ENGINEERING MATHEMATICS-20SC01T

UNIT : 1 - MATRICES AND DETERMINANTS

Matrices
Matrix: A matrix is a rectangular arrangement of a numbers in a rows and
columns with in a closed brackets called as matrix.
Matrices is a plural form of matrix.
1 2
Eg:[ ]
3 4
Order of a matrix:
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If number of rows and columns are represented by m and n then order of a
matrix can be defined as 𝑚 × 𝑛
Types of matrices:
1. Square matrix : A matrix in which the number of rows is equal to the
1 2
number of columns ,is called a square matrix .e.g.: [ ]
5 4
2. Row matrix : A matrix in which it having only one row but many
number of columns is called row matrix .e.g.: [5 4]
3. Column matrix: A matrix having only one column but many number
2
of rows is called column matrix e.g.: [ ]
4
4. Diagonal matrix :It’s a square matrix in which the non principal
2 0
diagonal elements are equal to zero .e.g.: [ ]
0 1
5. Scalar matrix : It’s a diagonal matrix in which the principal diagonal
2 0
elements are same other than 1. e.g.: [ ]
0 2
6. Identity (unit) matrix : It’s a scalar matrix in which the principal
1 0
diagonal elements are equal to 1. .e.g.: [ ]
0 1
7. Null matrix : A matrix in which all the elements are zero is called as
0 0
null (zero ) matrix .e.g.: [ ]
0 0
 Type equation here.

8. Symmetric matrix : A square matrix is said to be symmetric if it

remains same when rows are changed into columns or columns are
𝑎 𝑏 𝑐 𝑎 𝑏 𝑐
1
changed into rows .e.g.: 𝐴 = [𝑏 𝑐 𝑎] 𝐴 = [𝑏 𝑐 𝑎]
𝑐 𝑎 𝑏 𝑐 𝑎 𝑏
9. Equal matrix : If two matrices are said to be equal
i) They have the same order and
ii) Corresponding elements are equal
𝑥 𝑦 5 3
Eg: [ ]=[ ] if x=5, y=3, w=2, z=4
𝑤 𝑧 2 4
Algebra of matrices:
Transpose of a matrices:
The matrix obtained by interchanging rows into columns or columns into
rows is called transpose of a matrix and it is represented by 𝐴1 (𝑜𝑟)𝐴𝑇 .
1
e.g.: 𝐴 = [1 2]1×2 𝐴 𝑇 = [ ]
2 2×1

Addition of matrices

If A and B are two matrices of same order m  n then their sum A  B is also a
matrix of order m  n and is obtained by adding corresponding elements of A and B.

w x a b  w  a x  b 
Examples: (1) If X    and Y    then X  Y   
 y z c d   y  c z  d

5 3 3 4 5  3 3  4  8 7 
(2) If A  1 2 and B  0 1  then A  B  1  0 2  1  1 3
   
 2 4 1 5   2  1 4  5 3 9 

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 Type equation here.

Subtraction of matrices

If A and B are two matrices of same order m  n then their difference A  B is also
a matrix of order m  n and is obtained by subtracting elements of B from corresponding
elements of A.

w x a b  w  a x  b 
Examples: (1) If X    and Y    then X  Y   
 y z c d   y  c z  d

5 3 3 4  5  3 3  4   2 1
(2) If A  1 2 and B  0 2 then A  B  1  0 2  2   1 0 
   
 2 4 5 1   2  5 4  1   3 3 

Scalar Multiplication of a matrix

If A is a matrix and k is a scalar then the scalar multiplication kA is a matrix

obtained by multiplying each elements of A by scalar k .

w x  kw kx 
Examples: (1) If A    then kA   
 y z  ky kz 

 4 1 0  8 2 0
(2) If B    then 2 B   
 1 2 3   2 4 6

PROBLEMS

1 3 
Example 1: If A    then find 4A .
 4 2 

1 3 
Solution: Given A 
 4 2 

1 3 
 4A  4  
 4 2 

 4 1 43 
 4A   
 4  4 4  (2) 

 4 12 
 4A   
16 8

 1 0 3 
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Example 2: If B  
3 2 4 
 then find 2B .

Matrices and Determinants-1.4

e.in Page 3
 Type equation here.

1 0 3
Solution: Given B   
3 2 4 

1 0 3
 2B  2  
3 2 4 

 2 0 6
 2B   
6 4 8 

 3 1  2 0
Example 3: If X    and Y    then find X  Y .
 2 5 4 7

 3 1 2 0
Solution: Given X   and Y   
 2 5 4 7

 3 1  2 0 
Consider X Y    
 2 5  4 7 


3  2 1  0 
X Y   
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2  4 5  7


5 1 
X Y  
me.in

6 12

1 2 6 0 2 1 
Example 4: If A   5 3 7  and B   7 5 1 then find
  A B.
 0 1 4   2 6 4 

1 2 6  0 2 1 
Solution: Given  
A  5 3 7  and B  7 5 1
0 1 4  2 6 4 

1 2 6   0 2 1 
Consider A  B  5 3 7   7 5 1
0 1 4  2 6 4 

1  0 22 6 1 
 
A  B  5  7 35 7  (1) 
0  2 (1)  6 4  4 

 1 4 7
 A  B  12 8 6 
 2 5 8 

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 Type equation here.

2 1  5 1
Example 5: If A   6 2 and B   2 0  then find
  3B  2 A .
 4 3  1 4 

2 1 5 1
Solution: Given A  6 2 and B   2 0 
 
 4 3  1 4 

5 1 2 1
Consider  
3B  2 A  3  2 0   2 6 2 
1 4   4 3

15 3  4 2
  6 0   12 4 
 3 12   8 6 

15  4 3  2
 6  12 0  4 
 3  8 12  6 

 11 5
3B  2 A   6 4 
 5 6 

1 2   y 3  7 5
Example 6: If     then find values of x and y .
 4 x   2 5  6 5

 1 2   y 3  7 5 
Solution: Given    
 4 x   2 5  6 5

1  y 2  3 7 5
  4  2 x  5   6 5
   

1  y 5   7 5 Two matrices are said to be equal

  6 
 x  5 6 5 if they are of same order and their
corresponding elements are equal.
By equality of matrices, we get
1 y  7 And x  5  5

 y  7  1 And x  55

 y6 And x0

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 Type equation here.

 2 1 1 5 
Example 7: If A  B    and A  B   4 6 then find A .
3 4   

 2 1
Solution: Given A  B    --- (1)
3 4 

and A  B  
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1 5 
--- (2)

 4 6
Adding (1) and (2), we get

 2 1 1 5 
( A  B)  ( A  B)    
 3 4   4 6

 2  1 1  5
 A B A B   
3  4 4  6 

3 4 
 2A   
7 2 

1 3 4 
 A
2 7 2 

3 / 2 2 
 A 
7 / 2 1

ASSIGNMENT PROBLEMS

 3 1
1. If A    then find 3A .
2 5 

 3 0 2
2. If B    then find 2B .
 1 4 5 

 3 6
If A  15 9  then find A .
1
3.
3
12 3

 1 0 3
4. Find 2X given that X   2 5 7  .
 1 6 4

3 5 1 2 3 1
5. If X  6 2 4  and Y   4 1 0 then find X  Y .
 
1 7 0  7 5 1 

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 Type equation here.

 3 2 1   2 1 3 
6. If A    and B    then find 2A  B .
 4 0 1 5 3 6

 3 4 4 1
7. If A    and B    then find 3A  B .
 1 2  3 0

1 5   2 3
8. If A    and B    then verify that A  B  B  A .
7 6   2 5

 x 2  2 4  5 6
9. If     then find the values of x and y .
 3 5   1 y   2 7 

5 2  3 6 
10. If A  B    and A  B    then find matrix A and B
0 9  0 1

Multiplication of matrices:

If matrix A of order 𝑚 × 𝑛 and matrix B of order 𝑛 × 𝑝 then AB is defined as the matrix

of order 𝑚 × 𝑝 it is obtained by multiplying by elements of 1st row of matrix A by
corresponding elements of 1st column of matrix B and adding there products .

To perform multiplication of two matrices, we should make sure that the number of
columns in the 1st matrix is equal to the rows in the 2nd matrix. Therefore, the resulting
matrix product will have a number of rows of the 1st matrix and a number of columns of
the 2nd matrix. The order of the resulting matrix is the matrix multiplication order.

