Mathematics
Session
Matrices and Determinants - 3
Session Objectives
1. Singular and Non-singular Matrix
2. Adjoint of a Square Matrix and its Properties
3. Inverse of a Matrix and its Properties
4. Solution of Simultaneous Linear Equations
(Matrix Method)
5. Class Exercise
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
Let A= 1 3 3
5 3
3
A=1(9+9)+1(3+15)+3(3-15)
= 18+18-36
=0
A is a singular matrix .
Non-Singular Matrix
 1 1 1
Let B=  2 1 1
1 2 3
B= 1(-3+2)-1(6-1)+1(-4+1)
= -1 – 5 – 3
= -9
0
B is a non-singular matrix.
Example -1
Find the value of x for which the matrix
x 1 0 
A = 2 -1 1  is singular.
3 4 -2
Solution:
For matrix A to be singular
A =0
x
1
 2 -1
3
4
0
1 =0
-2
 x  2  4  - 1 -4 - 3  = 0
 -2x + 7 = 0  x =
7
2
Adjoint of a Square Matrix
The transpose of the matrix of cofactors of
elements of a square matrix A is called the
adjoint of A and is denoted by adjA.
adjA = [Cij]T  adjA = C ji 
Adjoint of a Square Matrix
a11
Let A = a21
a31
C11
adjA = C21
C31
a12
a22
a32
C12
C22
C32
a13 
a23  , then
a33 
C13 
C23 
C33 
T
C11
 C12
C13
C21
C22
C23
C31 
C32 
C33 
where Cij denotes the cofactor of aij in A.
Example - 2
Find the adjoint of matrix A  
a b

c d
Solution :
A = a b 
c d
C11 = d , C12 =-c
C21 =-b , C22 = a
adjA =  d -c 
-b a 
T
 d -b 
 adjA = 

-c a 
Example - 3
 1 2 1 
Find the adjoint of matrix A   3 1 0 
 0 1 1
Solution:
 1 2 1 
We have A   3 1 0 
 0 1 1
C11  (1  0)  1 , C12  ( 3  0)  3, C13  (3  0)  3
C21  ( 2  1)  3, C22  (1  0)  1 , C23  ( 1  0)  1
C31  (0  1)  1 , C32  (0  3)  3 , C33  (1  6)  5
Solution cont.
1 3 3 
 adjA  3 1 1 
 1 3 5 
T
1 3 1 
 adj A = 3 1 3 
3 1 -5
Properties
A (adjA) = |A| In = (adjA) A
Proof: Let A = [aij] be a square matrix and let Cij be
cofactor of aij in A, then (adjA) = [Cji] for all i, j = 1, 2, ..., n
 A , if i 
a
C

 ir rj 0, if i 

r 1
n
we know
j

j
Properties (Con.)
Therefore, each diagonal element of A (adjA) is
equal to |A| and all non-diagonal elements are equal
to zero.
| A | 0
 0 |A|
i.e. A (adjA) = 
 :
:

0
 0
Similarly, (adjA)A =
0
0
.. 0 
..0 
= A In
:
: 

0 ..| A |
n
 A , if
C
a
=
 ri rj O, if

r=1
Hence, A (adjA) = |A| In = (adjA)A
i = j

i  j
Properties (Con.)
2.
If A is a non-singular square matrix
of order n, then |adjA| = |A|n – 1
3.
If A and B are non-singular square
matrices of same order, then
adj AB = (adjB) (adjA)
4.
If A is a non-singular square matrix,
then adj (adjA) = |A|n–2 A.
Example-4
2
Compute the adjoint of matrix A= 1
3 -5 
and verify that A(adj A)=|A|I.
Solution:
2
We have A = 1
3 -5 
C11 = -5 , C12 = -3 , C21 = -2 , C22 =1
adjA = -5
-2
= -5
-3
-3
1 
T
-2 
1 
Solution (Con.)
L.H.S. = A  adj.A  = 1 2  -5 -2
3 -5 -3 1 
-2+2 =  -11 0 
=  -5-6
-15+15 -6-5   0 -11
A = -5-6 =-11
R.H.S.= A I=(-11) 1 0 =  -11 0 
0 1  0 -11
 L.H.S = R.H.S
Hence verified.
Example-5
p q
 r s , find det. {A(adjA)}.
If a matrix A = 
Solution:
We have A = p q
 r s 
T
 adjA =  s -r  =  s -q
-q p 
-r p 
A(adjA)=  p q  s -q = ps-qr -pq+pq
 r s  -r p   rs-rs -rq+sp 
=(ps-qr) 1 0
0 1
Now, det.{A(adj.A)}=(ps-qr)×1
= ps-qr
Inverse of a Matrix
If a matrix A is non - singular i.e. A  0,
then A-1 exists and is given by A-1 =
1
.(adjA)
A
Steps to find inverse of a matrix:
(i) Find out |A| and if A  0 , then the matrix is invertible.
(ii) Find out (adjA).
Then
A-1 =
1
.(adjA)
A
Example-6
-1 4 2 
Find the inverse of the Matrix A =  2 -1 4
 1 2 3 
Solution:
-1 4 2 
We have A =  2 -1 4
 1 2 3 
-1
 A= 2
1
4
2
-1 4 = -1 -3 - 8  - 4 6 - 4  + 2  4 +1 = 13  0
2
3
Solution cont.
C11 C12
adjA = C21 C22
C31 C32
T
T
C13 
-11 -2 5 
-11 -8 18
C23  =  -8 -5 6  =  -2 -5 8 
 18 8 -7
 5
C33 
6 -7 
-11 -8 18
1
1 

