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Inverse Of A Matrix And Its Application
MODULE - III
Sequences And
Series
Notes
5
INVERSE OF A MATRIX
AND ITS APPLICATION
LET US CONSIDER AN EXAMPLE:
Abhinav spends Rs. 120 in buying 2 pens and 5 note books whereas Shantanu spends
Rs. 100 in buying 4 pens and 3 note books.We will try to find the cost of one pen and the
cost of one note book using matrices.
Let the cost of 1 pen be Rs. x and the cost of 1 note book be Rs. y. Then the above
information can be written in matrix form as:
LM2 5OP LMxOP = LM120OP
N4 3Q NyQ N100Q
This can be written as A X = B
LM2 5OP , X = LMxOP and B = LM120OP
where A =
N4 3 Q N y Q
N100Q
LxO
Our aim is to find X = M P
N yQ
In order to find X, we need to find a matrix A−1 so that X = A−1 B
This matrix A-1 is called the inverse of the matrix A.
In this lesson, we will try to find the existence of such matrices.We will also learn to solve
a system of linear equations using matrix method.
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MODULE - III
Sequences And
Series
OBJECTIVES
After studying this lesson, you will be able to :
Notes
•
define a minor and a cofactor of an element of a matrix;
•
find minor and cofactor of an element of a matrix;
•
find the adjoint of a matrix;
•
define and identify singular and non-singular matrices;
•
find the inverse of a matrix, if it exists;
•
represent system of linear equations in the matrix form AX = B; and
•
solve a system of linear equations by matrix method.
EXPECTED BACKGROUND KNOWLEDGE
•
Concept of a determinant.
•
Determinant of a matrix.
•
Matrix with its determinant of value 0.
•
Transpose of a matrix.
•
Minors and Cofactors of an element of a matrix.
5.1
DETERMINANT OF A SQUARE MATRIX
We have already learnt that with each square matrix, a determinant is associated. For any given
matrix, say A =
LM2 5OP
N4 3Q
its determinant will be
2 5
. It is denoted by A .
4 3
LM1
Similarly, for the matrix A = 2
MM
N1
3
4
−1
OP
5P , the corresponding determinant is
7 QP
1
1 3 1
A = 2 4 5
1 −1 7
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MATHEMATICS
Inverse Of A Matrix And Its Application
A square matrix A is said to be singular if its determinant is zero, i.e. A = 0
A square matrix A is said to be non-singular if its determinant is non-zero, i.e. A ≠ 0
MODULE - III
Sequences And
Series
Example 5.1 Determine whether matrix A is singular or non-singular where
(a) A =
Solution:
LM1
(b) A = M0
MN1
LM−6 −3OP
N4 2Q
(a) Here, A =
OP
PP
Q
2 3
1 2
4 1
Notes
−6 −3
4
2
= (–6)(2) – (4)(–3)
= – 12 + 12 = 0
Therefore, the given matrix A is a singular matrix.
(b)
1 2 3
A = 0 1 2
1 4 1
Here,
=1
1 2
0 2
0 1
−2
+3
4 1
1 1
1 4
= −7 + 4 −3
= −6 ≠ 0
Therefore, the given matrix is non-singular.
Example 5.2 Find the value of x for which the following matrix is singular:
LM1
A= 1
MMx
N
MATHEMATICS
−2
2
2
OP
P
−3PQ
3
1
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Sequences And
Series
Solution: Here,
1 −2
A = 1 2
3
1
x
−3
2 1
1 1
1 2
+2
+3
2 −3
x −3
x 2
=1
Notes
2
= 1( −6 − 2) + 2( −3 − x ) + 3(2 − 2 x )
= − 8 − 6 − 2 x + 6 − 6x
= − 8 − 8x
Since the matrix A is singular, we have A = 0
A = − 8 − 8x = 0
or x = − 1
Thus, the required value of x is –1.
Example 5.3 Given A =
LM1 6OP . Show that A = A′ , where A′ denotes the
N3 2 Q
transpose of the matrix.
LM1 6OP
N3 2 Q
L1 3OP
This gives A′ = M
N6 2Q
Solution: Here, A =
Now,
A=
and
A' =
1 6
3 2
1 3
6 2
= 1 × 2 − 3 × 6 = −16
= 1 × 2 − 3 × 6 = −16
...(1)
...(2)
From (1) and (2), we find that A = A ′
5.2 MINORS AND COFACTORS OF THE ELEMENTS OF
SQUARE MATRIX
LMa
A= a
MMa
N
11
Consider a matrix
21
31
150
a12
a22
a32
a13
a23
a33
OP
PP
Q
MATHEMATICS
Inverse Of A Matrix And Its Application
MODULE - III
The determinant of the matrix obtained by deleting the ith row and jth
Sequences And
Series
column of A, is called the minor of aij and is denotes by Mij .
Cofactor Cij of aij is defined as
Cij = ( −1) i + j Mij
Notes
For example, M23 = Minor of a23
=
a11 a 12
a31 a 32
and C23 = Cofactor of a23
= ( −1)
2+3
M 23
= ( −1) M 23
5
= − M 23 = −
a11
a12
a31
a32
Example 5.4 Find the minors and the cofactors of the elements of matrix A =
Solution: For matrix A, A =
LM2 5OP
N6 3Q
LM2 5OP = 6 − 30 = − 24
N6 3Q
M 11 (minor of 2) = 3; C11 = ( −1)1+1 M11 = ( −1) 2 M11 = 3
M12 (minor of 5) = 6; C12 = ( −1)1+ 2 M12 = ( −1) 3 M12 = − 6
M 21 (minor of 6) = 5; C21 = ( −1) 2 +1 M 21 = ( −1) 3 M 21 = − 5
M 22 (minor of 3) = 2; C22 = ( −1) 2 + 2 M 22 = ( −1) 4 M 22 = 2
Example 5.5 Find the minors and the cofactors of the elements of matrix
LM−1
A= 2
MM 4
N
MATHEMATICS
OP
P
3 PQ
3 6
5 −2
1
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Series
Solution: Here, M11 =
1
3
2 −2
M12 =
4
Notes
3
M13 =
2 5
M 21 =
3 6
4 1
1 3
= 15 + 2 = 17; C11 = ( −1)1+1 M11 = 17
= 6 + 8 = 14; C12 = ( −1)1+ 2 M12 = − 14
= 2 − 20 = − 18; C13 = ( −1)1+ 3 M13 = − 18
= 9 − 6 = 3; C21 = ( −1) 2 +1 M 21 = − 3
M 22 =
−1 6
= ( −3 − 24) = − 27; C22 = ( −1) 2 + 2 M 22 = − 27
4 3
M 23 =
−1 3
= ( −1 − 12) = − 13; C23 = ( −1) 2 + 3 M 23 = 13
4 1
M 31 = 3
5
M 32 =
and M 33 =
5 −2
6 = ( − 6 − 30 ) = − 36; C = ( − 1) 3 +1 M = − 36
31
31
−2
−1 6
= (2 − 12) = − 10; C32 = ( −1) 3+ 2 M 32 = 10
2 −2
−1 3
= ( −5 − 6) = − 11; C33 = ( −1) 3+ 3 M 33 = − 11
2 5
CHECK YOUR PROGRESS 5.1
1.
(a)
152
Find the value of the determinant of following matrices:
L0 6OP
A=M
N2 5Q
(b)
LM 4
B = −1
MM 2
N
OP
1
P
2PQ
5 6
0
1
MATHEMATICS
Inverse Of A Matrix And Its Application
2.
Determine whether the following matrix are singular or non-singular.
(a)
3.
4.
L 3 2 OP
A=M
N−9 −6Q
3
5
OP
1
P
−1QP
2
LM3 −1OP
N7 4 Q
(a)
Find the minors of the elements of the 2nd row of matrix
2
0
−3
B=
LM0 6OP
N2 5Q
A=
(b)
Sequences And
Series
Notes
(a)
(b)
OP
P
1PQ
3
4
Find the minors of the elements of the 3rd row of matrix
LM 2
A= 5
MM−2
N
6.
−1
Find the minors of the following matrices:
LM 1
A = −1
MM−2
N
5.
(b)
LM1
Q= 2
MM4
N
MODULE - III
−1
4
0
OP
1
P
−3PQ
3
Find the cofactors of the elements of each the following matrices:
LM3
N9
−2
7
OP
Q
B=
LM 0 4OP
N−5 6Q
(a)
A=
(a)
Find the cofactors of elements of the 2nd row of matrix
LM 2
A = −1
MM 4
N
(b)
3
1
OP
0
P
−2QP
1
Find the cofactors of the elements of the 1st row of matrix
LM 2
A= 6
MM−5
N
MATHEMATICS
0
(b)
−1
4
−3
OP
−2
P
0 PQ
5
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MODULE - III
Sequences And
Series
7.
I
A=
f
(a)
Notes
LM2 3OP and B = LM−2 3OP , verify that
N4 5Q
N 7 4Q
A = A ′ and B = B ′
(b)
AB = A B = BA
5.3 ADJOINT OF A SQUARE MATRIX
Let A =
LM2 1OP be a matrix. Then
N5 7 Q
A =
2 1
5 7
Let Mij and Cij be the minor and cofactor of aij respectively. Then
M11 = 7 = 7; C11 = ( −1)1+1 7 = 7
M12 = 5 = 5; C12 = ( −1)1+2 5 = − 5
M21 = 1 = 1; C21 = ( −1) 2+1 1 = − 1
M22 = 2 = 2; C22 = ( −1) 2+2 2 = 2
We replace each element of A by its cofactor and get
B=
LM 7 −5OP
N−1 2 Q
...(1)
The transpose of the matrix B of cofactors obtained in (1) above is
B′ =
LM 7 −1OP
N−5 2 Q
...(2)
The matrix B′ obtained above is called the adjoint of matrix A. It is denoted by Adj A.
Thus, adjoint of a given matrix is the transpose of the matrix whose elements are the
cofactors of the elements of the given matrix.
Working Rule: To find the Adj A of a matrix A:
(a) replace each element of A by its cofactor and obtain the matrix of
cofactors; and
(b) take the transpose of the marix of cofactors, obtained in (a).
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MODULE - III
Example 5.6 Find the adjoint of
A=
LM−4 5 OP
N 2 −3Q
Solution: Here, A =
Then,
Sequences And
Series
−4
2
5
Let Aij be the cofactor of the element aij.
−3
A11 = ( −1)1+1 ( −3) = − 3
A21 = ( −1) 2 +1 (5) = − 5
A12 = ( −1)1+2 (2) = − 2
A22 = ( −1) 2 +2 ( −4) = − 4
Notes
We replace each element of A by its cofactor to obtain its matrix of cofators as
LM−3 − 2OP
N−5 − 4Q
...(1)
Transpose of matrix in (1) is Adj A.
Thus, Adj A =
LM−3 −5OP
N−2 −4Q
LM 1
A = −3
Find the adjoint of
MM 5
N
Example 5.7
OP
1
P
−1QP
−1
2
4
2
Solution: Here,
A=
1 −1
2
−3
4
1
5
2 −1
Let Aij be the cofactor of the element aij of A
Then
A11 = ( −1)1+1
A13 = ( −1)1+ 3
4
1
2 −1
= ( −4 − 2) = −6 ; A12 = ( −1)1+ 2
−3
1
5
−1
= − ( 3 − 5) = 2
−3 4
−1 2
= ( −6 − 20) = −26 ; A21 = ( −1) 2 +1
= ( −1 − 4 ) = 3
5 2
2 −1
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Series
A22 = ( −1) 2 + 2
A31 = ( −1) 3+1
1
2
5 −1
−1 2
4
1
= ( −1 − 10) = −11 ; A23 = ( −1) 2 + 3
= ( −1 − 8) = −9 ; A32 = ( −1) 3+ 2
Notes
A33 = ( −1)
and
3+ 3
1
−3
1 −1
5
1
2
2
−3 1
= − ( 2 + 5) = −7
= − (1 + 6) = −7
−1
= ( 4 − 3) = 1
4
Replacing the elements of A by their cofactors, we get the matrix of cofactors as
LM−6
MM 3
N −9
OP
P
1 PQ
LM −6
Thus, Adj A = M 2
MN−26
2 −26
−11 −7
−7
OP
P
1 PQ
3 −9
−11 −7
−7
If A is any square matrix of order n, then A(Adj A) = (Adj A) A = |A| In
where In is the unit matrix of order n.
Verification:
(1)
Consider A =
LM 2 4OP
N − 1 3Q
2 4
or |A| = 2 × 3 − ( − 1) × (4) = 10
−1 3
Then
| A| =
Here,
A11=3; A12=1; A21= − 4 and A22=2
Therefore,
Adj A =
LM3 −4OP
N1 2 Q
2
Now, A (AdjA) = 
 −1
(2)
LM3
Consider, A = M2
MN1
4  3
3  1
5
−3
1
−4  10
=
2  0
0 
1
=
10
0
10 

