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Chap1:-[Adjoint of Matrix]
Cofactor of an element:-
Cofactor of an element is given byAij = (-1)i+j
.Mij
Adjoint of a Matrix:-
A transpose of a Matrix of cofactor is called Adjoint.
Example:-
Ex :-1 ] If A= 1 1 02 1 −11 2 3
Find the all Cofactor elements and Find adjA.
SolLet
A= 1 1 02 1 −11 2 3
We know that Aij=(-1)i+j
.Mij
...A11 = (-1)
i+j. M11
= (-1)2
1 −12 3
= 1(3+2)`
= 5
A12=(-1)1+2.
M12
=(-1)3
2 −11 3
= -1(6+1)
= -7
A13 =(-1)1+3 .
M13
=(-1)4
2 11 2
= (4-1)
= 3
A21 =(-1)2+1 .
M21
=(-1)3
1 02 3
= -(3-0)
= -3
A22 =(-1)2+2 .
M22
=(-1)4
1 01 3
Page(2)
= 3
A23 =(-1)2+3 .
M23
=(-1)5
1 11 2
= -1
A31 =(-1)3+1.
M31
=(-1)4
1 01 −1
= -1
A32 =(-1)3+2 .
M32
=(-1)5
1 02 −1
= 1
A33 =(-1)3+3 .
M33
=(-1)6
1 12 1
= (1-2)
= -1
Matrix of Cofactor = 5 −7 3
−3 3 −1−1 1 −1
adj A = 5 −3 −1
−7 3 13 −1 −1
Ex:-2] If A= 1 1 −11 1 01 2 −1
Find minor and cofactor of a11 ,a21 ,a33 .
SolLet A= 1 1 −11 1 01 2 −1
Minor of a11 = 1 02 −1
= -1-0
= -0
∴Cofactor of a11 = A11 = (-1)1+1.
M11
= (-1)2.
1 02 −1
= -1
Minor of a21 = 1 −12 −1
= (-1+2)
Page(3)
= 1
∴Cofactor of a21 = A21=(-1)2+1.
M21
= (-1)3 .
(1)
= -1
Minor of a33 = 1 11 1
= 0
∴Cofactor of a33 =A33 = (-1)3+3.
M33
= (-1)6.(0)
= 0
Ex:-3] If A= −3 1 0
2 −2 1−1 −1 1
Show that A(adjA) is a Null matrix.
SolLet A= −3 1 0
2 −2 1−1 −1 1
Matrix of minor = −1 3 −41 −3 41 −3 4
∴Matrix of Cofactor = −1 −3 −4−1 −3 −41 3 4
adj A = −1 −1 1−3 −3 3−4 −4 4
A.(adj A)=
−3 1 02 −2 1
−1 −1 1 .
−1 −1 1−3 −3 3−4 −4 4
= 3 − 3 + 0 3 − 3 + 0 −3 + 3 + 0−2 + 6 − 4 −2 + 6 − 4 2 − 6 + 41 + 3 − 4 1 + 3 − 4 −1 − 3 + 4
= 0 0 00 0 00 0 0
A.(adj A) is a Null matrix.
