Matrices & Systems of Linear
Equations
Special Matrices
Zero Matrix
SquareMatrix
An n by n Matrix
Exam ples:
Exam ples:
0 0 
O22  
, O14  0 0 0 0

0 0 
5,
0 0 0 
0 
O21   , O33  0 0 0
0 
0 0 0
1 0
2 0

3 1

4  1
1 0
5 2,


1 8
1  8
5 4

2 0
1 0 6 
2  1 9 


3 5 4
Special Matrices
Diagonal Matrix
Identity Matrix
A square m atrixif
all entries that are not on the m ain
A is diagonal m atrixif
all entries that are on the m ain
diogonalare zeroes
diogonal are ones
Exam ples:
Exam ples:
5,
1
0

0

0
1 0
0 2  ,


0 0 0
7 0 0
0 5 0

0 0 8
1 0 0
0  1 0 


0 0 4
I11  1,
1
0
I 44  
0

0
1 0
I 22  
,

0 1 
0 0 0
1 0 0
0 1 0

0 0 1
1 0 0
I 33  0 1 0
0 0 1
Equality of Matrices
Two matrices are said
to be equal if they have
the same size and their
corresponding entries
are equal
Are the following m atricesequal?
1
1 2 3 4
2
5 6 7 8  , 


3
9 0 5 5

4
5 9
6 0
7 5

8 5
Are the following m atricesequal?
1 2
1 2
3 4 , 5 4




Equality of Matrices
Use the given equality
to find x, y and z
1.
1
1 2 3 4
5 6 7 8  5



9
9 0 5 5
2.
y 
1 2
x
3 4  3 2 z  5




2
x  2 4
y3
7
8 
5
0
5
2 z
Matrix Addition and Subtraction
Example (1)
1 2 3   0 5 1 
 4 5 6   2 4 6

 

7 8 9 8 1 3
1  0 2  5 3  1 
 4  2 5  4 6  6
 7  8 8  1 9  3
1 7 4
  6 9 12
15 9 12
Matrix Addition and Subtraction
Example (2)
1 2 3   0 5 1 
 4 5 6   2 4 6

 

7 8 9 8 1 3
1  0 2  5 3  1 
 4  2 5  4 6  6
 7  8 8  1 9  3
 1  3 2
  2
1 0
 1 7 6
Multiplication of a Matrix by a Scalar
Exam ple(1)
1 2 3  5(1) 5(2) 5(3) 
5 4 0 6  5(4) 5(0) 5(6)
7  1 9 5(7) 5(1) 5(9) 
 5 10 15
 20 0 30
35  5 45
Exam ple(2)
1
1 2 3
0 5
2 4 5 6  (1) 2 4
6 
7 8 9
8  1  3
 2 4 6   0  5  1
  8 10 12   2  4  6
14 16 18   8 1
3 
2  1 5 
 6 6 6 
6 17 21
Matrix Multiplication
(n by m) Matrix X (m by k) Matrix
The number of columns of the matrix on the left
=
number of rows of the matrix on the right
The result is a (n by k) Matrix
Matrix Multiplication
3x3 X 3x3
a1 a 2 a 3   x1 y1 z1 
b b b   x y z 
2
3  2
2
2
 1
 c1 c 2 c 3   x3 y3 z 3 
a1 x1  a 2 x2  a 3 x3 a1 y1  a 2 y2  a 3 y3
  b1 x1 b 2 x2 b 3 x3 b1 y1 b 2 y2 b 3 y3
 c1 x1 c 2 x2 c 3 x3 c1 y1 c 2 y2 c 3 y3
a1 z1  a 2 z 2  a 3 z3 
b1 z1 b 2 z 2 b 3 z3 
c1 z1 c 2 z 2 c 3 z3 
Matrix Multiplication
1x3 X 3x3→ 1x3
 x1 y1 z1 
a1 a 2 a 3   x2 y 2 z 2 
 x3 y 3 z 3 
 a 1 x1  a 2 x 2  a 3 x3 a 1 y1  a 2 y 2  a 3 y 3
a 1 z1  a 2 z 2  a 3 
Example (1)
 1 4  1 2 1
 0 0 2  0  1