0
= A I2
1 
OP
1P
2 PQ
7
Then, A = 3( − 6 − 1) − 5 (4 − 1) + 7 (2+3) = − 1
Here, A11 = − 7; A12= − 3; A13=5
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MODULE - III
A21= − 3; A22= − 1; A23=2
Sequences And
Series
A31=26; A32=11; A33= − 19
LM−7
−3
Adj A = M
MN 5
Therefore,
Now
LM3
(A) (Adj A) = M2
NM1
OP
− 1 11
P
2 − 19 PQ
−3
26
OP LM−7
PP M−3
2 Q MN 5
5 7
−3 1
1
Notes
OP
− 1 11
P
2 − 19 PQ
−3
26
LM−1 0 0 OP
LM1 0 0OP
= M 0 −1 0 P = ( − 1) M0 1 0P = |A|I
MN 0 0 −1PQ
MN0 0 1PQ
LM−7 −3 26 OP LM3 5 7OP
(Adj A) A = M −3 −1 11 P M2 −3 1P
MN 5 2 −19PQ MN1 1 2PQ
LM−1 0 0 OP
LM1 0 0OP
= M 0 −1 0 P = (_1) M0 1 0P = |A|I
MN 0 0 −1PQ
MN0 0 1PQ
3
Also,
3
Note : If A is a singular matrix, i.e. |A|=0 , then A (Adj A) = O
CHECK YOUR PROGRESS 5.2
1. Find adjoint of the following matrices:
(a)
LM2 −1OP
N3 6 Q
(b)
LMa b OP
Nc d Q
(c)
LM cos
N− sin 
sin 
cos 
OP
Q
2. Find adjoint of the following matrices :
(a)
LM 1
N2
OP L
Q MN
i
2
(b)
i
1
−i
i
OP
Q
Also verify in each case that A(Adj A) = (Adj A) A = |A|I2.
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Sequences And
Series
3. Verify that
A(Adj A) = (Adj A) A = |A| I3, where A is given by
LM 6
0
(a) M
MN−3
Notes
OP
PP
Q
LM2
0
(b) M
MN3
8 −1
5 4
2 0
LMcos
sin 
(c) M
MN 0
OP
PP
Q
− sin 
cos
0
0
0
1
LM 4
−1
(d) M
MN−4
7 9
−1 2
−7 4
−6
−1
11
OP
PP
Q
OP
1P
− 1 PQ
1
5.4 INVERSE OF A MATRIX
Consider a matrix A =
LMa b OP . We will find, if possible, a matrix
Nc d Q
LM x y OP such that AB = BA = I
Nu v Q
LMa b OP LM x y OP = LM1 0OP
Nc d Q Nu v Q N0 1Q
LMax + bu ay + bv OP = LM1 0OP
Ncx + du cy + dvQ N0 1Q
B=
i.e.,
or
On comparing both sides, we get
ax + bu = 1 ay + bv = 0
cx + du = 0
cy + dv = 1
Solving for x, y, u and v, we get
x=
d
−b
−c
a
, y=
,u=
,v=
ad − bc
ad − bc
ad − bc
ad − bc
a bO
L
provided ad − bc ≠ 0 , i.e. , M
Nc d PQ ≠ 0
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MATHEMATICS
Inverse Of A Matrix And Its Application
Thus,
or
LM d
ad − bc
B=M
MM −c
N ad − bc
MODULE - III
OP
PP
a
P
ad − bc Q
Sequences And
Series
−b
ad − bc
LM
N
d −b
1
B=
ad − bc − c a
Notes
OP
Q
It may be verified that BA = I.
It may be noted from above that, we have been able to find a matrix.
B=
LM
N
d −b
1
ad − bc − c a
OP
Q
=
1
Adj A
A
...(1)
This matrix B, is called the inverse of A and is denoted by A-1.
For a given matrix A, if there exists a matrix B such that AB = BA = I, then B is
called the multiplicative inverse of A. We write this as B= A-1.
Note: Observe that if ad − bc = 0, i.e., |A| = 0, the R.H.S. of (1) does not exist and
B (=A-1) is not defined. This is the reason why we need the matrix A to be non-singular
in order that A possesses multiplicative inverse. Hence only non-singular matrices possess
multiplicative inverse. Also B is non-singular and A=B-1.
Example 5.8 Find the inverse of the matrix
Solution :
4
L
A= M
N2
L4
A= M
N2
OP
Q
5O
−3PQ
5
−3
Therefore, |A| = − 12 − 10 = − 22 ≠ 0
∴A is non-singular. It means A has an inverse. i.e. A-1 exists.
−3 −5O
L
Now, Adj A = M
N−2 4 PQ
1
1  −3
A−1 = adj A =
A
−22  −2
MATHEMATICS
3 5 
−5  22 22 
=

