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Section 7.1
Matrices and Systems of Linear Equations
Matrices
 A matrix is a rectangular array of numbers
written within brackets
 Open books to page 562
Order of a Matrix
Order of a Matrix: Row x Column
 2 6 
 4  3


 9  8
2
 3

4

8
10
3
2
4
 2
1  8
2 3 
3 4

4 6
5 6
3 rows, 2 columns
3x2
4 rows, 4 columns
1 row, 3 columns
4x4
1x3
Augmented Matrices
Matrices can be used as shorthand for systems of equations. When
done so, they are called augmented matrices.
3x  4 y  1
x  2y  7
3 4 1


1

2
7


Each row is an equation
Vertical line represents the equal sign
First column is coefficients on the x
Second column is coefficients on the y
Constants to the right of the vertical line
Any variable not in the equation has an implied coefficient of 0
Write the system as an
augmented matrix
x  3 y  2z  7
 2x  4z  3  0
1 3 2 7



2
0
4

3


yxz 7
x yz 2
 1 1 1 7 


1

1

1
2


 0  1 1  1
z  y  1
Row Operations (Solving Systems)
 Interchange any 2 rows
Ri  R j
 Multiply a row by a nonzero constant
cRi  Ri
 Add a multiple of 1 row to another cRi  R j  R j
Perform the row operation
2  1 3


0 2 1
0 2 1


2  1 3
R1  R2
Perform the row operation
5
7 1 2 


3
2

4
1


 6  1 5  4
2R2  R2
5
7 1 2 




 6  1 5  4
Perform the row operation
1 0  2  3


0 1 2 3 
4 2  1 2 
R1  2R3  R1




0 1 2 3
4 2  1 2
Perform the row operation
 1 0  1  2


3

1
2
3


 0 1 3 1 
R1  3R2  R1




3  1 2 3
0 1 3 1
Row-Echelon Form of a Matrix
 Rows consisting entirely of 0’s are at the
bottom of the matrix
 For each row that does not consist entirely of
0’s, the first (leftmost) nonzero entry is 1
(called the leading 1)
 The leading 1 in each row must have all
zeros underneath it.
1
0

0

0
3 5 6 
1  2  1
0 1
5

0 0
0
Determine whether the matrices are in
Row-Echelon Form
1 2  2  5


0
1
4

6


0 0 1 7 
Yes
1 2  6  5 


0
1
5
1


1 0 4 7 
No
1 12  26  5


0
1
15

3


0 0
0 0 
Yes
1 7 8 10


0
4
6
2


0 0 1 7 
No
Rewrite the Matrix in Row Echelon
Form
 3 2 2


1

1
3


R1  R2
 1  1 3



3
2
2


3R1  R2  R2
1  1 3 


0

1
11


1  1 3 


0
1

11


 1R2  R2
Solve the system using
Gaussian Elimination
2 x  y  8
x  3y  6
Step 1: Write as an augmented matrix
2 1  8


1
3
6


1 3 6 

 R1  R2
2
1

8


Step 2: Use row operations to write
in row-echelon form.
1 3 6 
Need a 0 below the leading 1 in row 1


0

5

20

  2R1  R2  R2
…continued
1 3 6 


0

5

20


1 3 6


0
1
4


x  3y  6
y4
x  34  6
x  12  6
x  6
Need a leading 1 in row 2 (turn the -5 into a 1)
1
 R2  R2
5
Step 3: Write the augmented
matrix as a system of equations.
Step 4: Back substitute to find all
other variables.
x  6
y4
Solve the system
x  y  z  2
3 x  y  z  4
2 x  2 y  3z  3
1 1  1  2 


3
1

1

4


2  2 3 3 
1 1  1  2


0

2
2
2


0  4 5 7 
 3R1  R2  R2
and
 2 R1  R3  R3
1 1  1  2


0

2
2
2


0  4 5 7 
1 1  1  2


0
1

1

1


0  4 5 7 
1 1  1  2


0
1

1

1


0 0 1 3 
x  y  z  2
y  z  1
z 3

1
R2  R2
2
y  3  1
y2
4R2  R3  R3
x  1
x  2  3  2
x 1  2
x  1
y2
z 3
Infinitely Many and No Solutions
1  2 3 2 


0
1
6
4


0 0 0 0
Row 3 equation would say:
0x + 0y + 0z = 0
0=0
Infinitely Many Solutions on a line
1  2 3 2 


0
1
6
4


0 0 0 4
Row 3 equation would say:
0x + 0y + 0z = 4
0=4
No Solutions