𝑔 ℎ
E.g. let 𝐴 = [
𝑑
𝑎 𝑏
𝑒
𝑐
𝑓
] 𝐵 = [𝑖
𝑘
𝑗]
𝑙
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𝑎𝑔 + 𝑏𝑖 + 𝑐𝑘 𝑎ℎ + 𝑏𝑗 + 𝑐𝑙
Then 𝐴𝐵 = [ ]
𝑑𝑔 + 𝑒𝑖 + 𝑓𝑘 𝑑ℎ + 𝑒𝑗 + 𝑓𝑙

Problems

1 2 2 1
1. If 𝐴 = [ ] and 𝐵 = [ ] then find the product of two matrices.
3 4 3 4
1×2+2×3 1×1+2×4
Soln: AB = [ ]
3×2+4×3 3×1+4×4
2+6 1+8
AB = [ ]
6 + 12 3 + 16
8 9
AB = [ ] order of AB is 2 × 2
18 19
2.

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 Type equation here.

DETERMINANT
Definition:
A determinant is a real number associated with every square matrix.
The determinant of a square matrix A is denoted by "det A" or | A |.
1 2 1 2
Example: If A = [ ] , then det A = |A| = | |
3 6 3 6
Expansion of Determinant of a 2×2 Matrix:
a b a b
IF A = [ ] then det A = |A| = | |
c d c d
|𝐀| = ad – bc
1 2 1 2
Example: 1) Let A [ ] then |A| = | |
3 8 3 8
= (1) (8) - (2) (3)
=8–6
|𝐀| =2.
3 5] 3 5|
2) Let B = [ then |𝐵| = |
1 −4 1 −4
= (3) (-4) – (1) (5)
= -12-5
|𝑩| = -17

Expansion of Determinant of a 3×3 Matrix

𝑎1 𝑏1 𝑐1
Let C = [𝑎2 𝑏2 𝑐2 ]
𝑎3 𝑏3 𝑐3
𝒃 𝒄𝟐 𝒂𝟐 𝒄𝟐 𝒂𝟐 𝒃𝟐
Then, |𝑪|= 𝒂𝟏 | 𝟐 | - 𝒃𝟏 |𝒂 𝒄𝟑 | + 𝒄𝟏 |𝒂𝟑 |
𝒃𝟑 𝒄𝟑 𝟑 𝒃𝟑
+ − +
Note: Sign rule for expansion of order 3 determinant [− + −]
+ − +

Example:
2 3 4 2 3 4
Let A = [5 8 1] then |𝐴|=|5 8 1|
3 0 2 3 0 2
8 1 5 1 5 8
Value of |𝐴| = 2| |– 3 | | + 4| |
0 2 3 2 3 0

= 2(16 – 0) – 3(10 – 3) + 4(0 – 24)

=2 (16) – 3(7) + 4(– 24)

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 Type equation here.

= 32 – 21 – 96
|𝐀| = – 85

SINGULAR AND NON-SINGULAR MATRIX

Definition
A square matrix B is said to be singular if its determinant value is zero, i.e.|𝑩| = 𝟎,
otherwise non -singular.
1 2
Example: 1) Let B=[ ] then
3 6
|B| = |1 2|
3 6
= (1) (6) – (2) (3)
=6–6=0
|𝐁| = Zero Therefore B is singular.
10 20
2) Let A = [ ] then
2 4
|A| = |10 20|
2 4
= (10) (4) – (2) (20)
= 40 – 40
|𝐀| = 0, therefore A is singular.

3) Let A=[0
2 0
5
0
0] then
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0 0 0
2 0 0
|A| = |0 5 0|
0 0 0
= 2(0 – 0) – 0(0 – 0) + 0(0 – 5)
|𝐀| = 0 Therefore A is singular.
𝐒𝐎𝐋𝐕𝐄𝐃 𝐏𝐑𝐎𝐁𝐋𝐄𝐌𝐒
I. Check whether the following matrices are singular:
5 25
1) A=[ ]
1 5
5 25
Soln: |A|= | |
1 5
= (5) (5) – (1) (25)
= 25 – 25
|𝐀| = 0 therefore A is singular matrix.

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 Type equation here.

2 5
2) A = [ ]
1 5
2 5
Soln: |A|= |
1 5
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= (2) (5) – (1) (5)
= 10 – 5
|𝐀| = 5≠ 0 Therefore A is non-singular matrix.

4 12
3) A = [ ]
2 6
4 12
Soln: |A| = | |
2 6
= (4) (6) - (12) (2)
= 24 – 24
|𝐀| = 0 therefore A is singular matrix.
3 1 2
4) A=[6 2 4]
7 5 6
3 1 2
Soln: |A|=|6 2 4|
7 5 6
= 3(12-20)-1(36-28) +2(30-14)
= 3(-8)-1(8) +2(16)
= -24-8+32
= -32+32
|𝐀| = 0 therefore A is singular matrix.

II Find the value of 𝑥

1 5 7
5) If 2
| 𝑥 14| = 0
3 1 2
1 5 7
Soln: Given that |2 𝑥 14| = 0
3 1 2
Expanding the determinant, we get,
1 (2 𝑥 – 14) – 5 (4 – 42) + 7 (2 – 3 𝑥 ) = 0
2 𝑥 – 14 – 5 (– 38) + 14 – 21 𝑥 = 0
2 𝑥 – 14 + 190 +14 -21 𝑥 = 0
-19 𝑥 +190 = 0
19 𝑥 =190
190
𝑥= 19
𝒙 = 10

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 Type equation here.

1 2 3
6) If A = [𝑥 4 1] is Singular Matrix
3 6 5
Soln: Given A is singular i.e. |A| = 0
1 2 3
|A| = |𝑥 4 1| = 0
3 6 5
Expanding the determinant, we get,
1 (20 – 6) – 2 (5 𝑥 – 3) + 3(6 𝑥 – 12) = 0
1(14) -2 (5 𝑥 -3) +3(6 𝑥 -12) = 0
14-10 𝑥 +6+18 𝑥 -36 =0
-16+8 𝑥 = 0
8 𝑥 = 16
16
𝑥=
8
𝒙=2

I) EVALUATE THE FOLLOWING

1 2 3
1 2 3 −2
1) | | (2) | | (3) |3 2 1|
3 6 1 −1
2 3 1
1 −1 2 1 1 1
(4) |2 0 1| (5) |3 2 1|
3 2 1 2 3 1

(II) FIND THE VALUE OF ′𝑥′ IN THE FOLLOWING

𝑥 −1 𝑥 8
1) | |=0 (2) | |=0
2 1 8 𝑥
1 2 3 1 2 9
(3) |4 5 𝑥 |=0 (4) |2 𝑥 0 |=0
7 8 9 3 7 −6
2 𝑥 − 1 −3
(5) |1 −2 4 |=3x-1.
3 −1 5

1.5 : CRAMER’S RULE-TWO VARIABLES

SOLUTION OF SYSTEM OF TWO SIMULTANEOUS LINEAR
EQUATIONS BY CRAMER’S RULE (DETERMINANT METHOD)

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 Type equation here.

Consider the system of equations:

𝑎1 𝑥 + 𝑏1 𝑦 = 𝑐1 and 𝑎2 𝑥 + 𝑏2 𝑦 = 𝑐2

𝑎1 𝑏1
Δ=| | = 𝑎1 𝑏2 - 𝑎2 𝑏1 (det of coefficient)
𝑎2 𝑏2

𝑐1 𝑏1
𝛥1 = | | = 𝑐1𝑏2 - 𝑐2 𝑏1 (det obtained by replacing 1st column f Δ by
𝑐2 𝑏2

constants𝑐1 , 𝑐2)

𝑎1 𝑐1
𝛥2 = |𝑎 𝑐2 | = 𝑎1 𝑐2- 𝑎2 (det obtained by replacing 2nd column of Δ by
2

constants𝑐1 , 𝑐2)

𝛥1 𝛥
Therefore, x= and y = 𝛥2. Provaided (𝛥 ≠ 0 )
𝛥

 Cramer’s Rule can also be used to solve the simultaneous equations of ‘n’
variables

Example
Solve 2x +3y =1; 3x –y = – 2 by Cramer’s rule.

