1.
School of Management
By: Mr. Manoj Kumar Mishra
1/30/2023 Business Mathematics 1
UNIVERSITY OF STEEL TECHNOLOGY
AND MANAGEMENT
Matrices Algebra
For BBA & B.Com

2.
Topic To be Covered
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1. Solution of Simultaneous Linear Equations
2. Cramer Rule
3. Row echelon form of matrix
4. Gauss Jordan Method

3.
Singular Matrix
A square matrix A is said to be singular if lAl = 0 .
A is non-singular if 
A 0.
For Example:
1 1 3
1 3 3
5 3 3

 

 
 
 
Let A=
A is a singular matrix .
A=1(9+9)+1(3+15)+3(3-15)
= 18+18-36
= 0
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4.
Non-Singular Matrix
Let B=
1 1 1
2 1 1
1 2 3
 

 
 

 
B is a non-singular matrix.
B= 1(-3+2)-1(6-1)+1(-4+1)
= -1 – 5 – 3
= -9 0

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5.
Example -1
Find the value of x for which the matrix is singular.
x 1 0
A = 2 -1 1
3 4 -2
 
 
 
 
 
For matrix A to be singular
   
A = 0
x 1 0
2 -1 1 = 0
3 4 -2
-1 -4 - 3 = 0
7
-2x +7 = 0 x =
2
x 2 4

 
 
Solution:
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6.
Solution of Simultaneous Linear Equations
(Matrix Method)
Let the system of 3 linear equations be
1 1 1 1
2 2 2 2
3 3 3 3
a x +b y +c z = d
a x +b y +c z = d
a x +b y + c z = d
This system of linear equation can be written in matrix form as
1
1 1 1
2 2 2 2
3 3 3 3
d
a b c x
a b c y = d
a b c z d
 
   
 
   
 
   
 
   
 
   
 
AX = B ... i

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7.
Solution of Simultaneous Linear Equations
(Matrix Method)
1
X = (adjA)B
A

The matrix A is called the coefficient matrix of the
system of linear equations.
Multiplying (i) by A–1, we get
-1
If A 0 i.e. A is non - singular, then A exists.

 
1 1
A AX A B
 

 
1 1
A A X A B
 
 
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8.
Important Results
(i) If A is a non-singular matrix, then the system of equations given by
AX = B has a unique solution given by X = A–1B
(ii) If A is a singular matrix and (adjA)B = 0, then the system of
equations given by AX = B is consistent with infinitely many
solutions.
(iii) If A is a singular matrix and (adjA)B  0, then the system of
equations given by AX = B is inconsistent.
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9.
Question-1
Using matrix method, solve the following system of linear equations
x + 2y -3z = -4
2x + 3y + 2z = 2
3x - 3y - 4z = 11
Solution:
The given system of equations is
x + 2y - 3z = -4 ...(i)
2x + 3y + 2z = 2 …(ii)
3x -3y - 4z = 11 …(iii)
1 2 -3 x -4
or 2 3 2 y = 2
z
3 -3 -4 11
     
     
     
     
AX = B

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10.
Solution (Cont.)
1 2 -3 x -4
where A= 2 3 2 , X= y , B= 2
z
3 -3 -4 11
     
     
     
     
 
2 2 1 3 3 1
-1
1 2 -3
A = 2 3 2
3 -3 -4
1 0 0
= 2 -1 8 Applying C C -2C and C C +3C
3 -9 5
=1(-5+72)=67 0
A exists.
 


ij ij ij
Let C be the cofactor a in A = a , then
 
 
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11.
Solution Cont.
 
11 12 13
c =(-12+6) c =- -8-6 c =(-6-9)
=-6 =14 =-15
21 22 23
c =-(-8-9) c =(-4+9) c =-(-3-6)
=17 =5 = 9
31 32 33
c =(4+9) c =-(2+6) c =(3-4)
=13 = -8 = -1
31 32 33
c =(4+9) c =-(2+6) c =(3-4)
=13 = -8 = -1
-6 17 13
14 5 -8
-15 9 -1
 
 
 
 
T
-6 14 -15
adjA = 17 5 9 =
13 -8 -1
 
  
 
 
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12.
Solution (Con.)
-1
Now, X= A B
-6 17 13 -4
1
X= 14 5 -8 2
67 -15 9 -1 11
   
