 4


X=
7  1
4 2 


Second, multiply both sides of the equation
by A-1
 2 1  1 2
 2 1  7  1
1 
1 
3
x 3




3 4
4 2
 2
 2
2  
2  
Matrix Revolutions
B. Davis MathScience Innovation Center
1 2
3 4


X=
7  1
4 2 


Third, simplify both sides.
 2 1  1 2
 2 1  7  1
1 
1 
3
x 3




3
4
4
2
 2
 2


2  
2  
 2 1  7  1
1 0
0 1 x   3  1  4 2 
 2



2  
 2 1  7  1
1 
x 3


4
2
 2

2  
Matrix Revolutions
B. Davis MathScience Innovation Center
1 2
3 4


X=
7  1
4 2 


To simplify the right hand side, multiply the 2
matrices.
 2 1  7  1
1 
x 3


4
2
 2

2  
  2 * 7  1* 4
1
x  3
 2 * 7   2 * 4
 2 * 1  1* 2    10 4 
3
1    17
5
* 1   * 2 
 
2
2   2
2
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B. Davis MathScience Innovation Center
1 2
3 4



X=
7  1
4 2 


By Calculator:
1
1 2 7  1
x
*


3 4 4 2 
So, just enter A and B into the calculator.
Then on the home screen type [A] x-1 [B] enter.
Matrix Revolutions
B. Davis MathScience Innovation Center
1 2
3 4


X=
7  1
4 2 


  10 4 
5
x   17


 2
2 
Final Answer for x.
Matrix Revolutions
B. Davis MathScience Innovation Center
SO WHAT???????
We can use
this new skill–
solving
equations
using
matrices –
To solve
linear systems.
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Let’s learn how!
Basic idea comes from solving AX = B
Let’s write a system:
3x+2y=6
5 x - 9 y = 15
Now, let’s rewrite the system using matrices:
Matrix Revolutions
B. Davis MathScience Innovation Center
3x+2y=6
5 x - 9 y = 15
Re-writing the system using matrices:
•Make a matrix of coefficients.
•Make a matrix of variables.
•Make a matrix of constants.
3 2 
5  9


 x
 y
 
Matrix Revolutions
=
6
15
 
B. Davis MathScience Innovation Center
3x+2y=6
5 x - 9 y = 15
What size are these and can they be multiplied?
What size is the answer?
3 2 
5  9


2x2
 x
 y
 
2x1
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=
6
15
 
2x1
B. Davis MathScience Innovation Center
3x+2y=6
5 x - 9 y = 15
• We know we can multiply them and the
answer is a 2x1.
• We will use the same BIG IDEA:
If [A][x]=[B], then [x] = [A]
3 2 
5  9


 x
 y
 
Matrix Revolutions
=
–1
[B]
6
15
 
B. Davis MathScience Innovation Center
Remember BIG IDEA:
Here
If [A][x]=[B], then [x] =
we
[A] –1go!
[B]
1
3 2  3 2 
5  9 5  9


 
 x
 y
 
=
3 2 
5  9


1
Important!!! Notice order of multiplication
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6
15
 
We have identity matrix on left
1 0   x 
0 1   y 

  
Matrix Revolutions
=
3 2 
5  9


B. Davis MathScience Innovation Center
1
6
15
 
The identity times [x] is [x].
 x
 y
 
=
3 2 
5  9


1
6
15
 
Now just type [A]-1[B] on your TI
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Final answer
 x
 y
 
=
 84 
 37 
 15 
 
 37 
What does this mean?
 84 15 
,
The solution is the ordered pair  37 37 
Matrix Revolutions
B. Davis MathScience Innovation Center
Let’s try it !
2w – x + 5 y + z = -3
3w + 2x + 2 y – 6 z = -32
w + 3x + 3 y - z = -47
5w – 2 x - 3 y + 3 z = 49
We will need 3 matrices…
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Matrix of Coefficients
2w
3w
1w
5w
– 1 x + 5 y + 1z = -3
+ 2x + 2 y – 6 z = -32
+ 3x + 3 y -1 z = -47
– 2 x - 3 y + 3 z = 49
2
3

1

5
1
2
3
2
5
2
3
3
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1 
 6
 1

3 
B. Davis MathScience Innovation Center
Matrix of variables
2w – x + 5 y + z = -3
3w + 2x + 2 y – 6 z = -32
w + 3x + 3 y - z = -47
5w – 2 x - 3 y + 3 z = 49
Matrix Revolutions
B. Davis MathScience Innovation Center
 w
x
 
 y
 
z
Matrix of constants
2w – x + 5 y + z = -3
3w + 2x + 2 y – 6 z = -32
w + 3x + 3 y - z = -47
5w – 2 x - 3 y + 3 z = 49
Matrix Revolutions
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 3 
  32


  47


 49 
2
3

1

5
1
2
3
2
5
2
3
3
Matrix Revolutions
1 

 6
 1

3 
 w
x
 
 y
 
z
=
B. Davis MathScience Innovation Center
 3 
  32


  47


 49 
2
3

1

5
1
2
3
2
5
2
3
3
Matrix Revolutions
1 

 6
 1   w
 x
3   
 y
 
z
=
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 3 
  32


  47


 49 
2
3

1

5
1
2
3
2
Matrix Revolutions
5
1 

2
 6
w1 
3 
x 
 3  3 =
 y
 
z
B. Davis MathScience Innovation Center
 3 
  32


  47


 49 
2
3

1

5
1
2
3 w
x
 2 
 y
 
z
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5
2
3
3
1 

 6
 1

3 =
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 3 
  32


  47


 49 
 w
x
 
 y
 
z
=
2
3

1

5
1
2
3
2
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5
2
3
3
1 

 6
 1

3 
B. Davis MathScience Innovation Center
-1
 3 
  32


  47


 49 
 w
x
 
 y
 
z
=
 2 
 12


4


 1 
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Therefore,
The solution is
An ordered quadruplet:
(2,-12,-4, 1)
B. Davis MathScience Innovation Center
Way cool, huh?
Matrix Revolutions
B. Davis MathScience Innovation Center