2 3
Soln: Δ = | | = (2) (– 1) – (3) (3) = – 2 – 9 = – 11
3 −1

1 3
𝛥1 = | | = (1) (– 1) – (– 2) (3) = 1+6 = 5
−2 −1

2 1
𝛥2 = | | = (2) (– 2) – (1) (3) = – 4 – 3 = – 7
3 −2
𝜟𝟏 𝟓 𝜟𝟐 −𝟕 𝟕
Therefore, 𝒙 = = −𝟏𝟏 and 𝒚 = = −𝟏𝟏 = 𝟏𝟏
𝜟 𝜟

Worked Examples on Cramer’s Rule

1) Solve the equations 2x + y =1; 3x + 2y =1 by method of determinants.

Matrices and Determinants-1.4 Page 12

 Type equation here.

Soln: Given system of equation is 2x + y =1

3x +2y =1

2 1
𝑙𝑒𝑡 ∆= | |= 4−3 =1
3 2
www.mathswithme.in
1 1
∆1 = | |=2−1=1
1 2

2 1
∆2 = | | = 2 − 3 = −1
3 1

∆1 1 ∆2 −1
𝑛𝑜𝑤 𝑥 = = = 1 𝑎𝑛𝑑 𝑦 = = = −1
∆ 1 ∆ 1

∴ 𝒙 = 𝟏 𝒂𝒏𝒅 𝒚 = −𝟏

2) Solve the equations x + y = 3; 2x + 3y = 8 by Cramer's rule

Soln: Given system of equation is x+y=3

2x + 3y = 8

1 1
𝑙𝑒𝑡 ∆= | | = 3 − 2 = 1;
2 3

3 1
∆1 = | |=9−8=1 𝑎𝑛𝑑
8 3

1 3
∆2 = | | = 8−6 = 2
2 8

∆1 1 ∆2 2
𝑛𝑜𝑤 𝑥 = = = 1 𝑎𝑛𝑑 𝑦 = = =2
∆ 1 ∆ 1

∴ 𝒙 = 𝟏 𝒂𝒏𝒅 𝒚 = −𝟏

3) Solve 2x - 3y = 5, 7x – y = 8 by Cramer's rule.

Soln: Given system of the equations is 2x – 3y = 5

7x – y = 8

2 −3
𝑙𝑒𝑡 ∆= | | = −2 + 21 = 19
7 −1

5 −3
∆1 = | | = −5 + 2 = 19
8 −1

2 5
∆2 = | | = 16 − 35 = −19
7 8
Matrices and Determinants-1.4 Page 13
 Type equation here.

∆1 19 ∆2 −19
𝑛𝑜𝑤 𝑥 = = = 1 𝑎𝑛𝑑 𝑦 = = = −1
∆ 19 ∆ 19

∴ 𝒙 = 𝟏 𝒂𝒏𝒅 𝒚 = −𝟏

4) Solve 3u + 4v = 10; 2u – 3v = 1 by Cramer's rule

Soln: Given system of equation is 3 𝑢 + 4 𝑣 = 10

2𝑢–3𝑣 =1

3 4
𝑙𝑒𝑡 ∆ = | | = −9 − 8 = −17;
2 −3

10 4
∆1 = | | = −30 − 4 = −34
1 −3

3 10
∆2 = | | = 3 − 20 = −17
2 1
∆1 −34 ∆2 −17
𝑛𝑜𝑤 𝑢 = = = 2 𝑎𝑛𝑑 𝑢 = = =1
∆ −17 ∆ −17

∴ 𝑢 = 𝟐 𝐚𝐧𝐝 𝑣 = 𝟏

Applications of Cramer’s rule to mesh analysis

The following is an example to demonstrate the application of crammer’s rule to solve
mesh current analysis problems;
Mesh Current Analysis Circuit

One simple method of reducing the amount

of maths involved is to
Analyse the circuit using Kirchhoff’s Current Law equations to determine the
currents, I1 and I2 flowing in the two resistors. Then there is no need to calculate the
current I3 as it is just the sum of I1 and I2. So Kirchhoff’s second voltage law simply
becomes:
 Equation No 1 : 10 = 50I1 + 40I2
 Equation No 2 : 20 = 40I1 + 60I2

Matrices and Determinants-1.4 Page 14

 Type equation here.

The above equations can be solved using crammer’s rule, resulting in detection of I1 and
I2.

EXERCISE www.mathswithme.in
Solve by Cramer’s Rule:
1) 3x+y = 4, x +3y = 4
(2) 2x -3y =5, 7x –y = 8
(3) 5x+3y =1, 3x +5y = -9
(4) Y=4, x+3y=4
(5) 3x +4y-7=0, 7x-y-6=0
(6) R1+ 4R2 =70, 2R2 -3R1 = 0

1.6 : CRAMER’S RULE-THREE VARIABLES

SOLUTION OF SYSTEM OF THREE SIMULTANEOUS LINEAR

EQUATIONS BY CRAMER’S RULE (DETERMINANT METHOD)
Consider the equations a1 x + b1y + c1z = d1

a2 x + b2y + c2z = d2

a3 x + b3y + c3z = d3 ,

In three variables x, y and z.

𝑎1 𝑏1 𝑐1
Δ = |𝑎2 𝑏2 𝑐2 | (det of coefficient)
𝑎3 𝑏3 𝑐3

𝒅𝟏 𝑏1 𝑐1
Δ1 = |𝒅𝟐 𝑏2 𝑐2| (det obtained by replacing 1st column of Δ by constants)
𝒅𝟑 𝑏3 𝑐3

𝑎1 𝒅𝟏 𝑐1
Δ2 = |𝑎2 𝒅𝟐 𝑐2 | (det obtained by replacing 2nd column of Δ by constants)
𝑎3 𝒅𝟑 𝑐3

𝑎1 𝑏1 𝒅𝟏
Δ3 = |𝑎2 𝑏2 𝒅𝟐 | (det obtained by replacing 3rd column of Δ by constants)
𝑎3 𝑏3 𝒅𝟑
Matrices and Determinants-1.4 Page 15
 Type equation here.

𝚫𝒙 𝚫𝒚 𝚫𝒛
Therefore, x= , y= and z = provided 𝚫 ≠ 𝟎
𝚫 𝚫 𝚫

Example:

Solve using cramer’s rule: 5x – 2y – 3z =17 , 3x – y+z =15 and x+y – 6z = – 13

Solution: Given system Equation of

5x – 2y – 3z =17
3x – y+z =15
x+y – 6z = – 13
then
5 −2 −3
Δ= |3 −1 1 |
1 1 −6
=5(6 – 1)+ 2(– 18 – 1) – 3(3+1) = 25 – 38 – 12 = – 25.

𝟏𝟕 −2 −3
Δ1 = | 𝟏𝟓 −1 1 |
−𝟏𝟑 1 −6
= 17(6 – 1)+ 2(– 90+13) – 3(15 – 13) = 85 – 154 – 6 = – 75.

5 𝟏𝟕 −3
Δ2 = |3 𝟏𝟓 1|
1 −𝟏𝟑 −6
=5(– 90+13) – 17(– 18 – 1) – 3(– 39 – 15)
= – 385+323+162=100.
www.mathswithme.in
5 −2 𝟏𝟕
Δ3 = |3 −1 𝟏𝟓 |
1 1 −𝟏𝟑

=5(13 – 15)+ 2(– 39 – 15)+17(3+1)

= – 10 – 108+68= – 50.

Δ𝑥 −75 Δ𝑦 100 Δ𝑧 −50

x= = −25 = 3, y= = −25 = −4, z= = −25 = 2.
Δ Δ Δ

Therefore, x=3 , y=-4 and z=2

Matrices and Determinants-1.4 Page 16

 Type equation here.