Hence, A-1 = .(adjA) =
-2
-5
8

13 
A
 5
6 -7 
Properties
(i) A square matrix is invertible if it is non-singular.
(ii)
Every invertible matrix possesses a unique inverse.
Proof: Let A be an invertible matrix of
order n x n.
Let B and C be two inverses of A.
Then AB = BA = In
and AC = CA = In
Now AB = In
Multiplying by C
C(AB) = CIn

(CA)B = C In

In B = C I n

B=C
Hence, an invertible matrix
possesses a unique inverse.
Properties (Con.)
(iii)
(AB)–1 = B–1 A–1
or (ABC)–1 = C–1 B–1 A–1
(iv)
(AT)–1 = (A–1)T
Example-7
-3
2
Show that A = 2
3 4  satisfies the equation x - 6x + 17 = 0.


Hence, find A-1.
Solution:
We have A = 2 -3
3 4 
 A2 = A.A = 2 -3 2 -3 =  4-9 -6-12  =  -5 -18
3 4  3 4  6+12 -9+16 18 7 
A2 -6A+17I2 =  -5 -18 -6 2 -3 +17 1 0
18 7  3 4 
0 1
=  -5 -18 - 12 -18 + 17 0 
18 7  18 24   0 17
Solution (Cont.)
 0 0  O Hence, A satisfies the equation x2 - 6x + 17 = 0.
0 0
A2 - 6A + 17I = 0
Multiplying each side by A-1, we get
A-1A2 - 6(A-1A) + 17(A-1I) = A-1.0
(A-1A)A - 6I + 17A-1 = 0
IA - 6I + 17A-1 = 0,
( A1A  I, A1I  A1 , A1.0  0)
17A-1 = 6I - A
3
 4
1
1  6 0 2 -3  1  4 3  17 17 
-1
A =
(6I- A)=
=

=
17
17  0 6  3 4   17 -3 2  3
2

 17 17 
3
 4
 17 17 
-1

Hence, A = 
 3
2

 17 17 
Example-8
0 1 1
A2 -3I
-1
.
Find A-1 , if A = 1 0 1 . Also show that A =
2
1 1 0
Solution:
0 1 1
We have A = 1 0 1
1 1 0
0 1 1
 A = 1 0 1 =0(0-1)-1(0-1)+1(1-0) =1+1 = 2  0
1 1 0
Now,
C11 =-1,C12 =1,C13 =1
C21 =1,C22 =-1,C23 =1
C31 =1,C32 =1,C33 =-1
T
1
1
1
1
 -1
 -1
 adj.A=  1 -1
1 =  1 -1
1
 1 1 -1 
 1
1 -1 
Solution cont.
1
 1 1

2

 2 2
1
-1 1 1   1

1
1
1

Hence, A-1 =  adj A  =  1 -1 1  = 
2

2
2  1 1 -1  2
A


1
1
1

- 
 2
2
2


A2 -3I 1  0 1 1  0 1 1  1 0 0 
Also
=  1 0 1  1 0 1  -3 0 1 0  
2
2  1 1 0 1 1 0 0 0 1  


1  2 1 1 3 0 0 
=  1 2 1 - 0 3 0 
2  1 1 2 0 0 3 


1 -1 1 1 
=  1 -1 1  = A-1
2  1 1 -1
Example-9
2 1
4 5 
-1
If A =
and
B=
,
verify
that
AB
= B-1A-1.