4   1
2
−
11 11 
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MODULE - III
Sequences And
Series
Notes
Note: Verify that AA-1 = A-1A = I
Example 5.9 Find the inverse of matrix
LM3 2 −2OP
A = 1 −1 6
MM5 4 −5PP
N
Q
LM3 2 −2OP
A = 1 −1 6
Solution : Here,
MM5 4 −5PP
N
Q
∴
∴
|A| =
3(5 − 24) − 2( − 5 − 30) − 2(4 + 5)
=
3( − 19) − 2( − 35) − 2(9)
=
− 57 + 70 − 18
−5 ≠ 0
=
A − 1 exists.
Let Aij be the cofactor of the element aij.
Then,
A11 = ( −1)1+1
−1 6
= 5 − 24 = − 19 ,
4 −5
A12 = ( −1)1+ 2
1 6
= − ( −5 − 30) = 35 .
5 −5
A13 = ( −1)1+ 3
5
4
= 4−5 = 9,
A21 = ( −1) 2 +1
2 −2
= − ( −10 + 8) = 2
4 −5
A22 = ( −1) 2 + 2
3 −2
= − 15 + 10 = − 5 ,
5 −5
A23 = ( −1) 2 + 3
160
1 −1
3 2
5 4
= − (12 − 10) = − 2
MATHEMATICS
Inverse Of A Matrix And Its Application
2 −2
= 12 − 2 = 10 ,
−1 6
A31 = ( −1) 3+1
A = (−1)3 + 2
32
A33 = ( −1) 3+ 3
and
LM−19
2
Matrix of cofactors = M
MN 10
1
1
A
=
.
Adj
A
=
∴
A
−5
−1
MODULE - III
3
−2
1
6
Sequences And
Series
= − (18 + 2 ) = −20
Notes
3 2
= −3 − 2 = −5
1 −1
OP
PP
Q
LM
MM
N
OP
PP
Q
9
−19 2 10
−2 . Hence AdjA= 35 −5 −20
−20 −5
9 −2 −5
35
−5
LM−19
MM 359
N
2
−5
−2
19
5

=  −7
 −9

5
OP
−20
P
−5 PQ
10
−2
5
1

−2 

4

1

2
5
Note : Verify that A-1A = AA-1 = I3
Example 5.10
1 0O
−2 1 O
L
L
If A = M
N2 −1PQ and B= MN 0 −1PQ ; find
(i) (AB)
−1
−1
(ii) B A
−1
(iii) Is (AB)
−1
−1
=B A
−1
?
1 0 O L −2 1 O
L
Solution : (i) Here, AB = M
N2 −1PQ MN 0 −1PQ
L−2 + 0 1 + 0OP = LM−2 1OP
= M
N−4 + 0 2 + 1Q N−4 3Q
∴
−2 1
|AB|
=
−4 3
= − 6 + 4 = − 2 ≠ 0.
Thus, ( AB ) −1 exists.
Let us denote AB by Cij
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Sequences And
Series
Notes
Let Cij be the cofactor of the element cij of |C|.
Then,
C11= ( − 1)1+1 (3) = 3
C21 = ( − 1)2+1 (1) = − 1
C12= ( − 1)1+2 ( − 4) = 4
C22 = ( − 1)2+2 ( − 2) = − 2
3
L
Hence, Adj (C) = M
N4
OP
−2 Q
−1
 −3
−1  
= 2
−2  
 −2
1
1 3
C =
Adj ( C ) =
C
−2  4
−1
−1
C
(ii)
= (AB)
To find B
N
∴
o
w
,
B
−1
=
−1
 −3

=2
 −2
1
2

1
1
2

1
−1
A , first we will find B-1.
LM−2 1 OP ∴ |B| = LM−2 1 OP = 2 −0 = 2 ≠ 0
N 0 −1Q
N 0 −1Q
B-1 exists.
Let Bij be the cofactor of the element bij of |B|
then
B11 = ( − 1)1+1 ( − 1) = − 1
B12 = ( − 1)1+2 (0) = 0 and
Hence, Adj
∴ B
−1
B22 = ( − 1)2+2 ( − 2) = − 2
−1 −1O
L
B= M
N 0 −2PQ
L−1 − 1 OP
= 1 . AdjB = 1 M
2 N0 − 2 Q
B
Also, A =
B21 = ( − 1)2+1 (1) = − 1
 −1
= 2