 2 1 1  1 1
 1(2)  4(0)  (1)(1)
  (0)(2)  (0)(0)  (2)(1)
(2)(2)  (1)(0)  (1)(1)
 1  4 2
  2
2 4
 3  2 3
0
1 
2
1(1)  4(1)  (1)(1)
1(0)  4(1)  (1)(2) 
(0)(1)  (0)(1)  (2)(1) (0)(0)  (0)(1)  (2)(2) 
(2)(1)  (1)(1)  (1)(1) (2)(0)  (1)(1)  (1)(2)
Example (2)
(1X3) X (3X3) → 1X3
0  1 3 
2 3 4 1  2 10
2 3
4 
 2(0)  3(1)  4(2) 2(1)  3(2)  4(3) 2(3)  3(10)  4(4)
 11 4 52
Example (3)
(3X1) X (1X2) → 3X2
5 
1 6 7
 
0
5(6)
 1(6)
0(6)
30
  6
 0
5(7) 
1(7) 
0(7)
35
7 
0 
Example (4)
1 2  1
1  1 3 


0 4 5 
2
1 2  1 1 2  1
 1  1 3  1  1 3 
0 4 5  0 4 5 
3  4 0 
 0  1 11
4 16 37
Transpose of Matrix
a1
A  b1
 c1
a2
b2
c2
a3 
b3 
c3 
 a1
AT  a2
 a3
b1
b2
b3
c1 
c2 
c3 
Exam ple(1)
T
1 2 3 
1 4 0 
 4 5 6   2 5 7 




0 7 8
3 6 8
Exam ple(2)
1 2 3 
1 2  3
4 5 6  34  5 6   2 I
3




0 7 8
0 7
8 
T
6
 9
1 4 0   3
1 0 0
 2 5 7   12  15 18   2 0 1 0
3 6 8   0 21 24 
0 0 1
 4 10  9 2 0 0
 14  10 25   0 2 0
 3 27 32  0 0 2
 2 10  9
 14  12 25 
 3 27 30 
Exam ple(3)
T
 1 2 3 T 
1 2 3


  4 5 6    4 5 6
 



 0 7 8 
0 7 8


Properties of the Transpose
1. ( A )  A
T
T
2. ( AB)  B A
T
T
T
Matrix Reduction
Definitions (1)
1. Zero Row:
A row consisting entirely of zeros
2. Nonzero Row:
A row having at least one nonzero entry
3. Leading Entry of a row:
The first nonzero entry of a row.
Matrix Reduction
Definitions (2)
Reduced Matrix: A matrix satisfying the following:
1. All zero rows, if any, are at the bottom of the matrix
2. The leading entry of a row is 1
3. All other entries in the column in which the leading entry is located
are zeros.
4. A leading entry in a row is to the right of a leading entry in any row
above it.
Examples of Reduced Matrices
1.
1 0 0 
0 1 0 


0 0 1
2.
1 0 0 
0 1 0 


0 0 0
3.
1 0 5 
0 1 7 


0 0 0 
Examples matrices that are not
reduced
1 0 0
0 6 0 
1.


0 0 1
The leading entry in row 2 is not 1
0 1 0 
1 0 0
2.


0 0 0
The leading entry in row 2 is not to the right of the leading
entry in the row aboveit.
1 0 0
0 0 0 
3.


0 1 0
Row 2 is a zero row
but not at the bottomof the m atrix.
1 3 0
0 1 2 
4.