Problems on 3 linear equations using cramer’s rule:

1)solve the following equation using Cramer’s rule

x +y +z = 7; 2x +3y +2z = 17; 4x +9y +z = 37
Solution : Given system of equation is
x +y +z=7
2x +3y +2z=17
4x +9y +z=37
1 1 1
Let ∆ =|2 3 2|= 1 (3-18)-1 (2-8)+1 (18-12)= -15+6+6= -3
4 9 1
7 1 1
∆1= |17 3 2|=7(3-18)-1(17-74)+1(153-111)
37 9 1
= -105+57+42= -6
1 7 1
∆2= |2 17 2|1(17-74)-7(2-8)+1(74-68)= -57+42+6= -9
4 37 1
1 1 7
∆3= |2 3 17|=1(111-153)-1(74-68)+7(18-12)
4 9 37
= -42-6+42= -6
∆1 −6 ∆2 −9 ∆3 −6
𝑛𝑜𝑤 𝑥 = = = 2;y = = = 3 𝑎𝑛𝑑 𝑧 = = =2
∆ −3 ∆ −3 ∆ −3
∴ 𝑥 = 2, 𝑦 = 3, 𝑧 = 2
2)solve the following equation by Cramer's rule
2x +y –z = 3; x +y +z = 1; x -2y -3z = 4
Solution :Given system of equation is 2x +y –z = 3;
x +y +z = 1
x -2y -3z = 4
2 1 −1
Let ∆=|1 1 1 | =2(-3+2)-1(-3-1)-1(-2-1)=-2+4+3 = 5
1 −2 −3
3 1 −1
∆1 = |1 1 1 |=3(-3+2)-1(-3-4)-1(-2-4)= -3+7+6= 10
4 −2 −3
2 3 −1
∆2 = |1 1 1 |=2(-3-4)-3(-3-1)-1(4-1)= -14+12-3= -5
1 4 −3

Matrices and Determinants-1.4 Page 17

 Type equation here.

2 1 3
∆3 = |1 1 1|=2(4+2)-1(4-1)+3(-2-1)= 12-3-9= 0
1 −2 4
∆1 100 ∆2 −5 ∆3 0
𝑛𝑜𝑤 𝑥 = = = 2; 𝑦 = = = −1 𝑎𝑛𝑑 𝑧 = = =0
∆ 5 ∆ 5 ∆ 5
∴ 𝑥 = 2, 𝑦 = −1, 𝑧 = 0
Applications of cramer’s rule to mesh analysis
The following is an example to demonstrate the application of cramer’s rule to solve
mesh current analysis problems;
Mesh Current Analysis Circuit

One simple method of reducing the amount

of math’s involved is to
analyse the circuit using Kirchhoff’s Current Law equations to determine the
currents, I1 and I2 flowing in the two resistors. Then there is no need to calculate the
current I3 as its just the sum of I1 and I2. So Kirchhoff’s second voltage law simply
becomes:
 Equation No 1 : 10 = 50I1 + 40I2
 Equation No 2 : 20 = 40I1 + 60I2
The above equations can be solved using cramer’s rule, resulting in detection of I 1 and I2.
The following are the problems for practice:
I)Evaluate the following :
1 2 3
1 2 3 −2
1) | | (2) | | (3) |3 2 1|
3 6 1 −1
2 3 1
1 −1 2 1 1 1
(4) |2 0 1| (5) |3 2 1|
3 2 1 2 3 1
(II)Find the value of ‘x’ in the following:
1 2 3 1 2 9 2 𝑥 − 1 −3
𝑥 −1 𝑥 8
1) | |=0 (2) | |=0 (3) |4 5 𝑥 |=0 (4) |2 𝑥 0 |=0 (5) |1 −2 4 |=3x-
2 1 8 𝑥
7 8 9 3 7 −6 3 −1 5
1.

Matrices and Determinants-1.4 Page 18

 Type equation here.

(III) solve by cramer’s rule:

1) 3x+y =4, x+3y=4
(2)2x -3y =5 , 7x –y =8
(3) 5x+3y =1 , 3x +5y =-9
4) x+y+z=7 , 2x+3y+2z=17 , 4x+9y+z = 37.
5) x+y+z=7, x+2y+3z=16 , x+3y+4z=22
6) 2x+y=1, y+2z=7 , 3z-2x=11

(IV) a) Find the value of R1 and R2 cramer’s rule:

R1+ 4R2 =70, 2R2 -3R1 = 0
b)In an electrical network, currents i1, i2 , i3 are given by,
3i1 + i2+ i3 =8,
2i1 - 3i2 - 2i3 =-5 ,
7i1 + 2i2 - 5i3 =0.
Calculate the current i2 using cramer’s rule.
Mcq’s:

1)Which among the following matrices can be expanded as a determinant.

2
1 2 4 2 4
(a)[ ] (b)[ ] (c)[5] (d)[3 5 7]
6 3 5 6 7
6
Ans: option (b).

2)which among the following statements is true?

(a) A det is a square matrix of order n xn.
(b) A det is a real number associated with a square matrix.
(c)A det is always singular.
(d)A det is a matrix with equal number of rows and columns.
Ans: option (b).
2 0 0
3) The det value of |0 5 0| is
0 0 3
(a) 3 (b) 15 (c) 30 (d) 10.
Ans: option (c).

4) If matrix ‘ A ‘ is singular matrix , then

(a) |𝐴| = 100 (𝑏)|𝐴| = 0 (c)|𝐴| is plural (𝑑)|𝐴|is any real number

Matrices and Determinants-1.4 Page 19

 Type equation here.

Ans: option (b)

1 2
5) If A = [ ] is singular , then x=
3 𝑥
(a) 6 (b) -6 (c) 16 (d) 0.
Ans: option (a).

MCQ’S ON CRAMER’S RULE

1)Which among the following is a method to solve system of
simultaneous equations:
(a)chain rule (b) Pythagoras rule
(c) cramer’s rule (d) rule of matrix
soln: option (c).
2) To find value of ‘x’ in the system of equations : x+ y =1 and 2x +3y =2
how many determinants are required in Cramer’s rule
(a)1 (b) 2 (c) 3 (d) 4
Soln: option (b)
3) To find value of ‘x’ and ‘y’ in the system of equations :
x+ y =1 and 2x +3y =2
how many determinants are required in Cramer’s rule
(a)1 (b) 2 (c) 3 (d) 4
Soln: option (c)
4) The solution set of linear equations: x+ y =1 and 2x +3y =2 is
(a)x=1, y=0 (b) x=0, y=1 (c) x=-1, y=0 (d) x=0, y=0
Soln: option (a) ∆= 1, ∆1 = 1 , ∆2 = 0
∆1 1 ∆2 0
𝑥= = = 1 𝑎𝑛𝑑 𝑦 = = =0
∆ 1 ∆ 1
5)The value of ‘x’ that satisfies the system of equations
3x+4y=7 and 7x -y=6 is
(a)x=-2 (b) x=10 (c) x=1 (d) x=0
∆1 1
Soln: option (c) ∆= −31, ∆1 = −31 , 𝑥 = =1=1
∆

Consider the system of equations:

𝑎1 𝑥 + 𝑏1 𝑦 = 𝑐1 and 𝑎2 𝑥 + 𝑏2 𝑦 = 𝑐2

𝑎1 𝑏1
Δ =| | = 𝑎1 𝑏2 - 𝑎2 𝑏1 ( det of coefficient)
𝑎2 𝑏2

𝑐1 𝑏1
𝛥1 = | | = 𝑐1𝑏2 - 𝑐2 𝑏1 ( det obtained by replacing 1st column f Δ by
𝑐2 𝑏2

constants𝑐1 , 𝑐2)

Matrices and Determinants-1.4 Page 20

 Type equation here.

𝑎1 𝑐1
𝛥2 = |𝑎 𝑐2 | = 𝑎1 𝑐2- 𝑎2 ( det obtained by replacing 2nd column of Δ by
2

constants𝑐1 , 𝑐2)

𝛥1 𝛥2
Therefore, x= and y= . Provaided (𝛥 ≠ 0 )
𝛥 𝛥

 Cramer’s Rule can also be used to solve the simultaneous equations of ‘n’
variables

Example
Solve 2x +3y =1 ; 3x –y = – 2 by Cramer’s rule.

2 3
Soln : Δ = | | = (2)(– 1) – (3)(3)= – 2 – 9 = – 11
3 −1

1 3
𝛥1 = | | = (1)(– 1) – (– 2)(3)= 1+6 = 5
−2 −1

2 1
𝛥2 = | | = (2)(– 2) – (1)(3)= – 4 – 3 = – 7
3 −2
𝜟𝟏 𝟓 𝜟𝟐 −𝟕 𝟕
Therefore, 𝒙 = = −𝟏𝟏 and 𝒚 = = −𝟏𝟏 = 𝟏𝟏
𝜟 𝜟

Worked Examples on Cramer’s Rule

1) Solve the equations 2x + y =1; 3x + 2y =1 by method of determinants.

Soln: Given system of equation is 2x + y =1

3x +2y =1

2 1
𝑙𝑒𝑡 ∆= |
3 2
|= 4−3 =1
www.mathswithme.in
1 1
∆1 = | |=2−1=1
1 2

2 1
∆2 = | | = 2 − 3 = −1
3 1

Matrices and Determinants-1.4 Page 21

 Type equation here.