0
−1 
2

−1 
LM1 0 OP Therefore, A = 1 0 = 1 −0 = −1
2 −1
N2 −1Q
≠0
Therefore, A-1 exists.
Let Aij be the cofactor of the element aij of |A|
then
162
A11 = ( − 1)1+1 ( − 1) = − 1
A21 = ( − 1)2+1 (0) = 0
A12 = ( − 1)1+2 (2) = − 2 and
A22 = ( − 1)2+2 (1) = 1
MATHEMATICS
Inverse Of A Matrix And Its Application
Hence, Adj A=
MODULE - III
LM −1 0OP
N − 2 1Q
Sequences And
Series
LM
N
OP LM
Q N
OP
Q
−1 0
1 0
⇒ A −1 = 1 Adj A = 1
=
−1 −2 1
A
2 −1
 −1
B A = 2

0
−1
Thus,
−1 
2

−1 
−1
1
2

0 
−1
1
 −1
−1 0 + 
= 2
2 =


0 − 2 0 + 1 
(iii)
Notes
 −3
2

 −2
1
2

1
From (i) and (ii), we find that
−1
−1
(AB) =B A
−1
 −3

== 2
 −2
−1
−1
Hecne, (AB) = B A
1
2

1
−1
CH ECK YOUR PROGRESS 5.3
1.
Find, if possible, the inverse of each of the following matrices:
(a)
2.
LM1 3OP
N2 5Q
(b)
LM−1 2 OP
N−3 −4Q
(c)
LM2 −1OP
N1 0 Q
Find, if possible, the inverse of each of the following matrices :
(a)
LM1
MM24
N
OP
P
2PQ
0 2
1 3
1
LM3
5
(b) M
MN1
−1
2
−3
OP
P
−2PQ
2
4
Verify that A−1 A = AA−1 = I for (a) and (b).
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MODULE - III
Sequences And
Series
3.
If
LM1
0
A= M
MN3
Notes
4.
5.
Find
( A ′ ) −1
LM0
1
If A= M
MN1
OP
PP
Q
LM
MM
N
OP
PP
Q
2 3
2 −1 0
−1
−1 4 and B= 1 4 3 , verify that
( AB ) = B −1 A−1
1 5
3 0 −2
LM 1
0
if A = M
MN−2
OP
PP
Q
−2 3
−1 4
2 1
OP
Lb + c
1M
1 and B = c − b
P
2M
MNb − c
0PQ
OP
a − bP
a + b PQ
1 1
c−a b− a
0
1
c+a
a−c
show that ABA−1 is a diagonal matrix.
6.
7.
8.
9.
10.
164
LMcos x − sin x 0OP
If φ( x ) = M sin x cos x 0P , show that  ( x )  =  ( − x ) .
MN 0 0 1PQ
cos 2 x − sin 2 x O
1
tan x O
L
L
A
A
=
′
If A = M
MN sin 2 x cos 2 x PQ
N− tan x 1 PQ, show that
LMa 1+bbc OP
If A = c
MN a PQ , show that aA =(a + bc + 1) I −aA
LM−1 2 0OP
−1 1 1 , show that A = A
If A = M
MN 0 1 0PPQ
−1
−1
−1
−1
LM
MM
N
2
2
OP
PP
Q
−8 1 4
1
4 −4 7 , show that −1
If A = 9
A = A′
1 −8 4
MATHEMATICS
Inverse Of A Matrix And Its Application
5.5 SOLUTION OF A SYSTEM OF LINEAR EQUATIONS.
In earlier classes, you have learnt how to solve linear equations in two or three unknowns
(simultaneous equations). In solving such systems of equations, you used the process of
elimination of variables. When the number of variables invovled is large, such elimination process
becomes tedious.
You have already learnt an alternative method, called Cramer’s Rule for solving such systems
of linear equations.
MODULE - III
Sequences And
Series
Notes
We will now illustrate another method called the matrix method, which can be used to solve the
system of equations in large number of unknowns. For simplicity the illustrations will be for
system of equations in two or three unknowns.
5.5.1 MATRIX METHOD
In this method, we first express the given system of equation in the matrix formAX = B, where
A is called the co-efficient matrix.
For example, if the given system of equation is a1x + b1y = c1 and a2 x + b2 y = c2, we express
them in the matrix equation form as :
LMa
Na
OP LM x OP = LMc OP
Q N y Q Nc Q
La b OP , X= LM xOP and B = LMc OP
A= M
Na b Q N y Q
Nc Q
1
2
Here,
b1
b2
1
2
1
1
1
2
2
2
If the given system of equations is a1 x + b1 y + c1 z = d1 and
a2 x + b2 y + c2 z = d2 and a3 x + b3 y + c3 z = d3,then this system is expressed in the
matrix equation form as:
LMa
MMaa
N
1
2
3
b1
b2
b3
LMa
a
Where, A = M
MNa
1
2
3
c1
c2
c3
b1
b2
b3
OPLM x OP LMd OP
PPMM yzPP = MMdd PP
QN Q N Q
1
2
3
OP
PP
Q
LM OP
MM PP
NQ
LM
MM
N
c1
x
d1
c2 , X = y and B= d 2
c3
z
d3
OP
PP
Q
Before proceding to find the solution, we check whether the coefficient matrixA is non-singular
or not.
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MODULE - III
Sequences And
Series
Note: If A is singular, then |A|=0. Hence, A-1 does not exist and so, this method does not
work.
a
L
Consider equation AX = B, where A = M
Na
1
2
Notes
OP
Q
LM OP
NQ
LM OP
NQ
b1
x
c1
,X=
and B =
b2
y
c2
When |A| ≠0, i.e. when a1b2_a2b1≠ 0, we multiply the equation AX = B with A-1 on both side
and get
A-1(AX) = A-1B
⇒
(A-1A) X = A-1B
⇒
IX = A-1B
⇒
X = A-1B
(  A-1A = I)
LM −b OP, we get
N aQ
LM b −b OPLMc OP
1
X=
a Q Nc Q
a b − a b N− a
LM x OP = 1 LM b c − b c OP
N y Q a b − a b N− a c + a c Q
−1
Since A =
b2
1
a1b2 − a2b1 − a2
1
1
2
1 2
∴
2
2 1
=
2 1
1
1
2 1
1 2
Hence, x =
1
2 1
2
1 2
1 2
 b2 c1 − b1 c2 
a b − a b 
 1 2 2 1 
 − a2 c1 + a1 c2 
 a b −a b 
 1 2 2 1 
b2 c1 − b1c2
a1c2 − a2 c1
and y =
a1b2 − a2b1
a1b2 − a2b1
Example 5.11 Using matrix method, solve the given system of linear equations.
U|
V
3 x + 7 y = −1|W
4 x − 3 y = 11
......(i)
Solution: This system can be expressed in the matrix equation form as
LM4 −3OP LM x OP = LM11OP
N3 7 Q N yQ N−1Q
166
......(ii)
MATHEMATICS
Inverse Of A Matrix And Its Application
MODULE - III
LM4 −3OP , X = LM xOP , and B = LM11OP
N3 7 Q N y Q
N−1Q
Here,
A=
so,
(ii) reduces to
AX = B
Sequences And
Series
......(iii)
Notes
4 −3
= 28 + 9 = 37 ≠ 0
3 7
Now,
| A| =
Since
|A| ≠ 0, A-1 exists.
Now, on multiplying the equation AX = B with A-1 on both sides, we get
A-1 (AX) = A-1B
(A-1 A)X = A-1B
IX = A-1B
i.e.
X = A-1B
X=
Hence,
LM x OP = 1 LM 7 3OP LM11OP
N yQ 37 N−3 4Q N−1Q
LM x OP = 1 LM 74 OP
N yQ 37 N−37Q
LM x OP = LM 2 OP
N yQ N−1Q
or
So,
1
( Adj A) B
A
=
LM
N
1 77 − 3
37 −33 − 4
OP
Q
x = 2, y = − 1 is unique solution of the system of equations.
Example 5.12 Solve the following system of equations, using matrix method.
2x − 3y =7
x + 2y = 3
Solution : The given system of equations in the matrix equation form, is
LM2 −3OP LM x OP = LM7OP
N1 2 Q N yQ N3Q
MATHEMATICS
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Inverse Of A Matrix And Its Application
MODULE - III
Sequences And
Series
or,
AX = B
...(i)
2 −3O
xO
7O
L
L
L
A =M
N1 2 PQ , X = MN yPQ and B = MN3PQ
where
Notes
A=
Now,
LM2
N1
OP = 2 × 2 − 1 × (−3)
2Q
−3
=4+3=7≠0
∴
A-1 exists
Since,
Adj (A) =
LM 2 3 OP
N−1 2 Q
1
1 2
A−1 = adj ( A ) = 
A
7  −1
From (i), we have
X=A
2
7
or, X = 
 −1
 7
x 
or,   =
 y 
Thus, x =
3
7