0 0 0
The leading entry in row 2 is 1,
but not all other entries in the colum n
in which it is located are zeros
( Notice the 3 in the colum n)
Elementary Row Operations
1. Interchanging two rows
2. Replacing a row by a nonzero multiple of
itself
3. Replacing a row by the sum of that row
and a nonzero multiple of another row.
Interchanging Rows
3 
0 2
3 6  12


5  2
2 
R1  R2

3 6  12
0 2

3


5  2
2 
Replacing a row by a nonzero
multiple of itself
3 6  12 1 R1
3
0 2

3



5  2
2 
1 2  4
0 2

3


5  2 2 
Replacing a row by the sum of that row and
a nonzero multiple of another row
1 2  4
0 2

3


5  2 2 
R3  ( 5 R1 )

2
 4
1
0

2
3


0  12 22 
Augmented Matrix Representing a System
of linear Equations
The system of linear equations:
2 y  3z  7
3 x  6 y  12z  3
5 x  2 y  2 z  7
Is represented by the augm entedm atrix:
0 2
3 7 


3
6

12

3


5  2
2  7 
Exam ple(1)
Re ducethe two by two m atrix on the right of the
following augm entedm atrixby applyingbasic row
operations:
1 2 5


3 4 6
Solution:
1 2 5


3
4
6


R2 3 R1
1 2 5 
 0  2  9 


R1  R2
1 0  4


0  2  9 
(  12 R2 )
1 0  4

9 
0
1
2 



Solving a System of Linear Equations
by Reducing its Augmented Matrix
Using Row Operations
Exam ple(1)
Solvethe following system of linear equations:
x  2 y  5

3x  4 y  6
Solution:
We constructthe augm entedm atrix:
operations:
1 2 5


3
4
6


We reducethis m atrix arriving at the m atrix:
1 0  4

9 
0
1
2 

We concludethat :
x  4 &
y
9
2
Exam ple(2)
Solvethe following system of linear equations:
2 y  3z  7
3 x  6 y  12z  3
5 x  2 y  2 z  7
by reducingits augm entedm atrix:
0 2
3 7


3
6

12

3


5  2 2  7 
Solution
0 2
3 7 


3
6

12

3


5  2
2  7 
1
R1
3

1
R2
2

3 6  12  3


0
2
3
7


5  2
2  7 
R1  R 2
1 2  4  1 


3 7 
0 2
5  2 2  7 
1
2
 4  1

3
7 
0
1
2
2 

0  12 22  2

R3  ( 5 R1 )

R1  ( 2 R 2 )

1
2
 4  1


2
3 7 
0
0  12 22  2
1
0
 7  8

3
7 
0
1
2
2 

0  12 22  2
1
0
 7  8

3
7 
0
1
2
2 

0  12 22  2
1
R3
40

1 0  7  8

3
7 
0
1
2
2 

0 0 1 1 
R2  (  32 R3 )

R3 12 R2

R1  7 R3

1 0 0  1


0
1
0
2


0 0 1 1 
1 0  7  8 

3
7 
0
1
2
2 

0 0 40 40 
1 0 0  1

3 7 
0
1
2 2 

0 0 1 1 
Solution of the System
1 0 0  1


0 1 0 2 
0 0 1 1 
Thus,
x  1
y2
z 1
The Idea behind the Reduction
Method
The system of linear equations:
2 y  3z  7
3x  6 y  12z  3
5 x  2 y  2 z  7
The augm entedm atrix:
0 2
3 7


3
6

12

3


5  2 2  7
Interchanging the First & the Second Row
Intercangingt the places
Betweem Eq(1) and Eq(2) :
2 y  3z  7
3x  6 y  12z  3
5 x  2 y  2 z  7

3x  6 y  12z  3
2 y  3z  7
5 x  2 y  2 z  7
0 2
3 7


3
6

12

3


5  2 2  7
R1  R2

3 6  12  3


0
2
3
7


5  2 2  7
Multiplying the first Equation by 1/3
3x  6 y  12z  3
2 y  3z  7
3 6  12  3


0
2
3
7


5  2
2  7 
5 x  2 y  2 z  7

x  2 y  4 z  1
2 y  3z  7
5 x  2 y  2 z  7
1
R1
3

1 2  4  1 


0
2
3
7


5  2 2  7 
Subtracting from the Third Equation 5 times
the First Equation
x  2 y  4 z  1
2 y  3z  7
5 x  2 y  2 z  7
1 2  4  1 