∆1 1 ∆2 −1
𝑛𝑜𝑤 𝑥 = = = 1 𝑎𝑛𝑑 𝑦 = = = −1
∆ 1 ∆ 1

∴ 𝒙 = 𝟏 𝒂𝒏𝒅 𝒚 = −𝟏

2) Solve the equations x + y = 3; 2x + 3y = 8 by Cramer's rule

Soln: Given system of equation is x+y=3

2x + 3y = 8

1 1
𝑙𝑒𝑡 ∆= | | = 3 − 2 = 1;
2 3

3 1
∆1 = | |=9−8=1 𝑎𝑛𝑑
8 3

1 3
∆2 = | | = 8−6 = 2
2 8

∆1 1 ∆2 2
𝑛𝑜𝑤 𝑥 = = = 1 𝑎𝑛𝑑 𝑦 = = =2
∆ 1 ∆ 1

∴ 𝒙 = 𝟏 𝒂𝒏𝒅 𝒚 = −𝟏

3) Solve 2x - 3y = 5, 7x – y = 8 by Cramer's rule.

Soln: Given system of the equations is 2x – 3y = 5

7x – y = 8

2 −3
𝑙𝑒𝑡 ∆= | | = −2 + 21 = 19
7 −1

5 −3
∆1 = | | = −5 + 2 = 19
8 −1

2 5
∆2 = | | = 16 − 35 = −19
7 8

∆1 19 ∆2 −19
𝑛𝑜𝑤 𝑥 = = = 1 𝑎𝑛𝑑 𝑦 = = = −1
∆ 19 ∆ 19

∴ 𝒙 = 𝟏 𝒂𝒏𝒅 𝒚 = −𝟏

4) Solve 3u + 4v = 10; 2u – 3v = 1 by Cramer's rule

Soln: Given system of equation is 3 𝑢 + 4 𝑣 = 10

Matrices and Determinants-1.4 Page 22

 Type equation here.

2𝑢–3𝑣 =1

3 4
𝑙𝑒𝑡 ∆ = | | = −9 − 8 = −17;
2 −3

10 4
∆1 = | | = −30 − 4 = −34
1 −3

3 10
∆2 = | | = 3 − 20 = −17
2 1

∆1 −34 ∆2 −17
𝑛𝑜𝑤 𝑢 = = = 2 𝑎𝑛𝑑 𝑢 = = =1
∆ −17 ∆ −17

∴ 𝑢 = 𝟐 𝐚𝐧𝐝 𝑣 = 𝟏

EXERCISE
Solve by Cramer’s Rule:
1) 3x+y = 4, x +3y = 4
(2) 2x -3y =5 , 7x –y = 8
(3) 5x+3y =1 , 3x +5y = -9
(4) y=4, x+3y=4
(5) 3x +4y-7=0, 7x-y-6=0
(6) R1+ 4R2 =70, 2R2 -3R1 = 0
Minor of an element of a matrix: Minor of an element aij of a matrix is the determinant

obtained by deleting 𝒊𝒕𝒉 row and 𝒋𝒕𝒉 column of matrix.

Steps to find Minor of an element

For each element of the matrix
Step (1): Delete the elements on the current row and column
Step (2): Calculate the determinant of remaining elements
a1 a 2
Example 1: Consider a 2 × 2 matrix A = [b b ] Follow the same method to find the Minor of
1 2 remaining
elements

To find Minor of a1
a1 a2
Step (1): [b b ]
1 2
Minor of a1 =
Step (2): |b2 | = b2
| b2 | = b2
Minor of a2 = |b1 | = b1

Matrices and Determinants-1.4 Page 23

 Type equation here.

Minor of b1 =|a2 |=a2

Minor of b2 =|a1 |=a1
a1 a2 a3
Example 2: Consider a 3 × 3 matrix A = [b1 b2 b3 ]
c1 c2 c3

Minor of To find Minor of a1

a1 =
a1 a2Follow
a3the same method to find the Minor of
Step (1): [b1 b2remaining
b3 ] elements
c1 c2 c3
b b3
Step (2): | 2 | = (b2 c3 − b3 c2 )
c2 c3

b2 b3
| | = (b2 c3 − b3 c2 )
c2 c3
b1 b3
Minor of a2 = | | = (b1 c3 − b3 c1 )
c1 c3
b1 b2
Minor of a3 = | | = (b1 c2 − b2 c1 )
c1 c2
a2 a3
Minor of b1 = |c c3 | = (a2 c3 − a3 c2 )
2
a1 a3
Minor of b2 = | c c3 | = (a1 c3 − a3 c1 )
1
a1 a2
Minor of b3 = |c c2 | = (a1 c2 − a2 c1 )
1
a2 a3
Minor of c1 = |b b3 | = (a2 b3 − a3 b2 )
2
a1 a3
Minor of c2 = |b b3 | = (a1 b3 − a3 b1 )
1
a1 a2
Minor of c3 = |b b2 | = (a1 b2 − a2 b1 )
1

Problems on finding minor of an element of a matrix

2 4
1: Find the minor of 2 and 4 from the matrix [ ]
6 8
2 4
Solution: Let 𝐴 = [ ]
6 8
Minor of 2 = |8| = 8
Minor of 4 = |6| = 6
1 −3
2: Find the minor of -2 and -5 from the matrix [ ]
−2 −5

Matrices and Determinants-1.4 Page 24

 Type equation here.

1 −3
Solution: Let 𝐴 = [ ]
−2 −5

Minor of − 2 = |−3| = −3
Minor of − 5 = |1| = 1
1 2 6
3: Find the minor of 1 from the matrix [3 4 7]
8 9 5
1 2 6
Solution: Let A = [3 4 7]
8 9 5
4 7
Minor of 1=| | = (4 × 5) − (7 × 9) = 20 − 63 = −43
9 5
2 5 6
4: Find the minor of 5 and -3 from the matrix [−1 4 −3]
0 9 7
2 5 6
Solution: Let A = [−1 4 −3]
0 9 7
−1 −3
Minor of 5 = | | = (−1 × 7) − (−3 × 0) = −7 − 0 = −7
0 7
2 5
Minor of -3 = | | = (2 × 9) − (5 × 0) = 18 − 0 = 18
0 9
Cofactor of an element: Cofactor of an element is minor of the element with + or – sign
given to it. If the element is in the 𝒊𝒕𝒉 row and 𝒋𝒕𝒉 column, the sign to be allocated is (−1)𝑖+𝑗 .

Steps to find Cofactor of an element

For each element of the matrix
+ − +
+ −
Step (1): Allocate the + or – sign by using (−1)𝑖+𝑗 or [ ] or [− + −]
− +
+ − +
Step (2): Delete the elements on the current row and column
Step (3): Calculate the determinant of remaining elements

𝐚𝟏 𝐚𝟐
Example 1: Consider a 𝟐 × 𝟐 matrix 𝐀 = [𝐛 𝐛𝟐 ]
𝟏

To find Cofactor of a1
Step (1): Allocate + or – sign by using (−1)𝑖+𝑗 or
a1 a2
Step (2): + [b b ]
1 2
Follow the same method to find
Step (3): +|b2 | = b2 cofactors of remaining elements
Matrices and Determinants-1.4 Page 25
 Type equation here.