    
   
   
x 201 3
1
y = -134 = -2
67
z 67 1
x=3 , y=-2 , z=1
     
      
     
     

-1
-6 17 13
1 1
A = .adj A = 14 5 -8
A 67 -15 9 -1
 
  
 
 
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13.
Practice Question
Using matrices, solve the following system of
equations
x + y + z = 6
x + 2y + 3z = 14
x + 4y + 7z = 30
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14.
Cramer’s Rule
• System of Linear Equations
• How to solve using Cramer’s Rule
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15.
Introduction
• Cramer’s Rule is a method for solving linear
simultaneous equations. It makes use of
determinants and so a knowledge of these is
necessary before proceeding.
• Cramer’s Rule relies on determinants
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16.
Coefficient Matrices
• You can use determinants to solve a system of
linear equations.
• You use the coefficient matrix of the linear
system.
• Linear System
ax+by=e cx+dy=f






d
c
b
a
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Coeff Matrix

17.
Using Cramer’s Rule
to Solve a System of Three Equations
Define  
   
11 12 13
21 22 23
31 32 33
1 1
2 2
3 3
a a a
A a a a
a a a
x b
x x and B b
x b
 
 
  
 
 
   
   
   
   
   
   
3
1 2
1 2 3
If 0, then the system has a unique solution
as shown below (Cramer's Rule).
, ,
D
D
D D
x x x
D D D
         
   
    
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18.
Using Cramer’s Rule
to Solve a System of Three Equations
where
11 12 13 1 12 13
12 22 23 1 2 22 23
13 32 33 3 32 33
11 1 13 11 12 1
2 12 2 23 3 12 22 2
13 3 33 13 32 3
a a a b a a
D a a a D b a a
a a a b a a
a b a a a b
D a b a D a a b
a b a a a b
  
  
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19.
Example 1
Consider the following equations:
    
 
1 2 3
1 2 3
1 2 3
2 4 5 36
3 5 7 7
5 3 8 31
where
2 4 5
3 5 7
5 3 8
x x x
x x x
x x x
A x B
A
  
   
   


 
 
 
 
 

 
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20.
Example 1
   
1
2
3
36
7
31
x
x x and B
x
   
   
   
   
   

   
2 4 5
3 5 7 336
5 3 8
D

   

1
36 4 5
7 5 7 672
31 3 8
D

  
 
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21.
Example 1
2
2 36 5
3 7 7 1008
5 31 8
D   
 
3
2 4 36
3 5 7 1344
5 3 31
D

   

1
1
2
2
3
3
672
2
336
1008
3
336
1344
4
336
D
x
D
D
x
D
D
x
D

  

   


  

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22.
Row Echelon Form
•To be in this form, a matrix must have the
following properties.
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23.
Examples – Row-Echelon Form
•Determine whether each matrix is in row-echelon
form. If it is, determine whether the matrix is in
reduced row-echelon form.
a. b.
c. d.
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24.
Examples – Row-Echelon Form
e. f.
•Solution:
•The matrices in (a), (c), (d), and (f) are in row-echelon form.
•The matrices in (d) and (f) are in reduced row-echelon form
because every column that has a leading 1 has zeros in every
position above and below its leading 1.
cont’d
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25.
Using Matrices to Solve Systems of Equations
The use of Elementary Row Operations is required when solving a system of
equations using matrices.
12
7
3
1
13
5
0
2
8
2
6
3



Elementary Row Operations
I. Interchange two rows.
II. Multiply one row by a nonzero number.
III. Add a multiple of one row to a different row.
8
2
6
3
13
5
0
2
12
7
3
1



16
4
12
6
13
5
0
2
12
7
3
1



16
4
12
6
13
5
0
2
12
7
3
1



16
4
12
6
11
19
6
0
12
7
3
1





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26.
Use matrices to solve the following systems of equations.
12
2
1
5
15
4
3
2
4
1
1
1










12
2
1
5
7
2
1
0
4
1
1
1





32
7
6
0
7
2
1
0
4
1
1
1





32
7
6
0
7
2
1
0
4
1
1
1




10
5
0
0
7
2
1
0
4
1
1
1




2
1
0
0
7
2
1
0
4
1
1
1




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