3 7 
0 2
5  2 2  7 

x  2 y  4 z  1
2 y  3z  7
 12 y  22z  2
R3  ( 5 R1 )

1
2
 4  1


2
3 7 
0
0  12 22  2
Subtracting from the First Equation 2 times
the Second Equation
x  2 y  4 z  1
3
7
y z 
2
2
 12 y  22z  2

x  7 z  8
3
7
y z 
2
2
 12 y  22z  2
1
2
 4  1

3
7 
0
1
2
2 

0  12 22  2
R1  ( 2 R2 )

1
0
 7  8

3
7 
0
1
2
2 

0  12 22  2
Adding to the Third Equation 12 times
the Second Equation
x  7 z  8
3
7
y z 
2
2
 12 y  22z  2
1
0
 7  8

3
7 
1
2
2 
0
0  12 22  2

x  7 z  8
3
7
y z 
2
2
40z  40
R3 12 R2

1 0  7  8

3
7 
0 1 2 2 
0 0 40 40 
Dividing the Third Equation by 40
x  7 z  8
3
7
y z 
2
2
40z  40
1 0  7  8

3
7 
0 1 2 2 
0 0 40 40 

x  7 z  8
3
7
y z 
2
2
z 1
1
R3
40

1 0  7  8

3
7 
0 1 2 2 
0 0 1 1 
Adding to the First Equation 7 times the
third Equation
x  7 z  8
3
7
y z 
2
2
z 1

x  1
3
7
y z 
2
2
z 1
1 0  7  8

3
7 
0 1 2 2 
0 0 1 1 
1 0 0  1
R1  7 R3

3 7 

0 1 2 2 
0 0 1 1 
Subtracting from the Second Equation 3/2
times the third Equation
x  1
3
7
y z 
2
2
z 1

x  1
y2
z 1
1 0 0  1

3 7 
0
1
2 2 

0 0 1 1 
1 0 0  1
R2  (  32 R3 )



0 1 0 2 
0 0 1 1 
Systems with infinitely many
Solutions
Exam ple(1) :
Solvethe following system :
x  2 y  3

2 x  4 y  6
Notice that the sec ond equationis the sam e
as the first (Why ?)
Thus, we have only one equation:
x  3 2y
By letting y be any real num berr , we get :
x  3  2r
As shown in the oppositetable:
y=r
x=3-2r
0
3
-1
5
1
1
10
-17
Let' s considerthe augm entedm atrix:
 1 2 3


 2 4 6
We reducethis m atrix arriving at the m atrix:
1 2 3
 

0 0 0 
Thus :
x  2y  3
00
 x  3 2y
R2 ( 2 R )1
Systems with infinitely many
Solutions
Exam ple(2) :
Solvethe following system :
x  2 y  z  0

2 x  y  5 z  0
x  y  4z  0

Notice that when subtracting Eq(1) from Eq(2), we get
Eq(3)
Thus, we have only two independent equations
Subtracting Eq(1) from Eq(3), we get :
3 y  3z  0  y   z
Substituting that in Eq(1), we get :
x  2 z  z  0  x  3 z
Thus letting z be any real num berr , we get : x  3r and y  r
As shown in the oppositetable:
z=r
x=-3r
y=-r
0
0
0
1
-3
-1
-10
30
10
1/3
-1 -1/3
Let ' s consider the augm entedm atrix:
1  2 1 0 


2

1
5
0


1 1 4 0
Let ' s reduce this m atrix arriving at the m atrix:
1 0 3 0


0
1
1
0


0 0 0 0
The last row does not contributeany inf orm ation: 0  0
The other rows gives us :
x  3z  0 and y  z  0
 x  3 z and y   z
By letting z be any real num berr , we get :
x  3r and y  r
Details of reduction
1  2 1 0 
1  2 1 0