𝐂𝟎𝐟𝐚𝐜𝐭𝐨𝐫 𝐨𝐟 𝐚𝟏 (𝐂 𝐨𝐟 𝐚𝟏 ) = +|b2 | = A1
𝐂 𝐨𝐟 𝐚𝟐 = −|b1 | = A2
𝐂 𝐨𝐟 𝐛𝟏 = −|a2 | = B1
𝐂 𝐨𝐟 𝐛𝟐 = +|a1 | = B2
A1 A2
∴ Cofcator matrix of A = [ ]
B1 B2

𝐚𝟏 𝐚𝟐 𝐚𝟑
Example 2: Consider a 𝟑 × 𝟑 matrix 𝐀 = [ 𝟏
𝐛 𝐛𝟐 𝐛𝟑 ]
𝐜𝟏 𝐜𝟐 𝐜𝟑

To find Cofactor of a1
Step (1): Allocate the + or – sign by using (−1)𝑖+𝑗 or
a1 a2 a3
Follow the same method to find
Step (2): + [b1 b2 b3 ]
c1 c2 c3 cofactors of remaining
elements
b2 b3
Step (3): + | | = +(b2 c3 − b3 c2 )
c2 c3

b b3
Cofactor of 𝐚𝟏 (𝐂 𝐨𝐟 𝐚𝟏 ) = + | 2 | = +(b2 c3 − c2 b3 )= b2 c3 − b3 c2 = A1
c2 c3
b1 b3
𝐂 𝐨𝐟 𝐚𝟐 = − | | = −(b1 c3 − b3 c1 )= −b1 c3 + b3 c1 = A2
c1 c3
b1 b2
𝐂 𝐨𝐟 𝐚𝟑 = + | | = +(b1 c2 − b2 c1 )= b1 c2 − b2 c1 = A3
c1 c2
a2 a3
𝐂 𝐨𝐟 𝐛𝟏 = − |c c3 | = −(a2 c3 − a3 c2 )=−a2 c3 + a3 c2 = B1
2
a1 a3
𝐂 𝐨𝐟 𝐛𝟐 = + |c c3 | = +(a1 c3 − a3 c1 )=a1 c3 − a3 c1 = B2
1
a1 a2
𝐂 𝐨𝐟 𝐛𝟑 = − |c c2 | = −(a1 c2 − a2 c1 )=-a1 c2 + a2 c1 = B3
1
a2 a3
𝐂 𝐨𝐟 𝐜𝟏 = + |b b3 | = +(a2 b3 − a3 b2 )=a2 b3 − a3 b2 = C1
2

Matrices and Determinants-1.4 Page 26

 Type equation here.

a1 a3
𝐂 𝐨𝐟 𝐜𝟐 = − | c c3 | = −(a1 c3 − a3 c1 )=−a1 c3 + a3 c1 = C2
1
a1 a2
𝐂 𝐨𝐟 𝐜𝟑 = + |b b2 | = +(a1 b2 − a2 b1 )= a1 b2 − a2 b1 = C3
1
A1 A2 A3
∴ Cofcator matrix of A = [B1 B2 B3 ]
C1 C2 C3

Problems on finding cofactor of an element of a matrix

3 5
1: Find the cofactor of 3 and 5 from the matrix [ ]
7 9
3 5
Solution: Let 𝐴 = [ ]
7 9
+ −
Note: use these signs [ ] to find the cofactor of an element
− +
𝐂𝐨𝐟𝐚𝐜𝐭𝐨𝐫 𝐨𝐟 𝟑 = +|9| = 9
𝐂𝐨𝐟𝐚𝐜𝐭𝐨𝐫 𝐨𝐟 𝟓 = −|7| = −7

1 3
2: Find the cofactor matrix of A = [ ]
2 5
1 3
Solution: Let 𝐴 = [ ]
2 5
𝐂𝐨𝐟𝐚𝐜𝐭𝐨𝐫 𝐨𝐟 𝟏(𝐂 𝐨𝐟 𝟏) = +|5| = 5
𝐂 𝐨𝐟 𝟑 = −|2| = −2
𝐂 𝐨𝐟 𝟐 = −|3| = −3
𝐂 𝐨𝐟 𝟓 = +|1| = 1
5 −2
∴ Cofcator matrix of A = [ ]
−3 1
3 2 5
3: Find the cofactor of 5 and 6 from the matrix [7 8 2]
1 6 4
3 2 5
Solution: Let 𝐴 = [7 8 2]
1 6 4
7 8
𝐂𝐨𝐟𝐚𝐜𝐭𝐨𝐫 𝐨𝐟 𝟓 = + | | = +(42 − 8) = +(34) = 34
1 6
3 5
𝐂𝐨𝐟𝐚𝐜𝐭𝐨𝐫 𝐨𝐟 𝟔 = − | | = −(6 − 35) = −(−29) = 29
7 2
5 1 3
4: Find the cofactor matrix of [4 2 6]
2 1 4
5 1 3
Solution: Let 𝐴 = [4 2 6]
2 1 4

Matrices and Determinants-1.4 Page 27

 Type equation here.

+ − +
Note: use these signs [− + −] to find the cofactor of an element
+ − +
2 6
𝐂𝐨𝐟𝐚𝐜𝐭𝐨𝐫 𝐨𝐟 𝟓 (𝐂 𝐨𝐟 𝟓) = + | | = +(8 − 6) = +(2) = 2
1 4
4 6
𝐂 𝐨𝐟 𝟏 = − | | = −(16 − 12) = −(4) = −4
2 4

4 2
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𝐂 𝐨𝐟 𝟑 = +| | = +(4 − 4) = 0
2 1
1 3
𝐂 𝐨𝐟 𝟒 = −| | = −(4 − 3) = −(1) = 1
1 4
5 3
𝐂 𝐨𝐟 𝟐 = +| | = +(20 − 6) = +(14) = 14
2 4
5 1
𝐂 𝐨𝐟 𝟔 = −| | = −(5 − 2) = −(3) = −3
2 1
1 3
𝐂 𝐨𝐟 𝟐 = +| | = +(6 − 6) = 0
2 6
5 3
𝐂 𝐨𝐟 𝟏 = −| | = −(30 − 12) = −(18) = −18
4 6
5 1
𝐂 𝐨𝐟 𝟒 = +| | = +(10 − 4) = +(6) = 6
4 2
2 −4 0
∴ Cofactor matrix of A = [1 14 −3]
0 −18 6
Model questions(4marks)
1 2 3
1. Find the cofactor matrix of [5 4 1]
4 2 3
3 −1 2
2. Find the cofactor matrix of [0 −3 9]
1 5 4
4 1 5
3. If 𝐴 = [−1 0 7 ] Find the cofactor matrix of A
2 −3 −2

Adjoint of a matrix: Let A be square matrix of order n. The adjoint of a matrix A is the
transpose of the cofactor matrix of A. It is denoted by adj A.
i.e. adj A = [cofactor matrix of A] T

Steps to find adjoint of a matrix

Step (1): Find the cofactors of every element in the given matrix.
Step (2): Write down the cofactor matrix.
Step (3): Find the transpose of the cofactor matrix.

Matrices and Determinants-1.4 Page 28

 Type equation here.

a1 a2
Example 1: Consider a 2 × 2 matrix A = [b b2 ]
1

A1 A2
Cofactor matrix of A =[ ]
B1 B2
NOTE: Where A1 , A2 , B1 and B2 are the cofactors of a1 , a2 , b1 and b2 respectively.
We know that, Adjoint of A(adjA)= [cofactor matrix of A] T
A1 A2 T
=[ ]
B1 B2
A1 B1
Therefore, Adjoint of A=[ ]
A2 B2
a1 a2 a3
Example 2: Consider a 3 × 3 matrix A = [b1 b2 b3 ]
c1 c2 c3
A1 A2 A3
Cofactor matrix of A =[B1 B2 B3 ]
C1 C2 C3
NOTE: Where A1 , A2 , A3 , B1 , B2 , B3 , C1 , C2 and C3 are the cofactors of a1 , a2 , a3 , b1 , b2 , b3 , c1 , c2 and c3
respectively.

We know that, Adjoint of A(adjA)= [cofactor matrix of A] T

A1 A2 A3 𝑇
=[B1 B2 B3 ]
C1 C2 C3
A1 B1 C1
Therefore, Adjoint of A = [A2 B2 C2 ]
A3 B3 C3

NOTE: For any square matrix of A, 𝐀(𝐚𝐝𝐣𝐀) = (𝐚𝐝𝐣𝐀)𝐀 = |𝐀|I

Problems on finding adjoint of a matrix

𝟒 𝟐
1: If 𝑨 = [ ] find Adjoint of matrix A.
𝟓 𝟏
4 2 + −
Solution: Given 𝐴 = [ ] [ ]
5 1 − +
𝐂𝐨𝐟𝐚𝐜𝐭𝐨𝐫 𝐨𝐟 𝟒 (𝐂 𝐨𝐟 𝟒) = +|1| = 1
𝐂 𝐨𝐟 𝟐 = −|5| = −5
𝐂 𝐨𝐟 𝟓 = −|2| = −2
𝐂 𝐨𝐟 𝟏 = +|4| = 4
1 −5
∴ Cofactor matrix of A = [ ]
−2 4
We know that, adjA= [cofactor matrix of A] T

Matrices and Determinants-1.4 Page 29

 Type equation here.

1 −5 𝑇
adjA = [ ]
−2 4
1 −2
adjA = [ ]
−5 4

NOTE: Adjoint of 2 × 2 matrix can be found out by interchange the principal diagonal
elements (PDE) and interchange the signs of remaining elements.