2
7 
−1
2
3  7
=
2   −1
 7
3
7

2  .... (ii)
7 
B
7 
 
 
3 
 23 
7
 
 −1 
 7 
23
−1
,y=
is the solution of this system of equations.
7
7
Example 5.13 Solve the following system of equations, using matrix method.
U|
x − 2 y + z = 0V
2 x + 3y − z = 5 |
W
x + 2 y + 3 z = 14
168
MATHEMATICS
Inverse Of A Matrix And Its Application
Solution : The given equations expressed in the matrix equation form as :
LM1
MM21
N
OP LM x OP LM14OP
1
y = 0
P
M
P M P
−1PQ MN z PQ MN 5 PQ
2
3
−2
3
MODULE - III
Sequences And
Series
... (i)
Notes
which is in the form AX = B, where
LM1
1
A= M
MN2
∴
OP
PP
Q
LM OP
MM PP
NQ
LM
MM
N
2 3
x
14
−2 1 , X = y and B = 0
3 −1
z
5
OP
PP
Q
X = A-1 B
... (ii)
Here, |A| = 1 (2 − 3) − 2 ( − 1 − 2) +3 (3+4)
= 26 ≠ 0
∴
A-1 exists.
 −1
A = 3
7
Also, Adj
11
−7
1
8 
2 
−4 
−1
1
Hence, from (ii), we have X = A B = A AdjA. B
LM
MM
N
−1 11 8
1
X =
3 −7 2
26
7 1 −4
LM
MM
N
OP
PP
Q
LM xOP LM1OP
MM yzPP = MM23PP
NQ NQ
OP LM14OP
PP MM 05 PP
QN Q
LM OP
MM PP
NQ
26
1
1
=
52 = 2
26
78
3
or,
MATHEMATICS
169
Inverse Of A Matrix And Its Application
MODULE - III
Sequences And
Series
Thus, x =1, y = 2 and z = 3 is the solution of the given system of equations.
Example 5.14 Solve the following system of equations, using matrix method :
x+2y+z = 2
2x − y+3z = 3
Notes
x+3y − z = 0
Solution: The given system of equation can be represented in the matrix equation form as :
LM1
MM21
N
i.e.,
where
Now,
Hence,
Also,
2
−1
3
OP
P
−1PQ
1
3
LM x OP
MM yzPP
NQ
=
LM2OP
MM03PP
NQ
 (1)
AX = B
LM1 2 1 OP LM xOP
LM2OP
2 −1 3 , X = y and B = 3
A= M
MM0PP
MN1 3 −1PPQ MMN z PPQ
NQ
LM1 2 1 OP
| A| = 2 −1 3 = 1(1 9) 2 ( 2 3) +1 (6 +1) = 9 ≠ 0
MM1 3 −1PP − − − −
N
Q
A-1 exists.
LM−8
5
Adj A = M
MN 7
A −1
OP
−2 −1
P
−1 −5PQ
L−8
1M
1
=
5
=
Adj A
9M
| A|
MN 7
5
7
OP
P
−5PQ
5 7
−2 −1
−1
From (i) we have X = A-1 B
i.e.,
170
LM x OP 1 LM−8
MM yzPP = 9 MM 75
NQ N
5
−2
−1
OP LM2OP
−1 3
PMP
−5PQ MN0PQ
7
MATHEMATICS
Inverse Of A Matrix And Its Application
MODULE - III
LM− 1 OP
LM−1OP M 9 P
= 1M 4P=M 4 P
9
MN11 PQ MM119 PP
MN 9 PQ
so, x = −
5.6
Sequences And
Series
Notes
1
4
11
,y= ,z =
is the solution of the given system.
9
9
9
CRITERION FOR CONSISTENCY OF A SYSTEM OF
EQUATIONS
Let AX = B be a system of two or three linear equations.
Then, we have the following criteria :
(1)
If |A | ≠ 0, then the system of equations is consistent and has a unique
solution, given by X=A-1B.
(2)
If |A| = 0, then the system may or may not be consistent and if consistent,
it does not have a unique solution. If in addition,
(a)
(Adj A) B ≠ O, then the system is inconsistent.
(b)
(Adj A) B = O, then the system is consistent and has infinitely
many solutions.
Note : These criteria are true for a system of 'n' equations in 'n' variables as well.
We now, verify these with the help of the examples and find their solutions wherever possible.
5x + 7y = 1
(a)
2 x − 3y = 3
This system is consistent and has a unique solution, because
equation is
LM5 7 OP LM x OP = LM1OP
N2 −3Q N yQ N3Q
i.e.
where, A =
AX = B
5
7
≠
Here, the matrix
2 −3
.... (i)
LM5 7 OP , X = LM xOP and B = LM1OP
N2 −3Q N yQ
N3Q
MATHEMATICS
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Inverse Of A Matrix And Its Application
MODULE - III
Sequences And
Series
Notes
Here, |A| = 5 × ( − 3) − 2 × 7 = − 15 − 14 = − 29 ≠ 0
and A
−1
=
1
A
Adj A =
LM−3 − 7OP
−29 N−2 5 Q
1
.... (ii)
From (i), we have X = A −1 B
i.e.,
LM 24 OP
LM x OP = 1 LM−3 − 7OPLM1 OP = M 29 P
N y Q −29 N−2 5 QN3Q M− 13 P
N 29 Q
Thus, x =
[From (i) and (ii)]
24
−13 is the unique solution of the given system of equations.
, and y =
29
29
3x + 2 y = 7
(b)
6x + 4y = 8
This system is incosisntent i.e. it has no solution because
3 2 7
= ≠
6 4 8
In the matrix form the system can be written as
LM3 2OP LM x OP = LM7OP
N6 4 Q N y Q N8 Q
or,
where
Here,
AX = B
3 2O
xO
7O
L
L
L
A= M
N6 4PQ , X= MN yPQ and B = MN8PQ
|A| = 3 × 4 − 6 × 2 = 12 − 12 = 0
4 −6O
L
Adj A = M
N−6 3 PQ
Also,
172
4
(Adj A) B = 
 −6
−6  7   −20 
=
≠0
3  8   −18 
MATHEMATICS
Inverse Of A Matrix And Its Application
Thus, the given system of equations is inconsistent.
(c)
UV
W
3x − y = 7
This system is consistent and has infinitely
9 x − 3 y = 21
many solutions, because
3 −1 7
=
=
9 −3 21
MODULE - III
Sequences And
Series
Notes
In the matrix form the system can be written as
LM3 −1OP LM x OP = LM 7 OP
N9 −3Q N yQ N21Q
or,
AX =B, where
A=
LM3 −1OP ; X = LM xOP and B = LM 7 OP
N9 −3Q N yQ
N21Q
Here, |A| =
3 −1
= 3 × ( − 3) − 9 x ( − 1)
9 −3
= − 9+9 = 0
Adj A =
LM−3 1OP
N−9 3Q
 −3
Also, (Adj A)B = 
 −9
∴
1
3
7  0
 21 = 0  = 0
   