Ex:-4] If A= −1 −2 −22 1 −22 −2 1
Show that adjA=3A'
SolLet A= −1 −2 −22 1 −22 −2 1
Page(4)
matrix of minor = −3 6 −6−6 3 66 6 3
∴matrix of Cofactor= −3 −6 −66 3 −66 −6 3
adjA= −3 6 6−6 3 −6−6 −6 3
---------------------- 1
now,
3A' = 3 −1 2 2−2 1 −2−2 −2 1
= −3 6 6−6 3 −6−6 −6 3
---------------------- 2
From eqn (1) & (2), Hence proved that adj A= 3A'
Ex:-5] If A= 1 2 47 2 00 1 2
Show that A(adjA)= 𝐴 I
SolLet A= 1 2 47 2 00 1 2
Matrix of minor = 4 14 70 2 1
−8 −28 −12
Matrix of Cofactor= 4 −14 70 2 −1
−8 28 −12
adj A= 4 0 −8
−14 2 287 −1 −12
A (adj A)= 1 2 47 2 00 1 2
. 4 0 −8
−14 2 287 −1 −12
= 4 − 28 + 28 0 + 4 − 4 −8 + 56 − 4828 − 28 + 0 0 + 4 − 0 −56 + 56 − 00 − 14 + 14 0 + 2 − 2 0 + 28 − 24
= 4 0 00 4 00 0 4
---------------------- 1
Now,
Page(5)
𝐴 = 1 2 47 2 00 1 2
= 1 (4-0) -2 (14-0) +4 (7-0)
= 4 – 28 + 28
𝐴 = 4
∴ 𝐴 𝐼= 4. 1 0 00 1 00 0 1
= 4 0 00 4 00 0 4
---------------------- 2
From Eqn (1) and Eqn(2) ,
Hence proved A(adj A) = 𝐴 𝐼
Ex:-6]If A= 1 0 12 1 10 1 2
Show that adj(2A) = 22
adj A .
SolLet A= 1 0 12 1 10 1 2
∴2A = 2. 1 0 12 1 10 1 2
= 2 0 24 2 20 2 4
Matrix of Minor= 4 16 8
−4 8 4−4 −4 4
∴ Matrix of Cofactor= 4 −16 84 8 −4
−4 4 4
∴Adj 2A = 4 4 −4
−16 8 48 −4 4
---------------------- 1
Now,A = 1 0 12 1 10 1 2
Matrixof Minor = 1 4 2
−1 2 1−1 −1 1
Page(6)
∴Matrix of Cofactor = 1 −4 21 2 −1
−1 1 1
∴Adj A = 1 1 −1−4 2 1 2 −1 1
Now, 22 adj A = 4 adj A
= 4 1 1 −1−4 2 1 2 −1 1
= 4 4 −4−16 8 4 8 −4 4
---------------------- 2
From eqn (1) & (2),Hence proved that adj(2A) = 22
adj A
Ex:-7] If A = 2 −13 −2
, B = 5 01 2
Show that adj AB = adj B.adj A
Sol AB = 2 −13 −2
. 5 01 2
= 10 − 1 0 − 215 − 2 0 − 4
AB = 9 −2
13 −4
Matrix of minor of AB= −4 13−2 9
∴Matrix of Cofactorof AB = −4 −132 9
∴Adj AB = −4 2−13 9
---------------------- 1
Let B = 5 01 2
Matrix of minor of B = 5 01 2
∴Matrix of Cofactorof B = 2 −10 5
∴Adj B = 2 0
−1 5
Let A = 2 −13 −2
Matrix of minor of A = −2 3−1 2
∴Matrix of Cofactorof A = −2 −31 2
Page(7)
∴Adj A = −2 1−3 2
Now,
adj B .adj A =
2 0−1 5
× −2 1−3 2
= −4 2−13 9
---------------------- 2
From eqn (1) & (2),Hence proved that adj AB = adj B .adj A
Theorem:-1:-For any Square matrix A, show that A(adjA) = (adjA)A= 𝐴 𝐼
SolLet,
A=
𝑎11 𝑎12 𝑎13−−−−−− 𝑎1𝑛
𝑎21 𝑎22 𝑎23−−−−− 𝑎2𝑛
𝑎31 𝑎32 𝑎33−−−−−− 𝑎3𝑛
− − − − − − − −− − − − − − − −𝑎𝑛1 𝑎𝑛2 𝑎𝑛3 𝑎𝑛𝑛
be a given matrix.
