Autar Kaw
Benjamin Rigsby
http://nm.MathForCollege.com
Transforming Numerical Methods Education for STEM Undergraduates
http://nm.MathForCollege.com
1.
add, subtract, and multiply matrices, and
2.
apply rules of binary operations on matrices.
Two matrices and can be added only if they are the same size. The addition is then
shown as
[C ]  [ A]  [ B]
where
cij  aij  bij
Add the following two matrices.
5 2 3
[A]  

1
2
7


6 7  2
[B]  

3
5
19


.
[C ]  [ A]  [ B]
5 2 3 6 7  2




1 2 7 3 5 19 
5  6 2  7 3  2 


1

3
2

5
7

19


11 9 1 


4
7
26



Blowout r’us store has two store locations and , and their sales of tires are given by make
(in rows) and quarters (in columns) as shown below.
25 20 3 2 
[A]   5 10 15 25
 6 16 7 27
20 5 4 0 


[B]   3 6 15 21
 4 1 7 20
where the rows represent the sale of Tirestone, Michigan and Copper tires respectively
and the columns represent the quarter number: 1, 2, 3 and 4. What are the total tire sales
for the two locations by make and quarter?
G  Area of trapezoi d ABCD
[C ]  [ A]  [ B]
25 20 3 2  20 5 4 0 
  5 10 15 25   3 6 15 21
 6 16 7 27  4 1 7 20
25  20
  5  3
 6  4
20  5 3  4 2  0 
10  6 15  15 25  21
16  1 7  7 27  20
The answer then is,
45 25 7 2 
  8 16 30 46
10 17 14 47
So if one wants to know the total number of Copper tires sold in quarter 4 at the
two locations, we would look at Row 3 – Column 4 to give c34  47.
Two matrices [ A] and [B]can be subtracted only if they are the same size. The
subtraction is then given by
[ D]  [ A]  [ B]
Where
d ij  aij  bij
Subtract matrix
[B]
from matrix [ A].
5 2 3
[A]  

1 2 7
6 7  2
[B]  

3 5 19 
[ D]  [ A]  [ B]
5 2 3 6 7  2




1
2
7
3
5
19

 

(5  6) (2  7) (3  (2)) 


 (1  3) (2  5) (7  19) 
1  5 5 



2

3

12



Blowout r’us store has two store locations and , and their sales of tires are given by make
(in rows) and quarters (in columns) as shown below.
25 20 3 2 
[A]   5 10 15 25
 6 16 7 27
20 5 4 0 
[B]   3 6 15 21
 4 1 7 20
where the rows represent the sale of Tirestone, Michigan and Copper tires respectively
and the columns represent the quarter number: 1, 2, 3 and 4. What are the total tire sales
for the two locations by make and quarter?
[ D]  [ A]  [ B ]
25 20 3 2  20 5 4 0 
  5 10 15 25   3 6 15 21
 6 16 7 27  4 1 7 20
20 
25  20 20  5 3  4
  5  3 10  6 15  15 25  21
 6  4 16  1 7  7 27  20
The answer then is,
5 15  1 2
 2 4 0 4
2 15 0 7
So if you want to know how many more Copper tires were sold in quarter 4 in store A than
store B, d 34  7. Note that d13  1 implies that store A sold 1 less in Michigan tire than
store B in quarter 3.
Two matrices [A] and [B] can be multiplied only if the number of columns of [A] is
equal to the number of rows of [B] to give
[C ] mn  [ A] m p [ B ] pn
If [ A] is a m  p matrix and [B] is a p  n matrix, the resulting matrix [C ]
is a m n matrix.
So how does one calculate the elements of [C ] matrix?
p
cij   aik bkj
k 1
 ai1b1 j  ai 2 b2 j    aip b pj
for each i  1, 2,, m and j  1, 2, , n
i th row and j th column of the [C ] matrix in
th
is calculated by multiplying the i row of [ A] by the j th
To put it in simpler terms, the
[C ]  [ A][ B]
column of [B ] .
Given the following two matrices,
5 2 3
[A]  

1
2
7


Find their product,
C  AB
3  2 
[B]  5  8 
9  10
c12
be found by multiplying the first row of [ A] by the second column of [B],
2
c12  5 2 3  8 
 10
 (5)( 2)  (2)( 8)  (3)( 10)
 56
Similarly, one can find the other elements of [C ] to give
52  56
[C ]  