The given system has an infinite number of solutions.
Till now, we have seen that
(i)
if |A| ≠ 0 and (Adj A) B ≠ O, then the system of equations has a non-zero
unique solution.
(ii)
if |A| ≠ 0 and (Adj A) B = O, then the system of equations has only trivial
solution x = y = z = 0
(iii)
if |A| = 0 and (Adj A) B = O, then the system of equations has infinitely many
solutions.
(iv)
if |A| = 0 and (Adj A) B ≠ O, then the system of equations is inconsistent.
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MODULE - III
Sequences And
Series
Let us now consider another system of linear equations, where |A| = 0 and (Adj A) B ≠ O.
Consider the following system of equations
x + 2y + z = 5
2x + y + 2z = − 1
x − 3y + z = 6
Notes
In matrix equation form, the above system of equations can be written as
LM1
MM21
N
i.e.,
where
Now,
Also,
2
OP LM x OP LM 5 OP
2 y = −1
PM P M P
1PQ MN z PQ MN 6 PQ
1
1
−3
AX = B
LM1
2
A= M
MN1
2
1
−3
OP LM xOP
LM 5 OP
2 , X = y and B= −1
P MM z PP
MM 6 PP
1PQ
NQ
N Q
1
1 2 1
2 1 2 =0
|A| =
1 −3 1
LM 7
0
(Adj A ) B = M
MN−7
LM 58 OP
0
= M
P
MN−58PQ
Since
(C1 = C3 )
OP LM 5 OP
0
−1 [Verify (Adj A) yourself]
P
M
P
−3PQ MN 6 PQ
−5
3
0
5
≠O
|A| = 0 and (Adj A ) B ≠ O,
LM xOP 1
MM yzPP = | A| ( Adj
NQ
174
A) B
MATHEMATICS
Inverse Of A Matrix And Its Application
MODULE - III
LM 58 OP
0
M
P
= M−58P which is undefined.
N Q
Sequences And
Series
0
Notes
The given system of linear equation will have no solution.
Thus, we find that if |A| = 0 and (Adj A) B ≠ O then the system of equations will have no
solution.
We can summarise the above finding as:
(i)
If |A| ≠ 0 and (Adj A) B ≠ O then the system of equations will have a non-zero,
unique solution.
(ii)
If |A| ≠ 0 and (Adj A) B = O, then the system of equations will have trivial
solutions.
(iii)
If |A| = 0 and (Adj A) B = O, then the system of equations will have infinitely
many solutions.
(iv)
If |A| = 0 and (Adj A) B ≠ O, then the system of equations will have no solution
Inconsistent.
Example 5.15 Use matrix inversion method to solve the system of equations:
2 x − y + 3z = 1
6x + 4 y = 2
(i)
Solution: (i)
9x + 6y = 3
(ii)
x + 2 y − 3z = 2
5 y − 5z = 3
The given system in the matrix equation form is
LM6 4OP LM x OP = LM2OP
N9 6Q N yQ N3Q
i.e.,
AX = B
where,
A=
Now,
A =
∴
LM6 4OP, X = LM x OP and B = LM2OP
N9 6 Q N y Q
N3Q
6 4
= 6 × 6 − 9 × 4 = 36 − 36 = 0
9 6
The system has either infinitely solutions or no solution.
MATHEMATICS
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MODULE - III
Sequences And
Series
Let x = k, then 6k + 4 y = 2 gives y =
1 − 3k
2
Putting these values of x and y in the second equation, we have
9k + 6
Notes
FG 1 − 3k IJ = 3
H 2 K
⇒ 18k + 6 − 18k = 6
⇒ 6 = 6, which is true.
∴
The given system has infinitely many solutions. These are
x = k, y =
(ii)
1 − 3k
, where k is any arbitrary number..
2
The given equations are
 (1)
2 x − y + 3z = 1
( 2)
x + 2y − z = 2
 ( 3)
5 y − 5z = 3
In matrix equation form, the given system of equations is
LM2
MM10
N
i.e.,
−1
2
5
OP LM x OP LM1OP
−1 y = 2
PMP MP
−5PQ MN z PQ MN 3PQ
3
AX = B
3 
x 
1



−1 , X =  y 
and B = 2
 z 
−5
3
5
2 −1
Now,
LM OP
MM PP
NQ
−1
2
2
A = 1
where,
0
A = 1
0
2
5
3
2 −1
1 −1
1 2
−1 = 2 ×
+ 1×
+ 3×
5 −5
0 −5
0 5
−5
= 2(–10+5)+1(–5–0)+3(5–0)
= –10–5+15 = 0
176
MATHEMATICS
Inverse Of A Matrix And Its Application
∴
The system has either infinitely many solutions or no solution
Let z = k. Then from (1), we have 2x–y = 1–3k; and from (2), we have x+2y = 2+k
MODULE - III
Sequences And
Series
Now, we have a system of two equations, namely
2 x − y = 1 − 3k
Notes
x + 2y = 2 + k
LM2
N1
L2
A=M
N1
OP LM x OP = LM1 − 3k OP
Q N yQ N 2 + k Q
−1O
xO
1 − 3k O
L
L
,
X
=
B
=
MN yPQ and MN 2 + k PQ
2 PQ
−1
2
⇒
Let
Then
∴
A =
2 −1
= 4 +1 = 5 ≠ 0
1 2
A–1 exists.
1
A
2
L
1 L2 1 O M 5
=
Adj A = 5 MN−1 2 PQ M 1
MN− 5
Here,
A −1 =
∴
The solution is X = A–1B
2
5
=
− 1
 5
∴
1
5

