Mathematics. Session Matrices and Determinants - 3.
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Transcript of Mathematics. Session Matrices and Determinants - 3.
Mathematics
Session
Matrices and Determinants - 3
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
Session Objectives
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 31 3 35 3 3
Let A=
A is a singular matrix .
A=1(9+9)+1(3+15)+3(3-15) = 18+18-36 = 0
Non-Singular Matrix
Let B=1 1 12 1 11 2 3
B is a non-singular matrix.
B= 1(-3+2)-1(6-1)+1(-4+1)
= -1 – 5 – 3
= -9 0
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:
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.
Tij jiadjA = [C ] adjA = C
Adjoint of a Square Matrix
11 12 13 11 21 31
21 22 23 12 22 32
31 32 33 13 23 33
C C C C C C
adjA = C C C C C C
C C C C C C
T
where Cij denotes the cofactor of aij in A.
11 12 13
21 22 23
31 32 33
a a a
Let A = a a a ,
a a a
then
Example - 2
Find the adjoint of matrix
a bA
c d
Solution :
11 12
21 22
T
a bA= c d
C =d , C =-c
C =-b , C =a
d -cadjA= -b a
d -badjA =
-c a
Example - 3
Find the adjoint of matrix
1 2 1
A 3 1 0
0 1 1
Solution:
11 12 13
21 22 23
31 32 33
1 2 1
We have A 3 1 0
0 1 1
C (1 0) 1 , C ( 3 0) 3, C (3 0) 3
C ( 2 1) 3, C (1 0) 1 , C ( 1 0) 1
C (0 1) 1 , C (0 3) 3 , C (1 6) 5
Solution cont.
T1 3 3
adjA 3 1 1
1 3 5
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
n
ir rjr 1
A , if i jwe know a C
0, if i j
Properties (Con.)
Therefore, each diagonal element of A (adjA) is equal to |A| and all non-diagonal elements are equal to zero.
n
| A | 0 0 .. 0
0 | A | 0 ..0i.e. A (adjA) = = A I
: : : :
0 0 0 ..| A |
n
ri rjr=1
A , ifi = jSimilarly, (adjA)A = C a =
O, ifi j
Hence, A (adjA) = |A| In = (adjA)A
Properties (Con.)
2. If A is a non-singular square matrixof 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
Compute the adjoint of matrix A= 1 23 -5
and verify that A(adj A)=|A|I.
Solution:
1 2We have A = 3 -5
11 12 21 22
T
C =-5 , C =-3 , C =-2 , C =1
-5 -3adjA= -2 1
-5 -2= -3 1
Solution (Con.)
Hence verified.
1 2 -5 -2L.H.S. =A adj.A = 3 -5 -3 1
-5- 6 -2+2 -11 0= =-15+15 -6-5 0 -11
A = -5- 6=-11
1 0 -11 0R.H.S.= A I=(-11) =0 1 0 -11
L.H.S =R.H.S
Example-5
If a matrix p qA= r s , find det. {A(adjA)}.
Solution:
Ts -r s -qadjA= = -r p-q p
p q s -q ps- qr -pq+pqA(adjA)= =r s -r p rs- rs -rq+sp
1 0=(ps- qr) 0 1
Now, det.{A(adj.A)}=(ps- qr)×1
=ps- qr
p qWe have A= r s
Inverse of a Matrix
Steps to find inverse of a matrix:
-1 -1 1then A is given by A = .(adjA)
Aexists and
I f matrix A is non - singular i.e. A 0,a
(i) Find out |A| and if , then the matrix is invertible.
(ii) Find out (adjA).
Then -1 1
A = .(adjA)A
A 0
Example-6
Find the inverse of the Matrix
-1 4 2
A = 2 -1 4
1 2 3
-1 4 2
A = 2 -1 4 = -1 -3 - 8 - 4 6 - 4 +2 4+1 =13 0
1 2 3
Solution:
-1 4 2
We have A = 2 -1 4
1 2 3
Solution cont.
T T11 12 13
21 22 23
31 32 33
C C C -11 -2 5 -11 -8 18
adjA = C C C = -8 -5 6 = -2 -5 8
C C C 18 8 -7 5 6 -7
-1-11 -8 18
1 1Hence, A = .(adjA) = -2 -5 8
13A5 6 -7
Properties
(i) A square matrix is invertible if it is non-singular.
(ii) Every invertible matrix possesses a unique inverse.
Hence, an 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 In
B = C
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
Show that 2 -3A = 3 4
satisfies the equation x2 - 6x + 17 = 0.
Hence, find A-1.
Solution:
2 2 -3 2 -3 4- 9 -6-12 -5 -18A =A.A= = =3 4 3 4 6+12 -9+16 18 7
22
-5 -18 2 -3 1 0A - 6A+17I = - 6 +1718 7 3 4 0 1
2 -3We have A= 3 4
-5 -18 12 -18 17 0= - +18 7 18 24 0 17
Solution (Cont.)
A2 - 6A + 17I = 0
Multiplying each side by A-1, we getA-1A2 - 6(A-1A) + 17(A-1I) = A-1.0
1 1 1 1( A A I, A I A , A .0 0) (A-1A)A - 6I + 17A-1 = 0
IA - 6I + 17A-1 = 0, 17A-1 = 6I - A
-1
4 317 171 1 16 0 2 -3 4 3A = (6I - A)= - = =0 6 3 4 -3 217 17 17 3 2-17 17
-1
4 317 17
Hence, A =3 2
-17 17
0 0 O0 0
Hence, A satisfies the equation x2 - 6x + 17 = 0.