Let , Matrix of cofactor =
𝐴11 𝐴12 𝐴13−−−−−− 𝐴1𝑛
𝐴21 𝐴22 𝐴23−−−−− 𝐴2𝑛
𝐴31 𝐴32 𝐴33−−−−−− 𝐴3𝑛
− − − − − − − −− − − − − − − −𝐴𝑛1 𝐴𝑛2 𝐴𝑛3 𝐴𝑛𝑛
adj A =
𝐴11 𝐴21 𝐴31−−−−−− 𝐴𝑛1
𝐴12 𝐴22 𝐴32−−−−− 𝐴𝑛2
𝐴13 𝐴23 𝐴33−−−−−− 𝐴𝑛3
| | | || | | |
𝐴1𝑛 𝐴2𝑛 𝐴3𝑛 𝐴𝑛𝑛
A (adj A)=
𝑎11 𝑎12 𝑎13−−−−−− 𝑎1𝑛
𝑎21 𝑎22 𝑎23−−−−− 𝑎2𝑛
𝑎31 𝑎32 𝑎33−−−−−− 𝑎3𝑛
− − − − − − − −− − − − − − − −𝑎𝑛1 𝑎𝑛2 𝑎𝑛3 𝑎𝑛𝑛
𝐴11 𝐴21 𝐴31−−−−−− 𝐴𝑛1
𝐴12 𝐴22 𝐴32−−−−− 𝐴𝑛2
𝐴13 𝐴23 𝐴33−−−−−− 𝐴𝑛3
| | | || | | |
𝐴1𝑛 𝐴2𝑛 𝐴3𝑛 𝐴𝑛𝑛
Page(8)
=
𝑎1𝑗
𝑛
𝑗 =1
𝐴1j 𝑎1𝑗
𝑛
𝑗 =1
𝐴2j 𝑎1𝑗
𝑛
𝑗 =1
𝐴3j
−−−−−−
𝑎1𝑗
𝑛
𝑗 =1
𝐴nj
𝑎2𝑗
𝑛
𝑗 =1
𝐴1j 𝑎2𝑗
𝑛
𝑗 =1
𝐴2j 𝑎2𝑗
𝑛
𝑗 =1
𝐴3j
−−−−−−
𝑎2𝑗
𝑛
𝑗 =1
𝐴nj
𝑎3𝑗
𝑛
𝑗 =1
𝐴1j 𝑎3𝑗
𝑛
𝑗 =1
𝐴2j 𝑎3𝑗
𝑛
𝑗 =1
𝐴3j
−−−−−−
𝑎3𝑗
𝑛
𝑗 =1
𝐴nj
− − − − − − − − − − − − − − − − − −
𝑎𝑛𝑗
𝑛
𝑗 =1
𝐴1j 𝑎𝑛𝑗
𝑛
𝑗 =1
𝐴2j 𝑎𝑛𝑗
𝑛
𝑗 =1
𝐴3j − − − − 𝑎𝑛𝑗
𝑛
𝑗 =1
𝐴nj
We Know that ,
1. 𝑎𝑛𝑗 𝐴𝑛𝑗 = 𝐴
2. 𝑎𝑛𝑗 𝐴𝑘𝑗 = 0
Above eqn becomes,
A (adj A) =
𝐴 0 0 − − − 00 𝐴 0 − − 00 0 𝐴 − − 0
− − − − − − − − −− − − − − − − − −
0 0 0 − − − 𝐴
= 𝐴
1 0 0 − − − 00 1 0 − − − 00 0 1 − − − 0
− − − − − − − − − −0 0 0 − − − − 1
A (adj A) = 𝐴 I---------------------- 1
Similarly We can show that
(adj A) A = 𝐴 I---------------------- 2
From eqn (1) & (2), Hence proved that A (adjA) = (adjA) A= 𝐴 𝐼
Inverse of a Matrix:-If A and Bare the two matrices such that
AB = BA = I
then B is called inverse of A and we write B = A-1 .
Theorem:-2] State and prove Necessary and sufficient condition for existence
of inverse.
