Matrices
An Overview
A. Matrices: Used as a tool in solving systems of linear equations.
1. System of linear equations in standard form
2. Re-written as an augmented matrix
x  2 y  3z  4
 1 2 3
3 0 1

 1 3 4
z5
3x
x  3y  4z  0
B. The goal is to obtain a pattern of
1. A “1” in the first column, first row,
2. A“1” in the second column, second row,
3. A “1” in the third column, third row,
4. And so forth with zeroes below the ‘1’s.
 a11
a
a21
 31
a
12
a
22
a
32
a
13
a
23
a
33
4
5

0 
a 
14
a 
24

a 
34
1 0 0 a14 
0 1 0 a 
24


0 0 1 a34 
C. Use row operations to accomplish the above objective.
1. Two rows may be interchanged
a. R 2  R 1
2.
A row may be multiplied by a non- zero number, replacing the original row with the result
( n  non-zero number )
a.
3.
n  R1  R2
A row may be replaced by the sum of itself and another row
a. R 1 + R 2  R 1
4. A row may be replaced by the sum of itself and a multiple of another row
a. R 1 + (n  R 2 )  R 1
D. Comprehensive Example
Solve using an augmented matrix.
x  2 y  3z  4
3x
z5
x  3y  4z  0
1.
First, write the augmented matrix that represents this system.
 1 2 3 4 
 3 0 1 5


 1 3 4 0 
2.
 1 2 3 4 


Multiply row one by -3, add to row 2, and replace row 2 with the result: 3  R  R  R  0 6 10 7 
 1 3 4 0 
The ‘3’ in row 2 ( a21 ) needs to be a ‘0’.
1
3.
The ‘-1’ in row 3 ( a31 ) needs to be a ‘0’.
Add row one to row three and replace row three with the result:
R R R
1
4.
3
3
The ‘-6’ in row 2 ( a22 ) needs to be a ‘1’.
Switch row two and row three and multiply the new row two by -1:
1 2 3 4 
R  R  R  0 1 1 4 


 0 6 10 7 
2
3
2
1 2 3 4 


1  R  R  0 1 1 4 
 0 6 10 7 
2
2
2
2
1 2 3 4 
 0 6 10 7 


 0 1 1 4 
5.
The ‘-6’ in row three ( a32 ) needs to be a ‘0’, so multiply row two by ‘6’ and add to row three:
6R  R  R
2
6.
3
2
1
0

1 4 

4 31
3 4
The ‘4’ in row three ( a33 ) needs to be a ‘1’ so multiply row three by ‘¼’ (the same as dividing by ‘4’):
1
4
7.
3
1
0

0


1 2 3 4 


 0 1 1 4 
 0 0 1 31 



4 
R  R
3
Therefore, z 
3
31
. We now have some choices.
4
a.
We could substitute z =
into the second original equation, 3x + z = 5, and solve for x. Then substitute
4
both our z and x and solve for y using the original equations.
31
5
i. 3 x 
4
ii. 12 x  31  20 (clear fractions by multiplying by ‘4’)
17
iii. x 
4
17
31
 2 y  3(
)4
iv.
4
4
v. 17  8 y  93  16 (clear fractions by multiplying by ‘4’)
vi.
b.
c.
8.
So, x =
31
y
47
4
Or we could convert row two of our matrix:
31
i. y –z = -4 and substitute z =
4
31
ii. y – (
) = -4
4
47
iii. y =
4
Do the same type of substitution into row one of the matrix to solve for x.
i. x + 2y -3z = 4
47
31
ii. x + 2(
)–3(
)= 4
4
4
94
93
iii. x +
+
=4
4
4
1
17
iv. x = 4, x =
4
4
17
4
,y=
47
4
,z=
31
(some ugly numbers!)
4
Denise Hammond Fall 2004