2
5 
1
5
2
5
OP
PP
Q
4

−k + 

1 − 3k 
5

2 + k  = 
3

 k + 

5 
4
3
x = − k + , y = k + , where k is any number..
5
5
Putting these values of x, y and z in (3), we get
FG
H
5 k+
⇒
IJ
K
3
− 5( k ) = 3
5
5k + 3 − 5k = 3 ⇒ 3 = 3, which is true.
MATHEMATICS
177
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MODULE - III
Sequences And
Series
∴
The given system of equations has infinitely many solutions, given by
4
3
x = − k + , y = k + and z = k, where k is any number..
5
5
5.7 HOMOGENEOUS SYSTEM OF EQUATIONS
Notes
A system of linear equations AX = B with matrix, B = O, a null matrix, is called homogeneous
system of equations.
Following are some systems of homogeneous equations:
(i)
2 x + 5 y − 3z = 0
x + 2y = 0
−2 x + 3 y = 0
(ii)
x − 2y + z = 0
3x − y − 6 z = 0
2 x + y − 3z = 0
(iii)
x − 2y + z = 0
3x − y − 2 z = 0
Let us now solve a system of equations mentioned in (ii).
Given system is
2 x + 5 y − 3z = 0
x − 2y + z = 0
3x − y − 6z = 0
In matrix equation form, the system (ii) can be written as
LM2
MM13
N
i.e.,
5
−2
−1
AX = O
2
A = 1
where
3
Now
OP LM x OP LM0OP
1
y = 0
P
M
P MM0PP
−6PQ MN z PQ
NQ
−3
5
−2
−1
−3 
x 
0



1  , X =  y  and B= 0 
 z 
0 
−6 
A = 2 (12 + 1) − 5(−6 − 3) − 3(−1 + 6)
= 26+45 − 15
= 56 ≠ 0
178
MATHEMATICS
Inverse Of A Matrix And Its Application
But
B = O ⇒ (Adj A) B = O.
LM x OP
MM y PP = 1A
Nz Q
Thus,
MODULE - III
Sequences And
Series
( Adj A) B
Notes
= O.
∴
x = 0, y = 0 and z = 0.
i.e.,
the system of equations will have trivial solution.
Remarks: For a homogeneous system of linear equations, if A ≠ 0 and (Adj A) B = O.
There will be only trivial solution.
Now, cousider the system of equations mentioned in (iii):
2 x + y − 3z = 0
x − 2y + z = 0
3x − y − 2 z = 0
In matrix equation form, the above system (iii) can be written as
LM2
MM13
N
i.e.,
OP LM x OP LM0OP
y = 0
P
M
P MM0PP
−2PQ MN z PQ
NQ
1 −3
−2 1
−1
AX = 0
where,
LM2
A= 1
MM3
N
Now,
LM2
A = 1
MM3
N
Also, B
1
−2
−1
OP LM x OP
LM0OP
1 , X = y and B = 0
P MM z PP
MM0PP
−2PQ
NQ
NQ
−3
OP
P
−2PQ
1 −3
−2 1
−1
= 2(4 + 1) − 1( −2 − 3) − 3( −1 + 6)
= 10 + 5 − 15
=0
= O ⇒ (Adj A) B = O
MATHEMATICS
179
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MODULE - III
Sequences And
Series
Notes
Thus,
⇒
LM x OP
MM yzPP =
NQ
LM OP
MM PP
NQ
0
1
1
=
0
(Adj A) B 0
A
0
The system of equations will have infinitely many solutions which will be non-trivial.
Considering the first two equations, we get
2 x + y = 3z
x − 2y = − z
Solving, we get x = z, y = z, let z = k, where k is any number.
Then x = k, y = k and z = k are the solutions of this system.
For a system of homogenous equations, if A = 0 and, (AdjA) B = O, there
will be infinitely many solutions.
CHECK YOUR PROGRESS 5.4
1.
Solve the following system of equations, using the matrix inversion method:
(a)
2 x + 3y = 4
(b)
x − 2y = 5
(c)
3x − 7 y = 11
3x + 4 y − 5 = 0
(d)
x − 2y + 6 = 0
2.
2 x − 3y + 6 = 0
6x + y − 8 = 0
Solve the following system of equations using matrix inversion method:
(a)
(c)
180
x+ y = 7
x + 2y + z = 3
(b)
2 x + 3 y + z = 13
2 x − y + 3z = 5
3x + 2 y − z = 12
x+y−z=7
x + y + 2z = 5
− x + 2 y + 5z = 2
(d)
2x + y − z = 2
2 x − 3 y + z = 15
x + 2 y − 3z = − 1
−x + y + z = − 3
5x − y − 2 z = − 1
MATHEMATICS
Inverse Of A Matrix And Its Application
MODULE - III
3. Solve the following system of equations, using matrix inversion method:
(a)
x+ y+z = 0
(b) 3x − 2 y + 3z = 0
2x − y + z = 0
2x + y − z = 0
x − 2 y + 3z = 0
4 x − 3y + 2z = 0
(c)
Sequences And
Series
x + y +1 = 0
y + z −1 = 0
z+x = 0
Notes
4. Determine whether the following system of equations are consistent or not. If consistent, find
the solution:
(a)
2 x − 3y = 5
(b)
x+ y = 7
(c)
3x + y + 2 z = 3
2 x − 3y = 5
4 x − 6 y = 10
(d)
−2 y − z = 7
x + 15 y + 3z = 11
x + 2 y − 3z = 0
4 x − y + 2z = 0
3x + 5 y − 4 z = 0
LET US SUM UP
•
A square matrix is said to be non-singular if its corresponding determinant is non-zero.
•
The determinant of the matrix A obtained by deleting the ith row and jth column of A, is
called the minor of aij. It is usually denoted by Mij.
•
The cofactor of aij is defined as Cij = ( − 1)i+j Mij
•
Adjoint of a matrix A is the transpose of the matrix whose elements are the cofactors of
the elements of the determinat of given matrix. It is usually denoted by AdjA.
•
If A is any square matrix of order n, then
A (Adj A) = (Adj A) A = A In where In is the unit matrix of order n.
•
For a given non-singular square matrix A, if there exists a non-singular square matrix B
such that AB = BA = I, then B is called the multiplicative inverse of A. It is written as
B = A–1.
•
Only non-singular square matrices have multiplicative inverse.
•
If a1 x + b1 y = c1 and a2 x + b2 y = c2, then we can express the system in the matrix
equation form as
MATHEMATICS
181
Inverse Of A Matrix And Its Application
MODULE - III
 a1
a
 2
Sequences And
Series
Notes
Thus,
L
A=M
Na
a1
i
f
2
X = A −1 B =
b1   x  c1 
=
b2   y  c2 
LMc OP
MNc PQ , then
OP, X = LM OP and B =
bQ
N yQ
b1
1
x
2
2
LM
N
b2
− b1
1
a1b2 − a2 b1 − a2 a1
OP LMc OP
Q Nc Q
1
2
•
A system of equations, given by AX = B, is said to be consistent and has a unique
solution, if |A| ≠ 0.
•
A system of equations, given by AX = B, is said to be inconsistent, if |A| = 0 and
(Adj A) B ≠ O.
•
A system of equations, given by AX = B, is said be be consistent and has infinitely
many solutions, if |A| = 0, and (Adj A) B = O.
•
A system of equations, given by AX = B, is said to be homogenous, if B is the null
matrix.
•
A homogenous system of linear equations, AX = 0 has only a trivial solution
x1 = x2 = ... = xn = 0, if |A| ≠ 0
•
A homogenous system of linear equations, AX = O has infinitely many solutions,
if |A| = 0
SUPPORTIVE WEB SITES
http://www.wikipedia.org
http://mathworld.wolfram.com
TERMINAL EXERCISE
1.
Find |A|, if
(a)
182
LM 1
A = −3
MM−2
N
2
−1
5
OP
0
P
4PQ
3
(b)
LM−1
A= 7
MM 0
N
OP
0
P
2PQ
3 4
5
1
MATHEMATICS
Inverse Of A Matrix And Its Application
MODULE - III
2.
Find the adjoint of A, if
LM−2
A = −1
MM −1
N
(a)
OP
5
P1P
Q
LM 1
A= 3
MM−2
N
3 7
4
0
(b)
OP
PP
Q
−1 5
1 2
1 3
Sequences And
Series
Notes
Also, verify that A(Adj A) = |A|I3 = (Adj A) A, for (a) and (b)
3.
Find A–1, if exists, when
LM3 6OP
N7 2Q
(a)
(b)
LM2 1OP
N3 5Q
(c)
LM 3 −5OP
N−4 2 Q
Also, verify that (A′)–1 = (A–1)′ , for (a), (b) and (c)
4.
Find the inverse of the matrix A, if
LM1
A= 3
MM5
N
(a)
5.
0
3
2
OP
0
P
−1PQ
0
(b)
LM1
A= 0
MM1
N
2
3
0
OP
−1
P
2 PQ
0
Solve, using matrix inversion method, the following systems of linear equations
x + 2y = 4
2 x + 5y = 9
(a)
(c)
2x + y + z = 1
3
x − 2y − z =
2
3 y − 5z = 9
(b)
(d)
6x + 4 y = 2
9x + 6y = 3
x−y+z = 4
2 x + y − 3z = 0
x+ y+z = 2
x + y − 2z = − 1
(e)
3x − 2 y + z = 3
2x + y − z = 0
6.
Solve, using matrix inversion method
2 3 10
4 6 5
8 9 20
+ + = 4; − + = 1; + −
=3
x y z
x y z
x y z
MATHEMATICS
183
Inverse Of A Matrix And Its Application
MODULE - III
Sequences And
Series
7.
Find non-trivial solution of the following system of linear equations:
3x + 2 y + 7 z = 0
4 x − 3y − 2z = 0
5x + 9 y + 23z = 0
Notes
8.
Solve the following homogeneous equations :
(a)
9.
x+ y−z = 0
x − 2y + z = 0
3x + 6 y − 5z = 0
(b)
x + 2 y − 2z = 0
2 x + y − 3z = 0
5x + 4 y − 9 z = 0
Find the value of ‘p’ for which the equations
x + 2 y + z = px
2 x + y + z = py
x + y + 2 z = pz
have a non-trivial solution
10.
Find the value of  for which the following system of equation becomes consistent
2 x − 3y + 4 = 0
5x − 2 y − 1 = 0
21x − 8 y +  = 0
184
MATHEMATICS
Inverse Of A Matrix And Its Application
MODULE - III
Sequences And
Series
ANSWERS
CHECK YOUR PROGRESS 5.1
1.
(a) –12
(b) 10
2. (a) singular (b) non-singular
Notes
3.
(a) M11 = 4; M12 = 7; M21 = –1; M22 = 3
(b) M11 = 5; M12 = 2; M21 = 6; M22 = 0
4.
(a) M21 = 11; M22 = 7; M23 = 1
(b) M31 = –13; M32 = –13; M33 = 13
5.
(a) C11 = 7; C12 = –9; C21 = 2; C22 = 3
(b) C11 = 6; C12 = 5; C21 = –4; C22 = 0
6.
(a) C21 = 1; C22 = –8; C23 = –2
(b) C11 = –6; C12 = 10; C33 = 2
CHECK YOUR PROGRESS 5.2
1.
(a)
2.
(a)
LM 6 1OP (b) LM d −bOP (c) LMcos
N−3 2Q
N− c a Q
N sin
LM 1 − 2 OP
LM i i OP
(b)
N −i i Q
N− 2 1 Q
− sin 
cos
OP
Q
CHECK YOUR PROGRESS 5.3
1.
2.
(a)
(a)
MATHEMATICS
LM−5 3 OP
N 2 −1Q
LM 15
MM− 8
MM 5
MN 25
−2
(b)
OP
PP
PP
1
− P
5Q
2
5
5
6 −1
5
5
1
5
 4
 − 10