Statement:- The Necessary and Sufficient condition for square matrix A to have
an inverse is that A is a non singular matrix.(i.e 𝐴 ≠ 0)
Page(9)
Proof:-Necessary part:-
Suppose that the matrix A has inverse say B
∴By Definition,
AB=BA=I
i.e. AB=I
taking determinant on both sides,
𝐴𝐵 = 𝐼
𝐴 B = 1 ≠ 0
⇨ 𝐴 ≠ 0and 𝐵 ≠ 0
∴A is a non Singular matrix.
Conversly,
Suppose that A is a non Singular matrix,
i.e. 𝐴 ≠ 0
We know that,
A(adj A)= (adj A)A = 𝐴 𝐼
Divide each term by 𝐴
We get,
A 1
𝐴 𝑎𝑑𝑗𝐴 =
1
𝐴 𝑎𝑑𝑗𝐴 𝐴 = 𝐼
∴ By Definition of Inverse ,
⇨1
𝐴 𝑎𝑑𝑗𝐴 = A−1
Ex:-1]Find the inverse of Following matrix
A = 1 2 −1
−1 1 22 −1 1
SolLet,
A = 1 2 −1
−1 1 22 −1 1
∴ 𝐴 = 1 2 −1
−1 1 22 −1 1
= 1(1+2)-2(-1-4)-1(1-2)
= 1(3)-2(-5)-1(-1)
= 3+10+1
= 14 ≠ 0
∴A−1exist.
Matrix of minor= 3 −5 −11 3 −55 1 3
∴Matrix of Cofactor = 3 5 −1
−1 3 55 −1 3
Page(10)
∴Adj A= 3 −1 55 3 −1
−1 5 3
∴A−1 =1
𝐴 𝑎𝑑𝑗𝐴
=1
14
3 −1 55 3 −1
−1 5 3
Ex:-2] If A = 1 24 −1
Find A-1
Sol Let A= 1 24 −1
∴ 𝐴 = 1 24 −1
= -1-8
= -9 ≠ 0
A-1
exist.
Matrix of minor = −1 42 1
∴Matrix of Cofactor = −1 −4−2 1
∴Adj A = −1 −2−4 1
We know that,
A-1
= 1
𝐴 adj A
A-1
= 1
−9 −1 −2−4 1
Ex:-3] Show that A= 4 −1
−3 2 Satisfy the eqn 𝐴2 − 6A + 5I = 0
Hence find A-1
Sol Let, A= 4 −1
−3 2
∴ A2= A
.A
= 4 −1
−3 2
4 −1−3 2
= 19 −6
−18 7
Consider, 𝐴2 − 6A + 5I= 19 −6
−18 7 − 6
4 −1−3 2
+ 5 1 00 1
= 19 −6
−18 7 −
24 −6−18 12
+ 5 00 5
Page(11)
∴𝐴2 − 6A + 5I= 0 00 0
∴𝐴2 − 6A + 5I= 0
Premultiplying by A-1
Then,
A-1
A2 - 6A
-1.A + 5 A−1 I = 0
A –6I+5A-1
= 0
5A-1
= 6 I– A
5A-1
= 6 1 00 1
− 4 −1
−3 2
= 6 00 6
− 4 −1
−3 2
5A-1
= 2 13 4
A-1
= 1
5 2 13 4
Theorem:-1]If A is a non-singular matrix then show that A−1 =1
𝐴
Proof:- We know that,
A A−1 = I
Taking the determinant on both sides,we get
A A−1 I
𝐴 A−1 = 1
A−1 =1
𝐴
Theorem:-2] If A & B are non –Singular matrices then show that AB is a non-
singular matrix 𝑎𝑛𝑑 (𝐴𝐵)−1 = 𝐵−1. 𝐴−1
Proof:-Since, A & B are non-singular matrix.
𝐴 0 𝑎𝑛𝑑 𝐵 ≠ 0
⇨ 𝐴 . 𝐵 ≠ 0
⇨ 𝐴𝐵 ≠ 0
AB is a non-singular matrix.