2 7 3 −7
For example, If A =[ ] then,
PDE adjA = [ ]
9 3 −9 2

𝟔 𝟓
2: Find the adjoint of the matrix 𝑨 = [ ]
−𝟏 𝟖
6 5 + −
Solution: Given 𝐴 = [ ] [ ]
−1 8 − +
𝐂 𝐨𝐟 𝟔 = +|8| = 8
𝐂 𝐨𝐟 𝟓 = −|−1| = +1
𝐂 𝐨𝐟 − 𝟏 = −|5| = −5
𝐂 𝐨𝐟 𝟖 = +|6| = 6
8 1
Cofactor matrix of A = [ ]
−5 6
We know that, adj A= [cofactor matrix of A] T
8 1𝑇
adjA = [ ]
−5 6
8 −5
adjA = [ ]
1 6
𝟑 𝟏 𝟓
3: If 𝑨 = [𝟔 𝟏 𝟐] find adjA.
𝟒 𝟐 𝟏
3 1 5 + − +
Solution: Given 𝐴 = [6 1 2] [− + −]
4 2 1 + − +
1 2
𝐂 𝐨𝐟 𝟑 = + | | = +(1 − 4) = +(−3) = −3
2 1
6 2
𝐂 𝐨𝐟 𝟏 = − | | = −(6 − 8) = −(−2) = 2
4 1
6 1
𝐂 𝐨𝐟 𝟓 = + | | = +(12 − 4) = +(8) = 8
4 2
1 5
𝐂 𝐨𝐟 𝟔 = − | | = −(1 − 10) = −(−9) = 9
2 1
3 5
𝐂 𝐨𝐟 𝟏 = + | | = +(3 − 20) = +(−17) = −17
4 1
3 1
𝐂 𝐨𝐟 𝟐 = − | | = −(6 − 4) = −(2) = −2
4 2
1 5
𝐂 𝐨𝐟 𝟒 = + | | = +(2 − 5) = +(−3) = −3
1 2

Matrices and Determinants-1.4 Page 30

 Type equation here.

3 5
𝐂 𝐨𝐟 𝟐 = − | | = −(6 − 30) = −(−24) = 24
6 2
3 1
𝐂 𝐨𝐟 𝟏 = + | | = +(3 − 6) = +(−3) = −3
6 1
−3 2 8
∴ Cofactor matrix of A = [ 9 −17 −2]
−3 24 −3
We know that, adj A= [cofactor matrix of A] T
−3 2 8 𝑇
adjA = [ 9 −17 −2]
−3 24 −3
−3 9 −3
adjA = [ 2 −17 24 ]
8 −2 −3
𝟑 −𝟏 𝟐
4: Find the adjoint of [𝟐 −𝟑 𝟏]
𝟎 𝟒 𝟐
3 −1 2 + − +
Solution: Given 𝐴 = [2 −3 1] [− + −]
0 4 2 + − +
−3 1
𝐂𝐨𝐟 𝟑 = + | | = +(−6 − 4) = +(−10) = −10
4 2
2 1
𝐂 𝐨𝐟 − 𝟏 = − | | = −(4 − 0) = −(4) = −4
0 2
2 −3
𝐂 𝐨𝐟 𝟐 = + | | = +(8 − 0) = +(8) = 8
0 4
−1 2
𝐂 𝐨𝐟 𝟐 = − | | = −(−2 − 8) = −(−10) = 10
4 2
3 2
𝐂 𝐨𝐟 − 𝟑 = + | | = +(6 − 0) = +(6) = 6
0 2
3 −1
𝐂 𝐨𝐟 𝟏 = − | | = −(12 − 0) = −(12) = −12
0 4
−1 2
𝐂 𝐨𝐟 𝟎 = + | | = +(−1 + 6) = +(5) = 5
−3 1
3 2
𝐂 𝐨𝐟 𝟒 = − | | = −(3 − 4) = −(−1) = 1
2 1
3 −1
𝐂 𝐨𝐟 𝟐 = + | | = +(−9 + 2) = +(−7) = −7
2 −3
−10 −4 8
∴ Cofactor matrix of A = [ 10 6 −12]
5 1 −7
We know that, adj A= [cofactor matrix of A] T
−10 −4 8 𝑇
adjA = [ 10 6 −12]
5 1 −7
−10 10 5
adjA = [ −4 6 1]
8 −12 −7

Matrices and Determinants-1.4 Page 31

 Type equation here.

Model questions (4marks)

1 2
1. Find the adjoint of the matrix [ ]
3 4
2 1
2. If 𝐴 = [ ] find adjA
−3 3
5 1
3. If 𝐴 = [ ] find Adjoint of A
6 8

Model questions (5marks)

1 2 3
1. Find the adjoint of the matrix [1 3 3]
2 4 3
1 1 −1
2. If 𝐴 = [ 1 2 1 ] find Adjoint of A
−1 1 −3
3 0 −2
3. If 𝐴 = [0 1 4 ] find adjA
0 2 3

Singular matrix: A square matrix ‘A’ is said to be singular if and only if |A| = 0
2 4
Example: A = [ ] is a singular matrix because |A| = 0
3 6

Non-singular matrix: A square matrix ‘A’ is said to be non-singular if and only

if |A| ≠ 0
1 2
Example: A = [ ] is a non-singular matrix because |A| ≠ 0
3 4

Inverse of a matrix: If a matrix ‘A’ is non-singular (|A| ≠ 0), the inverse of matrix A can
be defined.
𝟏
i.e. 𝐀−𝟏 = |𝐀| 𝐚𝐝𝐣𝐀

What is the Inverse of a Matrix?

This is the reciprocal of a number:

The Inverse of a Matrix is the same idea

but we write it A−1

Matrices and Determinants-1.4 Page 32

 Type equation here.

1
Why not ? Because we don't divide by a matrix!
A

Why Do We Need an Inverse?

Because with matrices we don't divide! Seriously, there is no concept of dividing by a matrix.

But we can multiply by an inverse, which achieves the same thing.

Steps to find inverse of a matrix
Step (1): Find the determinant of a given matrix (|A| ≠ 0).
Step (2): Find the adjoint of a given matrix.
1
Step (3): substitute |A| value and adjoint of A in the formula A−1 = |A| adjA

NOTE 1: If A and B are two non-singular matrices of same order then (AB)−1 = B −1 A−1
NOTE 2: 𝐀𝐀−𝟏 = 𝐀−𝟏 𝐀 = 𝐈

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Problems on finding inverse of a matrix
𝟏 𝟑
1: Find inverse of the matrix 𝑨 = [ ]
𝟐 𝟖
1 3
Solution: Given A = [ ]
2 8
1 3
Consider |A| = | | = (8 − 6) = 2 ≠ 0
2 8
Therefore A−1 exists.
To find adjA interchange the principal diagonal elements and interchange the signs of remaining
elements.
8 −3
From the matrix A, adjA=[ ]
−2 1
𝟏
We know that, 𝐀−𝟏 = |𝐀| 𝐚𝐝𝐣𝐀
1 8 −3
𝐴−1 = [ ]
2 −2 1
𝟏 −𝟏 𝟐
2: Find the inverse of the matrix 𝑨 = [𝟐 𝟏 𝟏]
𝟒 −𝟏 −𝟐
1 −1 2
Solution: Given 𝐴 = [2 1 1]
4 −1 −2

Matrices and Determinants-1.4 Page 33

 Type equation here.

