Section 6.3
Basic Matrix Operations
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Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall.
Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall.
Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall.
If A and B are both m × n matrices then the sum of A and B,
denoted A + B, is a matrix obtained by adding corresponding
entries of A and B. The difference of A and B, denoted A – B, is
obtained by subtracting corresponding entries of A and B.
1 2 2
 3 0 4 
A
and B  


0

1
3
2
1

4




 2  2 6 
A B  

2
0

1


 4 2 2 
A B  


2

2
7


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The Zero Matrix
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If A is an m × n matrix and s is a scalar, then we let kA denote the
matrix obtained by multiplying every element of A by k. This
procedure is called scalar multiplication.
1 2 2 
A

0

1
3


0 2 3
B

1
2
1


1 2 2  4 8 8 
(a) 4 A  4 



0

1
3
0

4
12

 

3 1 
C

0

4


1 

1
1
1 3 1  
3 
(b) C  



4
3
3 0 4  
0  

3 
1 2 2
0 2 3  3 2 0 
(c) 3A  2 B  3 
2




0

1
3
1
2
1

2

7
7



 

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Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall.
2


Find RC if R  1 2 4 and C   1
 3 
RC  1 2   2 1   43  2  2  12  16
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RC   40100   20 200   40050  4000  4000  20000  28000
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Find the product AB if
 3 2 1 
A

0
4

1


 2 4
B   1 3 
 3 1 
3(2)  2(1)  1(3) 3(4)  2(3)  1(1) 
AB  

0(2)

4(

1)

1(

3)
0(4)

4(3)

1(1)


 5 7
AB  


1
11


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 2 4
3 2 1 
 1 3 
Find the product BA if A  
B




0
4

1


 3 1 
 2 4
B   1 3 
 3 1 
 6 12  2


BA   3 14  4
 9 10  4
 3 2 1 
A

0
4

1


Recall from last example:
 5 7
AB  


1
11


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1 3
 2 0 
A
and B  


 2 7 
 3 4
12 
 1 3   2 0   7
AB  






2

7
3
4

17

28


 

3   2 6 
 2 0   1
BA  





 3 4  2 7   5 19
AB ≠ BA
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 2 4
3 2 1 
 1 3 
A
and
B




0
4

1


 3 1 
1 0 0 
3 2 1  
3 2 1 

(a) AI 3  
0 1 0  



0
4

1
0
4

1

 0 0 1  



1 0 3 2 1  3 2 1 
(b) AI 2  





0
1
0
4

1
0
4

1


 

 2 4
 2 4
1 0  



(c) BI 2   1 3 


1
3
 

0
1


 3 1 
 3 1 
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Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall.

1 
 1  
 3 1 
2 
1

Show that the inverse of A  
is
A

.


3 
 4 2 
 2

2 


1

 3 1
1
AA  


 4 2  2

1

1  

2  3
1
A A

3  4
2

2 
1
 
1 0 
2

3  0 1 
2 
1 1 0 


2  0 1 
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1 2
1 3  is nonsingular. Find its inverse.
R3= –2r1+ r3
2 1 
0 0  1 1 2 1 0 0 
 

1 0   0 1 3 0 1 0 
0 1  0 4 3 2 0 1 
R1= r1+ r2
R3= – 4r2+ r3


1
0

1
1

1
0
R2= – 1 1 2 1
0 0
r2

  0 1 3 0 1 0 
 0 1 3 0 1 0 


0 0 9 2 4 1 
0 4 3 2 0 1 
R3= 1/9r3
1
The matrix A  0
 2
1 1 2 1

A I 3   0 1 3 0
 2 2 1 0

 1 0 1 1

  0 1 3 0
0 0 1 2


9


7



1 0
1 0 0 9

2

1 0   0 1 0 

3

0 0 1
4 1
2



9 9

9
5
9
1
3
4
9

1
9 
1
3

1
9 
 7
5
1
R
 19= r1 + r93 9 
R
 2= 3r3 + r2 
2
A1   
 3

 2
 9
1
3
4
9
1
3

1
9 
 2 1 
Show that the matrix A  
has no inverse.

 4 2
R1 = –1/2r1
R2 = – 4r1 + r2




1 1
1 1
 2 1 1 0
1 
0
1 
0



2 2
2
2

 



 4 2 0 1
 4 2 0 1 
0 0 2 1 
We can see that we cannot get the identity on the left side of the
vertical bar. We conclude that A is singular and has no inverse.
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x  y  2z  1

Solve the system of equations:   y  3z  2

2 x  2 y  z  1
1 1 2
 x
1
A  0 1 3 X   y  B   2
 2 2 1 
 z 
 1
AX  B
X  A1 B
 7
 9

2
A1 B   
 3

 2
 9
5

9
1
3
4
9
1
 16 
9  1   9 



1    5 
2   


 3
3
  1 

1     11 

9 
 9 
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