To show that (𝐴𝐵)−1 = 𝐵−1. 𝐴−1
Consider, 𝐴𝐵 (𝐵−1. 𝐴−1) = 𝐴(𝐵. 𝐵−1)𝐴−1
= 𝐴 𝐼𝐴−1
= 𝐴 𝐴−1
Page(12)
= 𝐼
Also,
𝐵−1. 𝐴−1 𝐴𝐵 = 𝐵−1 𝐴−1. 𝐴 𝐵
= 𝐵−1𝐼 𝐵
= 𝐵−1𝐵
= 𝐼
Thus; 𝐴𝐵 𝐵−1 . 𝐴−1 = 𝐵−1. 𝐴−1 𝐴𝐵 = 𝐼
By Definition of Inverse,
(𝐵−1. 𝐴−1) = (𝐴𝐵)−1
i.e.(𝐴𝐵)−1 = (𝐵−1. 𝐴−1)
Theorem:-3] If A is a non-singular matrix , then show that
1. 𝑎𝑑𝑗𝐴 = 𝐴 𝑛−1
2. 𝑎𝑑𝑗 𝐴is a non-singular matrix.
Proof:-
1. We know that,
A(𝑎𝑑𝑗𝐴) = 𝐴 I
Taking determinant on both sides ,
𝐴(𝑎𝑑𝑗𝐴) = 𝐴 𝐼
𝐴 𝑎𝑑𝑗𝐴 = 𝐴 𝑛 − − − 𝐹𝑜𝑟𝑚𝑢𝑙𝑎: −[ 𝐴 𝐼 =
𝐴 𝑛 ]
𝑎𝑑𝑗𝐴 = 𝐴 𝑛
𝐴
𝑎𝑑𝑗𝐴 = 𝐴 𝑛−1---------------------- 1
2. Since, A is a non-singular matrix.
A 0
𝐴 𝑛−1 0
𝑎𝑑𝑗𝐴 0 − − − − − (𝐹𝑟𝑜𝑚 𝑒𝑞𝑛 1 )
adj A is non singular matrix.
Page(13)
Principle of Mathematical Induction:-
Let, P(n) denote the given statement such that
1) P(1) is true
2) P(m) is true
P(m+1) is true
Then we say that the statement is true for every natural numbers.
Theorem:-1 ] Show that (𝐴𝑛)−1 = (𝐴−1)𝑛 , ∀ 𝑛 ∈ 𝑁
Proof:-We prove this result by the principle of Mathematical induction.
Step-1 Let n=1
L.H.S = (𝐴𝑛)−1
= (𝐴1)−1
= 𝐴−1
R.H.S = (𝐴−1)𝑛
= (𝐴−1)1
= 𝐴−1
L.H.S = R.H.S
The theorem is true for n=1.
Step-2 Assume that Theorem is true for 𝑛 = 𝑚
We get,
(𝐴𝑚 )−1 = (𝐴−1)𝑚 − − − − − 1
Now, we show that for 𝑛 = 𝑚 + 1
Consider,
(𝐴𝑚+1)−1 = (𝐴𝑚𝐴)−1
= 𝐴−1. (𝐴𝑚 )−1
= 𝐴−1. (𝐴−1)𝑚 − − − − − 𝑏𝑦 𝑒𝑞𝑛 1
Page(14)
= (𝐴−1)𝑚+1
Theorem is true for n = m+1
By the principle of Mathematical induction Theorem is true for every natural
number.
Page(15)
Theorem:-2]If A is a non-singular matrix and k is any scalar then show that
(𝐾𝐴)−1 =1
𝑘𝐴−1
Proof:- Consider
𝐾𝐴 1
𝑘𝐴−1 = 𝑘.