 R2  (  R1 ) 

2

1
5
0





1
1
4
0




1 1 4 0
1 1 4 0
1  2 1 0 
3 0 9 0




3  (  R2 )
1  ( 2 R2 )
R
 1 1 4 0 R
 1 1 4 0
0 0 0 0
0 0 0 0
1 0 3 0 


 1 1 4 0
0 0 0 0
1
R1
3
1 0 3 0 


2  (  R1 )
R
 0 1 1 0
0 0 0 0
Systems with no Solution
Exam ple(1) :
Solvethe following system :
x  2 y  3

2 x  4 y  8
Notice that the sec ond equationcontradicts the first (Why ?)
1
Dividing the secon Eq by the num ber2 ( Multiplying by ) :
2
 x  2 y  4  (3)
But the first Eq claim sthat : x  2 y  3
By subtracting Eq(1) from Eq(3), we get the im possiblestatem ent:
0 1
Thus there are no num bersx and y satisfying both equations
Thus the system has no solution
Let' s considerthe augm entedm atrix:
1 2 3


2 4 8
We reducethis m atrix arriving at the m atrix:
1 2 3
 

0 0 2 
Thus :
x  2y  3
02
Which is im possible
R2 ( 2 R )1
Exam ple(2) :
Solve the following system :
x  2 y  4z  6

 y  2z  3
x  y  2z  1

Notice that when subtracting Eq(3) from Eq(1), we get
y  2 z  5 (4)
This equationcontradicts Eq(2), which states that : y  2 z  3
Subtracting Eq(2) from Eq(4), we get the im possiblestatem ent:
02
Thus, there are no real num bersx, y and z saisfying all the three
equations of the system.
 the system has no slution
Let ' s consider the augm entedm atrix:
1 2 4 6 


0
1
2
3


1 1 2 1
Let ' s reduce this m atrix arriving at the m atrix:
1 0 0 0 


0
1
2
0


0 0 0 1
This corresponds to the system( which has no solution)
x  0

 y  2z  0
0  1

Details of the reduction
1 2 4 6
1 2
4 6

 R3 (  R1 ) 

0
1
2
3





0
1
2
3




1 1 2 1
0  1  2  5
1 0
1 0 0 0 
0 0




1  ( 2 R2 )
3  ( R2 )
R
 0 1
2 3  R
 0 1 2 3 
0  1  2  5
0 0 0  2
1 0 0 0


 0 1 2 3
0 0 0 1
1
(  ) R3
2
1 0 0 0 


2  ( 3 R3 )
R
 0 1 2 0
0 0 0 1
A better approach:
1 2 4 6 
 0 1 2 5

 R1 (  R3 ) 

 0 1 2 3
0 1 2 3  
1 1 2 1
1 1 2 1
0 0 0 2 


1  (  R2 )
R

 0 1 2 3
1 1 2 1 
Which corrresponds to the system :
0  2

 y  2z  3
x  y  2z  1

which has no solution
Finding the Inverse of an nXn square Matrix A
1. Adjoin the In identity matrix to obtain
the Augmented matrix [A| In ]
2. Reduce [A| In ] to [In | B ] if possible
Then
B = A-1
Example (1)
2
1 0
Let A  4  2 1 
1 2  10
Solution:
Find A1
1 0
 2 1 0 0


4

2
1
0
1
0


1 2  10 0 0 1
( 4 R1 )  R2

(  R1 )  R3

( R2 )  R3

1 0
 2 1 0 0


0

2
9

4
1
0


1 2  10 0 0 1
1 0  2 1 0 0


0

2
9

4
1
0


0 2  8  1 0 1
1 0  2 1 0 0 


0

2
9

4
1
0
1


0 0
1  5 1 1 
( 2 R3 )  R1

( 9 R3 )  R2

 12 R2

1 0 0  9 2 2


0  2 9  4 1 0 
0 0 1  5 1 1 
1 0 0  9 2
2


0

2
0
41

8

9


0 0 1  5 1
1 
1 0 0  9 2 2

9
41
0
1
0

4
2
2

0 0 1  5 1 1 
Thus,
  9 2 2
A1   412 4 92 
  5 1 1 
1 2 3 
Let A  2 1 2
3 3 5
Solution:
Example (2)
Find A1
1 2 3 1 0 0 