76  88

Blowout r’us store has two store locations and , and their sales of tires are given
by make (in rows) and quarters (in columns) as shown below.
25 20 3 2 
[A]   5 10 15 25
 6 16 7 27
20 5 4 0 
[B]   3 6 15 21
 4 1 7 20
where the rows represent the sale of Tirestone, Michigan and Copper tires
respectively and the columns represent the quarter number: 1, 2, 3 and 4.
A
. The price matrix is
Find the per quarter sales of store
tire.
A if the following are the prices of each
Tirestone = $33.25
Michigan = $40.19
Copper = $25.03
The answer is given by multiplying the price matrix by the quantity of sales
of store A . The price matrix is 33.25 40.19 25.03 .
Therefore, the per quarter sales of store
columns of the row vector
C  1182.38
A dollars is given by the four
1467.38 877.81 1747.06
Remember since we are multiplying a 1 3 matrix by a 3 4 matrix, the
resulting matrix is a 1 4 matrix.
If [ A] is a n  n matrix and k is a real number, then the scalar product of
k and [ A] is another n  n matrix [B ], where bij  k aij .
Given the matrix,
2.1 3 2
[ A]  

5
1
6


Find
2[ A]
The solution to the product of a scalar and a matrix by the following method,
2.1 3 2
2[ A]  2

5
1
6


2  2.1 2  3 2  2


 2  5 2  1 2  6
4.2 6 4 


 10 2 12
If [ A1 ], [ A2 ],....., [ A p ] are matrices of the same size and k1 , k 2 ,....., k p are scalars,
then k1 [ A1 ]  k 2 [ A2 ]  ........  k p [ A p ] is called a linear combination of
[A 1 ], [ A2 ],....., [ A p ]
If
5 6 2
2.1 3 2
0 2.2 2
[ A1 ]  
, [ A2 ]  
, [ A3 ]  



3
2
1
5
1
6
3
3
.
5
6






then find
[ A1 ]  2[ A2 ]  0.5[ A3 ]
[ A1 ]  2[ A2 ]  0.5[ A3 ]
5 6 2
2.1 3 2
0 2.2 2

  2  5 1 6  0.5 3 3.5 6
3
2
1






5 6 2 4.2 6 4   0 1.1 1

   10 2 12  1.5 1.75 3
3
2
1

 
 

 9.2 10.9 5 


11
.
5
2
.
25
10


Commutative law of addition
If [A] and [B] are m×n matrices, then
[A]+[B]=[B]+[A]
Associative law of addition
If [A], [B] and [C] are m×n, n×p, and p×r size matrices, respectively, then
[A]+([B]+[C])=([A] +[B])+[C]
Associative law of multiplication
If [A], [B] and [C] are all m×n, n×p and p×r size matrices, respectively, then
[A]([B][C])=([A][B])[C]
and the resulting matrix size on both sides of the equation is m×r.
Distributive Law
If [A] and [B] are m×n matrices, and [C] and [D] are n×p size matrices
[A]([C]+[D])=[A][C]+[A][D]
([A]+[B])[C]=[A][C]+[B][C]
and the resulting matrix size on both sides of the equation is m×r.
Illustrate the associative law of multiplication of matrices using
1 2
 2 5
2 1


[ A]  3 5, [ B]  
, [C ]  


9
6
3
5




0 2
[ B][C ] 
 2 5   2 1



9 6 3 5
19 27


36 39 
1 2
19 27
[ A]([ B][C ])  3 5 
36 39 

0 2
 91 105 
 237 276


 72 78 
1 2
2 5


[ A][ B ]  3 5 

 9 6
0 2
20 17
  51 45


18 12
20 17 
2 1


([ A][ B])[C ]   51 45 
3 5

18 12 
 91 105 
 237 276
 72 78 
If [A][B] exists, number of columns of [A] has to be same as the number of rows
of [B] and if [B][A] exists, number of columns of [B] has to be same as the
number of rows of [A].
Now for [A][B]=[B][A], the resulting matrix from [A][B] and [B][A] has to be of
the same size. This is only possible if [A] and [B] are square and are of the same
size. Even then in general [A][B]≠[B][A].
Determine if
[A][B]=[B][A]
for the following matrices
6 3
  3 2
[ A]  
, [ B]  


2 5
 1 5
6 3  3 2
[ A][ B ]  
  1 5
2
5



 15 27



1
29


Therefore
[ A][ B]  [ B][ A]
 3 2 6 3
 2 5
1
5



[ B ][ A]  
 14 1 


16
28