1
𝑘 𝐴. 𝐴−1
= 1 (I)
= I
Also,
1
𝑘𝐴−1 𝐾𝐴 =
1
𝑘. 𝑘 . (𝐴−1. 𝐴)
= 1 (I)
= I
Thus 𝐾𝐴 1
𝐾𝐴−1 =
1
𝐾𝐴−1 𝐾𝐴 = 𝐼
By definition of inverse,
1
𝐾𝐴−1 = 𝐾𝐴 −1
i.e. 𝐾𝐴 −1 =1
𝐾𝐴−1
Theorem:-3]If A is a non-singular matrix and k is any scalar then show that
𝑎𝑑𝑗 𝐾𝐴 = 𝑘𝑛−1(𝑎𝑑𝑗 𝐴)
Proof:- We have
𝑎𝑑𝑗 𝐾𝐴 = 𝐾𝐴 𝐾𝐴 −1
= 𝑘𝑛 𝐴 . 1
𝐾𝐴−1 − − − 𝐾𝐴 = 𝑘𝑛 𝐴 & (𝐾𝐴)−1 =
1
𝐾𝐴−1
= 𝑘𝑛−1 𝐴 𝐴−1
= 𝑘𝑛−1𝑎𝑑𝑗𝐴
Page(16)
Theorem:-4]If A and B are non-singular matrices then show that𝑎𝑑𝑗 𝐴𝐵 =𝑎𝑑𝑗𝐵. 𝑎𝑑𝑗𝐴.
Proof:- We know that,
𝐴 𝑎𝑑𝑗𝐴 = 𝑎𝑑𝑗𝐴 𝐴 = 𝐴 𝐼 − − − − − − 1
𝐵 𝑎𝑑𝑗𝐵 = 𝑎𝑑𝑗𝐵 𝐵 = 𝐵 𝐼 − − − − − − 2
𝐴𝐵 𝑎𝑑𝑗𝐴𝐵 = 𝑎𝑑𝑗𝐴𝐵 𝐴𝐵 = 𝐴𝐵 𝐼 − − − − − − 3
Consider,
𝐴𝐵 𝑎𝑑𝑗𝐵 𝑎𝑑𝑗𝐴 = 𝐴 𝐵. 𝑎𝑑𝑗𝐵 𝑎𝑑𝑗𝐴
= 𝐴 𝐵 𝐼 𝑎𝑑𝑗𝐴 − − − −𝑓𝑟𝑜𝑚 𝑒𝑞𝑛 2
= 𝐴 . 𝑎𝑑𝑗𝐴 ( 𝐵 𝐼)
= 𝐴 𝐼 𝐵 𝐼 − − − − − 𝑓𝑟𝑜𝑚 𝑒𝑞𝑛 1
= 𝐴 𝐵 𝐼
= 𝐴𝐵 𝐼
𝐴𝐵 𝑎𝑑𝑗𝐵 𝑎𝑑𝑗𝐴 = 𝐴𝐵 𝑎𝑑𝑗𝐴𝐵
𝑎𝑑𝑗𝐵. 𝑎𝑑𝑗𝐴 = 𝑎𝑑𝑗𝐴𝐵
𝑎𝑑𝑗𝐴𝐵 = 𝑎𝑑𝑗𝐵. 𝑎𝑑𝑗𝐴
Theorem:-5]Inverse of matrix if exist.Show that it is unique
Proof:- Suppose that the matrix A has two inverses say B & C.
By Definition of inverse
𝐴𝐵 = 𝐵𝐴 = 𝐼 − − − − − − 1
𝐴𝐶 = 𝐶𝐴 = 𝐼 − − − − − − 2
Consider,
𝐵 = 𝐵. 𝐼
= 𝐵 𝐴𝐶 − − − − − 𝐹𝑟𝑜𝑚 𝑞𝑛 2
= 𝐵𝐴 𝐶
= 𝐼. 𝐶 − − − − − −𝐹𝑟𝑜𝑚 𝑒𝑞𝑛 1
= 𝐶
The matrix ‘A’ has unique inverse.
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