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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
corresponding
elements of
elements
of AA and
and B.
B.
add
these
add
these
add
these
add
these
add these
add these
111 222 222 
AA
A 11 22 22 
AA000 111 333 
 0  1 3  
0  1 3 
 3 0 4 
300
0 44
33
4 
B  



3
0
4


BB
 44
BB
222311 01
B 2
2 1
1 
44
44
 2 1  4

2


2
222
A

B



2

A

B



AAAABBBB 2
 22

2  6 
2222 666
0  1
0  
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A B  B  A
A  ( B  C )  ( A  B)  C
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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
 31 3 2 32 3  6 6
A
3A  










0

1
3
3
0
3

1
3
3
0

3
9



 

PROPERTIES OF SCALAR MULTIPLICATION
k  hA    kh  A
 k  h  A  kA  hA
k  A  B   kA  kB
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The m × n zero matrix, denoted 0, is the m × n
matrix whose elements are all zeros.
0
0 0 
0 0 


0 0
1×3
2×2
A0  A
A  (  A)  0
0 A  0
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The multiplication of matrices is easier shown than put
into words. You multiply the rows of the first matrix
with the columns of the second adding products
Find AB
3  2 1 
A

0 4  1
322 2211  1 3  5
3
 2 4


B    1 3
 3 1
First we multiply across the first row and down the
first column adding products. We put the answer in
the first row, first column of the answer.
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Find AB
3  2 1 
A

0 4  1
55
AB
AB  
 1
 2 4


B    1 3
 3 1
777
  00





244
2
4

4
442
3
3123

3003
4
311111 311
7 1
11
 
Notice the sizes of A and B and the size of the product AB.
Now we multiply across the first
second
rowrow
andand
down
down
the the
second
first
We multiplied across first row and down first column
column column
second
and we’lland
putwe’ll
the answer
put the in
answer
the second
first
in the
row,row,
second
second
firstrow,
so we put the answer in the first row, first column. USC
column.column.
second
To multiply matrices A and B
look at their dimensions
m n
n p
MUST BE SAME
SIZE OF PRODUCT
If the number of columns of A does not
equal the number of rows of B then the
product AB is undefined.
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Now let’s look at the product BA.
 2 4
3  2 1 
A



B    1 3
0 4  1
 3 1 3132121213123223114134034140104410

9
6414
2312
across
acrossthird
second
second
third
row
row
row
row
across
first
second
row
row
as
as
aswe
we
godown
down
we
as
we
go go
go
down
down
first
third
third
first
second
second
third
column:
column:
column:
column:
column:
second
column:
first
column:
column:
32
23
12
12
12
12 2
222
6666 12











BA
BA
BA
BA
BA
333 14
14
14 
4
4
4
 
 99 10
 4

10 
Commuter's Beware!
Completely different than AB!
AB  BA
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PROPERTIES OF MATRIX
MULTIPLICATION
ABC    AB C
AB  C   AB  AC
 A  B C  AC  BC
AB  BA
Is it possible for AB = BA ?,yes it is possible.
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What is AI?
 2  1 2
0 1 5   A


2  2 3
1 0 0


I 3  20 1 12 0
What is IA? 0 1 5  A



0
0
1


2

2
3
 1 2



2
A  0 1 5
2  2 3
Multiplying a
matrix by the
identity gives the
matrix back again.
1 0 0


I 3  0 1 0
0 0 1
an n  n matrix with ones on the main diagonal
and zeros elsewhere
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Can we find a matrix to multiply the first matrix by to
get the identity?
1

 3  1  1  2  1
 ? 3   
4

22
0




2 
1


1


3

1
1





2





3 4
2
0


2

2 

0

1
0

1
Let A be an n n matrix. If there exists a matrix B
such that AB = BA = I then we call this matrix the
inverse of A and denote it A-1.
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If A has an inverse we say that A is nonsingular.
If A-1 does not exist we say A is singular.
To
matrix A,
A, a
a
To find
find the
the inverse
inverse of
of a
a matrix
matrix we
we put
put the
the matrix
line
identity matrix.
matrix. We
We then
then perform
perform row
row
line and
and then
then the
the identity
operations
toturn
turnititinto
intothe
theidentity.
identity. We
We
operations on
on matrix
matrix A
A to
carry
right hand
hand side
side
carry the
the row
row operations
operations across
across and
and the
the right
will
will turn
turn into
into the
the inverse.
inverse.
3
1
3 1 0
1
 2  7 0 1


1 3 1 0

2r1+r2 
0  1 2 1 
A

 2  7 
r2
r1  r2
0
1 3 1
0 1  2  1


3
1 0 7
0 1  2  1
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
3
1
A

 2  7 
3
7
A 

 2  1
1
Check this answer by multiplying. We should
get the identity matrix if we’ve found the
inverse.
1 0
AA  

0 1
1
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We can use A-1 to solve a system of equations
x  3y  1
2x  5 y  3
To see how, we can re-write a
system of equations as matrices.
Ax  b
coefficient
matrix
variable
matrix
1 3
2 5


 x
 y
 
constant
matrix
1
  
3
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Ax  b
1
left multiply both sides
by the inverse of A
1
A Ax  A b
This is just the identity
1
Ix  A b
but the identity times a
matrix just gives us
back the matrix so we
This then gives us a formula have:
for finding the variable
matrix: Multiply A inverse
by the constants.
1
x A b
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x  3y  1
2x  5 y  3
1 3 1 0 
2 5 0 1


1 3 1 0 


-r2 0 1 2 1


1 3
A

2 5
-2r1+r2
r1-3r2 1
1 0
1 3
0  1  2 1 


0 5
0 1

 5 3  1  4 
1
A b
 



 2  1 3  1
find the inverse
x
y
2
3

 1
This is the
answer to
the system
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Your calculator can compute inverses and
determinants of matrices. To find out how, refer to
the manual or click here to check out the website.
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Acknowledgement
I wish to thank Shawna Haider from Salt Lake Community College, Utah
USA for her hard work in creating this PowerPoint.
www.slcc.edu
Shawna has kindly given permission for this resource to be downloaded
from www.mathxtc.com and for it to be modified to suit the Western
Australian Mathematics Curriculum.
Stephen Corcoran
Head of Mathematics
St Stephen’s School – Carramar
www.ststephens.wa.edu.au
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