3
10
(b)
2
10 

1
−
10 
−
LM− 13
MM− 7
MM 12
17
NM 24
LM 0 1OP
N−1 2Q
(c)
1
3
1
3
−1
3
OP
1 PP
12 P
− 11 PP
24 Q
1
3
185
Inverse Of A Matrix And Its Application
MODULE - III
Sequences And
Series
Notes
4.
OP
2
P
−1PQ
−8 −2
7
−4
CHECK YOUR PROGRESS 5.4
1.
2.
3.
4.
186
(A)–1
LM−9
8
= M
MN−5
23
−6
, y=
7
7
(a)
x=
(b)
x = 6, y = 1
(c)
7
23
x=− , y=
5
10
(d)
x = 27 , y = 13
30
5
(a)
x=
(b)
x = 2, y = 3, z = 0
(c)
x = 2, y = − 3, z = 2
(d)
x = 1, y = 2, z = 2
(a)
x = 0, y = 0, z = 0
(b)
x = 0, y = 0, z = 0
(c)
x = 0, y = 0, z = 0
(a)
Consistent; x =
(b)
Consistent; infinitely many solutions
(c)
Inconsistent
(d)
Trivial solution, x = y = z = 0
58
2
21
, y=− , z=−
11
11
11
26
9
, y=
5
5
MATHEMATICS
Inverse Of A Matrix And Its Application
MODULE - III
Sequences And
Series
TERMINAL EXERCISE
1.
2.
(a)
–31
(b)
–24
LM 4
MM−44
N
(a)
LM 1
MM−513
N
(b)
3.
(a)
(b)
(c)
4.
LM− 181
MM 7
N 36
LM 57
MM− 3
N 7
LM− 17
MM− 2
N 7
(a)
(b)
MATHEMATICS
OP
PP
Q
−3 −13
5
3
−3 −5
1
OP
P
4 PQ
1
6
− 1
12
OP
PP
Q
Notes
8 −7
13 13
−1
7
2
7
OP
PP
Q
− 5
14
− 3
14
OP
PP
Q
LM 1 0 0OP
MM−1 1 0 PP
MM 3 PP
MN 3 23 − 1PQ
LM 23 −1 − 12 OP
MM− 1 1 1 PP
MM 4 2 4 PP
MN − 43 12 43 PQ
187
Inverse Of A Matrix And Its Application
MODULE - III
Sequences And
Series
5.
(a)
x = 2, y = 1
(b)
x = k, y =
1 3
− k
2 2
(c)
x = 1, y =
1
3
, z=−
2
2
(d)
x = 2, y = − 1, z = 1
(e)
x=
Notes
1
1
1
, y=− , z=
2
2
2
6.
x = 2, y = 3, z = 5
7.
x = − k , y = − 2k , z = k
8.
(a) x = k , y = 2 k , z = 3k
(b) x = y = z = k
188
9.
p = 1, − 1, 4
10.
 = −5
MATHEMATICS