Example-8
Find A-1 , if0 1 1
A= 1 0 11 1 0
. Also show that A-1=2A - 3I
.2
Solution:
11 12 13
21 22 23
31 32 33
Now,C =-1,C =1,C =1
C =1,C =-1,C =1
C =1,C =1,C =-1
0 1 1A = 1 0 1 =0(0-1)-1(0-1)+1(1- 0) =1+1= 2 0
1 1 0
0 1 1We have A= 1 0 1
1 1 0
T-1 1 1 -1 1 1
adj.A= 1 -1 1 = 1 -1 11 1 -1 1 1 -1
Solution cont.
-1
11 1- 22 2
1-1 1 11 1 1 1Hence, A = adj A = 1 -1 1 = - 2A 2 2 21 1 -1
1 1 1-
2 2 2
2 0 1 1 0 1 1 1 0 0A - 3I 1Also = 1 0 1 1 0 1 - 3 0 1 0
2 2 1 1 0 1 1 0 0 0 1
2 1 1 3 0 01= 1 2 1 - 0 3 0
2 1 1 2 0 0 3
-1-1 1 11
= 1 -1 1 = A2 1 1 -1
Example-9
Solution:
2 1 4 5(AB) = 5 3 3 4
11 14= 29 37
-1 -1 -12 1 4 5If A = and B= , verify that AB = B A .
5 3 3 4
T
-1 37 -29 37 -14LHS = AB = =-14 11 -29 11
Solution (Cont.)
T-1 4 -3 4 -5B = =-5 4 -3 4
T-1 3 -5 3 -1A = =-1 2 -5 2
-1 -1 4 -5 3 -1 37 -14RHS =B A = = =LHS-3 4 -5 2 -29 11
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
11 1 1
2 2 2 2
3 3 3 3
da b c x
a b c y = d
a b c z d
AX = B ... i
Solution of Simultaneous Linear Equations (Matrix Method)
1X = (adjA)B
A
The matrix A is called the coefficient matrix of the system of linear equations.
Multiplying (i) by A–1, we get
-1I f A 0 i.e. A is non - singular, then A exists.
1 1A AX A B
1 1A A X A B
1IX A B
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-10Using matrix method, solve the following system of linear equationsx + 2y -3z = -42x + 3y + 2z = 23x - 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 -4or 2 3 2 y = 2
z3 -3 -4 11
AX = B
Solution (Cont.)
1 2 -3 x -4where A= 2 3 2 , X= y , B= 2
z3 -3 -4 11
2 2 1 3 3 1
-1
1 2 -3A = 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 ijLet C be the cofactor a in A = a , then
Solution Cont.
11 12 13c =(-12+6) c =- -8- 6 c =(-6- 9)=- 6 =14 =-15
21 22 23c =- (-8- 9) c =(-4+9) c =-(-3- 6)=17 =5 = 9
31 32 33c =(4+9) c =-(2+6) c =(3- 4)=13 = - 8 =-1
31 32 33c =(4+9) c =-(2+6) c =(3- 4)=13 = - 8 =-1
-6 17 1314 5 -8-15 9 -1
T-6 14 -15
adjA = 17 5 9 =13 -8 -1
Solution (Con.)
-1Now, X=A B
-6 17 13 -41X= 14 5 -8 2
67 -15 9 -1 11
x 201 31y = -134 = -2
67z 67 1
x=3 , y=-2 , z=1
-1-6 17 131 1
A = .adj A = 14 5 -8A 67 -15 9 -1
Example-11
Using matrices, solve the following system of equationsx + y + z = 6x + 2y + 3z = 14x + 4y + 7z = 30
Solution:
x + y + z = 6 …(i)
x + 2y + 3z = 14 …(ii)
x + 4y + 7z = 30 …(iii)
1 1 1 x 6or 1 2 3 y = 14
z1 4 7 30
AX = B
The given system of linear equations is
Solution (Cont.)
1 1 1Now, A = 1 2 3 =1(14-12)-1(7- 3)+1(4- 2)
1 4 7
=2- 4+2=0
1 1 1 x 6where A= 1 2 3 ; X= y ; B= 14
z1 4 7 30
11 12 13c =(14-12) c =-(7- 3) c = 4- 2=2 =- 4 =2
ij ij ijLet C be the cofactor a in A = a , then
Solution cont.
The given system of equations is consistent with infinitely many solutions.
21 22 23c =-(7- 4) c =(7-1) c =-(4-1)=- 3 = 6 =- 3
31 32 33c =(3- 2) c =-(3-1) c =(2-1)= 1 =- 2 =1
T2 -4 2 2 -3 1
adjA = -3 6 -3 = -4 6 -21 -2 1 2 -3 1
2 -3 1 6 0and (adjA)B= -4 6 -2 14 = 0 =0
2 -3 1 30 0
Solution (Con.)
Putting z = k in first two equations, we get
x + y = 6 - k
x + 2y = 14 - 3k
1 1 x 6 - k= AX = B
1 2 y 14 - 3kor
-1
1 1A = = 2 - 1 = 1 0
1 2
A exists.
T2 - 1 2 - 1adjA = =
-1 1 -1 1
Solution (Con.)
-1 2 - 11A = adjA =
| A | -1 1
-1 x 2 - 1 6 - kNow, X = A B =
y -1 1 14 - 3k
x 12 - 2k - 14 + 3k -2 + k= =
y -6 + k +14 - 3k 8 - 2k
x = -2 + k and y = 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