5 3
3 4
Solution:
(AB) = 2 1 4 5 
5 3 3 4
= 11 14 
29 37
LHS =  AB 
-1
T
=  37 -29 =  37 -14
-14 11 
-29 11 
Solution (Cont.)
T
B =  4 -3 =  4 -5
-5 4 
-3 4 
-1
T
A =  3 -5 =  3 -1
-1 2 
-5 2 
-1
RHS =B-1 A-1 =  4 -5  3 -1 =  37 -14 =LHS
-3 4  -5 2  -29 11 
Solution of Simultaneous Linear
Equations (Matrix Method)
Let the system of 3 linear equations be
a1x +b1y +c1z = d1
a2 x +b2 y +c2 z = d2
a3 x +b3 y + c3 z = d3
This system of linear equation can be written in matrix form as
 a1 b1
a b
2
 2
a3 b3
c1  x  d1 
 
c2  y  = d2 
c3  z  d3 
 AX = B
... i
Solution of Simultaneous Linear
Equations (Matrix Method)
The matrix A is called the coefficient matrix of the
system of linear equations.
If A  0 i.e. A is non - singular, then A-1 exists.
Multiplying (i) by A–1, we get
A 1  AX  A 1B


 A 1A X  A 1B
B1 A  XI 
 X=
1
(adjA)B
A
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.
Example-10
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)
2 -3 x  -4 
1
or 2 3 2  y  =  2 
3 -3 -4  z  11
 AX = B
Solution (Cont.)
2 -3
1
x 
-4 
where A = 2 3 2  , X = y  , B=  2 
3 -3 -4
 z 
11
1 2 -3
A= 2 3 2
3 -3 -4
1 0 0
= 2 -1 8
3 -9 5
 Applying C2  C2 - 2C1 and C3  C3 +3C1 
=1(-5+72)= 67  0
 A-1 exists.
Let Cij be the cofactor aij in A = aij  , then
Solution Cont.
c11 =(-12+6)
= -6
c31 =(4+9)
=13
c21 =-(-8-9)
=17
c31 =(4+9)
=13
c12 =- -8-6 
=14
c13 =(-6-9)
=-15
c32 =-(2+6)
= -8
c22 =(-4+9)
=5
c32 =-(2+6)
= -8
T
 -6 14 -15
 adjA = 17 5
9  =
13 -8 -1 
c33 =(3- 4)
= -1
c23 =-(-3-6)
=9
c33 =(3- 4)
= -1
 -6 17 13
 14 5 -8 
-15 9 -1 
Solution (Con.)
1
1  -6 17 13
 A = .adj A =
 14 5 -8 
A
67 -15 9 -1 
-1
Now, X = A-1 B
1  -6 17 13 -4 
 X=
 14 5 -8   2 
67 -15 9 -1  11
x  1  201   3 
 y  =
-134 = -2
 z  67  67   1 
 x =3 , y =-2 , z =1
Example-11
Using matrices, solve the following
system of equations
x+y+z=6
x + 2y + 3z = 14
x + 4y + 7z = 30
Solution:
The given system of linear equations is
x+y+z=6
…(i)
x + 2y + 3z = 14
…(ii)
x + 4y + 7z = 30
…(iii)
1 1 1 x   6 
or 1 2 3 y  = 14
1 4 7  z  30
 AX = B
Solution (Cont.)
1 1 1
x 
6
where A = 1 2 3 ; X = y  ; B= 14 
1 4 7
 z 
30
1 1 1
Now, A = 1 2 3 =1(14-12)-1(7-3)+1(4-2)
1 4 7
=2- 4+2= 0
Let Cij be the cofactor aij in A = aij  , then
c11 =(14-12)
=2
c12 =-(7-3)
=-4
c13 =  4-2 
=2
Solution cont.
c21 =-(7- 4)
=-3
c22 =(7-1)
=6
c31 =(3-2) c32 =-(3-1)
=1
= -2
c23 =-(4-1)
=-3
c33 =(2-1)
=1
T
2
 2 -4
 2 -3 1
 adjA =  -3 6 -3  =  -4
6 -2 
 1 -2
 2 -3 1
1
 2 -3 1  6  0
and (adjA)B = -4 6 -2 14  = 0 = 0
 2 -3 1 30 0
The given system of equations is consistent with infinitely many solutions.
Solution (Con.)
Putting z = k in first two equations, we get
x+y=6-k
x + 2y = 14 - 3k
1
or 
1
A=
1  x  6 - k 
 =
  AX = B
2 y  14 - 3k 
1
1
1
2
= 2-1=1  0
 A-1 exists.
2
 adjA = 
-1
- 1

1
T
2
=
-1
- 1

1
Solution (Con.)
-1
A
2
1
=
adjA = 
|A|
-1
- 1

1
x   2
Now, X = A B    = 
y  -1
-1
x  12 - 2k - 14 + 3k 
   =
=
y  -6 + k + 14 - 3k 
 x = -2 + k and y = 8 - 2k
- 1 6 - k 


1 14 - 3k 
-2 + k 


8
2k


These values of x, y and z = k also satisfy (iii) equation.
Hence, x = -2 + k, y = 8 - 2k and z = k, where k  R.
Thank you