1 −1 2
Consider |𝐴| = |2 1 1 | = 1(−2 + 1) + 1(−4 − 4) + 2(−2 − 4) = -21≠ 0
4 −1 −2
Therefore 𝐴−1 exists.
To find adjoint of matrix A
1 1
𝐂 𝐨𝐟 𝟏 = + | | = +(−2 + 1) = +(−1) = −1
−1 −2
2 1
𝐂 𝐨𝐟 − 𝟏 = − | | = −(−4 − 4) = −(−8) = 8
4 −2
2 1
𝐂 𝐨𝐟 𝟐 = + | | = +(−2 − 4) = +(−6) = −6
4 −1
−1 2
𝐂 𝐨𝐟 𝟐 = − | | = −(2 + 2) = −(4) = −4
−1 −2
1 2
𝐂 𝐨𝐟 𝟏 = + | | = +(−2 − 8) = +(−10) = −10
4 −2
1 −1
𝐂 𝐨𝐟 𝟏 = − | | = −(−1 + 4) = −(3) = −3
4 −1
−1 2
𝐂 𝐨𝐟 𝟒 = + | | = +(−1 − 2) = +(−3) = −3
1 1
1 2
𝐂 𝐨𝐟 − 𝟏 = − | | = −(1 − 4) = −(−3) = 3
2 1
1 −1
𝐂 𝐨𝐟 − 𝟐 = + | | = +(1 + 2) = +(3) = 3
2 1
We know that, adjA= [cofactor matrix of A] T
−1 8 −6 𝑇
adjA = [−4 −10 −3]
−3 3 3
−1 −4 −3
adjA = [ 8 −10 3 ]
−6 −3 3
𝟏
We know that, 𝐀−𝟏 = |𝐀| 𝐚𝐝𝐣𝐀

1 −1 −4 −3
𝐴−1 = − [ 8 −10 3]
21
−6 −3 3
𝟏 𝟐 𝟒
3: If 𝑨 = [𝟑 −𝟏 𝟐 ] find 𝐀−𝟏
𝟓 𝟏 −𝟐
1 2 4
Solution: Given 𝐴 = [3 −1 2 ]
5 1 −2
1 2 4
Consider |𝐴| = |3 −1 2 | = 1(2 − 2) − 2(−6 − 10) + 4(3 + 5) = 64 ≠ 0
5 1 −2
Therefore 𝐴−1 exists.
To find adjoint of matrix A
−1 2
𝐂 𝐨𝐟 𝟏 = + | | = +(2 − 2) = +(0) = 0
1 −2
Matrices and Determinants-1.4 Page 34
 Type equation here.

3 2
𝐂 𝐨𝐟 𝟐 = − | | = −(−6 − 10) = −(−16) = 16
5 −2
3 −1
𝐂 𝐨𝐟 𝟒 = + | | = +(3 + 5) = +(8) = 8
5 1
2 4
𝐂 𝐨𝐟 𝟑 = − | | = −(−4 − 4) = −(−8) = 8
1 −2
1 4
𝐂 𝐨𝐟 − 𝟏 = + | | = +(−2 − 20) = +(−22) = −22
5 −2
1 2
𝐂 𝐨𝐟 𝟐 = − | | = −(1 − 10) = −(−9) = 10
5 1
2 4
𝐂 𝐨𝐟 𝟓 = + | | = +(4 + 4) = +(8) = 8
−1 2
1 4
𝐂 𝐨𝐟 𝟏 = − | | = −(2 − 12) = −(−10) = 10
3 2
1 2
𝐂 𝐨𝐟 − 𝟐 = + | | = +(−1 − 6) = +(−7) = −7
3 −1
We know that, adjA= [cofactor matrix of A] T
0 16 8 𝑇
adjA = [8 −22 10 ]
8 10 −7
0 8 8
adjA = [16 −22 10 ]
8 10 −7
𝟏
We know that, 𝐀−𝟏 = |𝐀| 𝐚𝐝𝐣𝐀

1 0 8 8
𝐴−1 = [16 −22 10 ]
64
8 10 −7
Model questions (5marks)
1 2
1. Find the inverse of the matrix [ ]
3 −1
3 5
2. If 𝐴 = [ ] find the inverse of A
2 4

Model questions (6marks)

2 3 1
1. Find the inverse of the matrix [1 2 1]
5 4 3

Matrices and Determinants-1.4 Page 35

 Type equation here.

1 2 −1
2. If 𝐴 = [−1 1 2 ] find the inverse of A
2 −1 1
1 0 3
3. If 𝐴 = [2 1 4] find 𝐴−1
3 −1 5

Characteristic equation of matrix: Let A be square matrix of order n then |𝐀 − 𝛌𝐈| = 𝟎

is called characteristic equation of matrix A. Where I is identity matrix of order n and 𝜆 is
constant.

Characteristic roots of matrix: Let A be square matrix of order n then the roots of the
characteristic equation |𝐀 − 𝛌𝐈| = 𝟎 are called characteristic roots or eigen values.

NOTE: The sum of the eigen values is equal to the sum of the principal diagonal elements of the
matrix.

Problems on finding characteristic equation and eigen values of a

matrix
1: Find the characteristic equation of 𝑨 = [
𝟐 −𝟏
]
𝟑 𝟐
2 −1
Solution: Given 𝐴 = [ ]
3 2
C.E is given by |A − λI| = 0
2 −1 1 0 1 0
|[ ] − λ[ ]| = 0 (Where I = [ ] Identity matrix)
3 2 0 1 0 1
2 − λ −1
| |=0
3 2−λ
(2 − λ)(2 − λ) + 3 = 0
4 − 2λ − 2λ + λ2 + 3 = 0
λ2 − 4λ + 7 = 0 is required characteristic equation.
𝟏 𝟐
2: Find the characteristic equation of 𝑨 = [ ]
𝟑 𝟒
1 2
Solution: Given 𝐴 = [ ]
3 4
C.E is given by |A − λI| = 0
1 2 1 0
|[ ] − λ[ ]| = 0
3 4 0 1
1−λ 2
| |=0
3 4−λ
(1 − λ)(4 − λ) − 6 = 0
4 − λ − 4λ + λ2 − 6 = 0

Matrices and Determinants-1.4 Page 36

 Type equation here.

λ2 − 5λ − 2 = 0 is required characteristic equation.

𝟑 𝟐
3: Find the characteristic equation and eigen values of 𝑨 = [ ]
𝟒 𝟓
3 2
Solution: Given 𝐴 = [ ]
4 5
C.E is given by |A − λI| = 0
3 2 1 0
|[ ] − λ[ ]| = 0
4 5 0 1
3−λ 2
| |=0
4 5−λ
(3 − λ)(5 − λ) − 8 = 0
15 − 3λ − 5λ + λ2 − 8 = 0
λ2 − 8λ + 7 = 0
λ2 − 7λ − 1λ + 7 = 0
λ(λ − 7) − 1(λ − 7) = 0
(λ − 7)(λ − 1) = 0
λ − 7 = 0 or λ − 1 = 0
λ = 7 or λ = 1
λ = 7,1 are the eigen values of given matrix.

𝟑 −𝟏
4: If 𝑨 = [ ] find the eigen values
𝟎 −𝟐
3 −1
Solution: Given 𝐴 = [ ]
0 −2
C.E is given by |A − λI| = 0
3 −1 1 0
|[ ] − λ[ ]| = 0
0 −2 0 1
3−λ −1
| |=0
0 −2 − λ
(3 − λ)(−2 − λ) − 0 = 0 www.mathswithme.in
−6 − 3λ + 2λ + λ2 − 0 = 0
λ2 − λ − 6 = 0
λ2 − 3λ + 2λ − 6 = 0
λ(λ − 3) + 2(λ − 3) = 0
(λ − 3)(λ + 2) = 0
λ − 3 = 0 or λ + 2 = 0
λ = 3 or λ = −2
λ = 3,-2 are the eigen values of given matrix.
𝟏 −𝟏
5: Find the characteristic equation and its roots of 𝑨 = [ ]
−𝟔 −𝟐

Matrices and Determinants-1.4 Page 37

 Type equation here.

1 −1
Solution: Given 𝐴 = [ ]
−6 −2
C.E is given by |A − λI| = 0
1 −1 1 0
|[ ] − λ[ ]| = 0
−6 −2 0 1
1−λ −1
| |=0
−6 −2 − λ
(1 − λ)(−2 − λ) − 6 = 0
−2 − 1λ + 2λ + λ2 − 6 = 0
λ2 + 1λ − 8 = 0
The above equation is the form of quadratic equation ax 2 + bx + c = 0
−b±√b2 −4ac
By using formula X =
2a

−1±√(1)2 −4(1)(−8)
We get λ =
2(1)

−1±√1+32
λ=
2
−1±√33)
λ=
2
−1+√33) −1−√33)
λ= , are the characteristic roots of given matrix.
2 2

Model questions (6marks)

1 4
1. Find the characteristic equation and roots for the matrix [ ]
2 3
2 −1
2. Find the characteristic equation and eigen values for the matrix [ ]
−3 1
−1 2
3. Find the characteristic equation and eigen values for the matrix [ ]
3 4
3 2
4. Find the eigen values for the matrix [ ]
0 1

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Matrices and Determinants-1.4 Page 38