2
1
2
0
1
0


3 3 5 0 0 1
( 2 R1 )  R 2

( 3 R1 )  R3

1 2
3 1 0 0


0

3

4

2
1
0


3 3
5 0 0 1
1 2
3 1 0 0


0

3

4

2
1
0


0  3  4  3 0 1
1 2
3 1
0 0



0

3

4

2
1
0


0 0
0  1  1 1
Thus A has no inverse
(  R 2 )  R3
AnotherWay :
1 2 3 1

2 1 2 0
3 3 5 0
1
(  R1 )  R2


3
3
0 0

1 0
0 1
2 3 1 0 0

3 5 1 1 0
3 5 0 0 1
1 2 3 1 0 0



3 3 5 1 1 0 
0 0 0  1  1 1
Thus A has no inverse
(  R2 )  R3
Inverse Matrix
The formula for the inverse of a 2X2 Matrix
Exam ple:
Let
a b 
A

c
d


&
det A  ad  bc
If det A  0, Then :
1 d b
A 
det A  c a
1
 2 5
LeT : A  

 3 9
2 5
det A 
 18  15  3  0
3 9
1  9  5 1  9  5
A 

det A  3 2  3  3 2 
5

3


3

2 
 1

3 

1
Check that :
Checking
AA1  I 2  A 1 A
5
5


3

3

 2 5 
3   1 0  
3   2 5
 

 3 9 
 


   1 2  0 1    1 2   3 9 
3 
3 


Using the Inverse Matrix
to Solve System of Linear Equations
Solvethe following system of linear equations:
2 x  5 y  9

3x  9 y  15
Re writing in m atrix form :
2 x  5 y   9 
3x  9 y   15

  
 2 5  x   9 

    
3 9  y  15
5
5


3

3

2
5
x

  
3
3 9 


2  3 9  y  
2  15
 1

 1

3 
3 


5


3
(
9
)

(

)(
15
)

 x 
3
 I2    

 y  (1)(9)  ( 2 )(15)
3


 x   2
  
 y  1 
Thus,
The Solutionis :
x  2 and y  1
Problem
Solvethe following system of linear equations:
2 x  y  z  1

3x  2 y  z  2
2 x  y  2 z  1

If given that :
1
2 1 1 
 3  1  1
3 2 1    4 2 1 




2 1 2
  1 0 1 
2 x  y  z  1

3 x  2 y  z  2
 2 x  y  2 z  1

Re writing in m atrix form :
2 x  y  z  1  1 
 3x  2 y  z    2 

  
 2 x  y  2 z   1
2 1 1  x   1 
 3 2 1   y    2 
2 1 2  z   1
 3  1  1 2 1 1   x   3  1  1  1 
  4 2 1  3 2 1   y    4 2 1   2 
  1 0 1  2 1 2  z    1 0 1   1
 x  2 
 I 3  y     1 
 z   2
 x  2 
  y     1 
 z   2
Thus,
The Solutionis :
x2 ,
y  1 and z  2
Homework
6.1
6.2
6.3
6.5
6.6
1.
Examples: 3
Exercises: 17 - 20
Examples: 1, 2, 4, 5 and 6
Exercises: odd numbered:1—17 and 25,29,35,37,39,41
Examples: 1, 2, 3, 4, 5, 7, 10, 11, 12 and 13.
Exercises:19, 21, 23, 25, 27, 31, 33, 37, 51, 53, 57, 59, 61
Examples: 2, 3.a and 4.
See given exercises.
Examples: 1 - 6
Exercises: odd numbered:1—15, 21, 27, 29, 35 and 37.