Summary Class notes English Income Tax: Law

1.APPLICATIONS OF MATRICES AND DETERMINANTS THEOREM 1: For every square matrix A of order n, 𝑨(𝒂𝒅
 If |𝐴| ≠ 0, then the square matrix A is called a non-singular matrix. |𝑨|𝑰𝒏
 If |𝐴| = 0, then the square matrix A is called a singular matrix. Proof:
 ADJOINT OF A SQUARE MATRIX: Let A be the square matrix of order n, then For simplicity, we prove the theorem for n=3 only.
the adjoint matrix of A is defined as the transpose of the matrix of cofactors 𝑎11 𝑎12 𝑎13 𝐴11 𝐴1
𝑻 𝑻 Consider 𝐴 = [𝑎21 𝑎22 𝑎23 ] 𝑎𝑛𝑑 𝑎𝑑𝑗 𝐴 = [𝐴21 𝐴2
of A. It is denoted by adj A. 𝒂𝒅𝒋 𝑨 = [𝑨𝒊𝒋 ] = [(−𝟏)𝒊+𝒋 𝑴𝒊𝒋 ]
𝑎31 𝑎32 𝑎33 𝐴31 𝐴3
 𝐴(𝑎𝑑𝑗𝐴) = (𝑎𝑑𝑗𝐴)𝐴 = |𝐴|𝐼𝑛 𝑎11 𝑎12 𝑎13 𝐴11 𝐴12 𝐴13 |𝐴|
 INVERSE MATRIX: Let A be a square matrix of order n. If there exists a 𝐴(𝑎𝑑𝑗 𝐴) = [𝑎21 𝑎22 𝑎23 ] [𝐴21 𝐴22 𝐴23 ] = [ 0
square matrix B of order n such that 𝐴𝐵 = 𝐵𝐴 = 𝐼𝑛 , then the matrix B is 𝑎31 𝑎32 𝑎33 𝐴31 𝐴32 𝐴33 0
called an inverse of A. 1 0 0
 𝐴𝐴−1 = 𝐴−1 𝐴 = 𝐼𝑛 Namma Kalvi = |𝐴| [0 1 0] = |𝐴|𝐼3 → (1)
 A singular matrix has no inverse. www.nammakalvi.org 0 0 1
 If A is non-singular, then 𝐴11 𝐴12 𝐴13 𝑎11 𝑎12 𝑎13 |𝐴|
(𝑖)|𝐴−1 | =
1 1
(𝑖𝑖) (𝐴𝑇 )−1 = (𝐴−1 )𝑇 (𝑖𝑖𝑖)(𝜆𝐴)−1 = 𝐴−1 (𝑎𝑑𝑗 𝐴)𝐴 = [𝐴21 𝐴22 𝐴23 ] [𝑎21 𝑎22 𝑎23 ] = [ 0
|𝐴|
1
𝜆 𝐴31 𝐴32 𝐴33 𝑎31 𝑎32 𝑎33 0
 If A is non-singular, then (λA)−1 −1
= A , where 𝛌 is a non-zero scalar. 1 0 0
λ
 −1
(AB) = B A −1 −1 = |𝐴| [0 1 0] = |𝐴|𝐼3 → (2)
 If A is non-singular, then A−1 is also non-singular and (A−1 )−1 = A 0 0 1
From (1) & (2), we get A(adj A) = (adj A)A = |A|In
 If A is non-singular square matrix of order n, then
1 THEOREM 2: If a square matrix has an inverse, then it is
(𝑖)(𝑎𝑑𝑗𝐴)−1 = 𝑎𝑑𝑗(𝐴−1 ) = 𝐴 ⟹ 𝑎𝑑𝑗 𝐴 = |𝐴|𝐴−1
|𝐴| Proof:
(𝑖𝑖)|𝑎𝑑𝑗𝐴| = |𝐴|𝑛−1 (𝑖𝑖𝑖)𝑎𝑑𝑗(𝑎𝑑𝑗𝐴) = |𝐴|𝑛−2 𝐴 Let A be a square matrix of order n such that an inverse o
2
(𝑖𝑣)𝑎𝑑𝑗(𝜆𝐴) = 𝜆𝑛−1 𝑎𝑑𝑗(𝐴) (𝑣)|𝑎𝑑𝑗(𝑎𝑑𝑗𝐴)| = |𝐴|(𝑛−1) there be two inverse B and C of A.
(𝑣𝑖)(𝑎𝑑𝑗𝐴)𝑇 = 𝑎𝑑𝑗(𝐴𝑇 ) (𝑣𝑖𝑖)𝑎𝑑𝑗(𝐴𝐵) = (𝑎𝑑𝑗𝐵)(𝑎𝑑𝑗𝐴) By definition, 𝐴𝐵 = 𝐵𝐴 = 𝐼𝑛 𝑎𝑛𝑑 𝐴𝐶 = 𝐶𝐴 = 𝐼𝑛
 |𝑎𝑑𝑗 𝐴| = |𝐴|2 ⟹ |𝐴| = ±√|𝑎𝑑𝑗 𝐴| 𝐶 = 𝐶𝐼𝑛 = 𝐶(𝐴𝐵) = (𝐶𝐴)𝐵 = 𝐼𝑛 𝐵 = 𝐵 ⟹ 𝐵 = 𝐶
 A−1 = ±
1
adjA Hence it is proved.
√|adjA| THEOREM 3: Let A be a square matrix of order n. Then 𝑨
1
 A=± adj(adjA) is non-singular.
√|adjA|
 If A and B are any two non-singular square matrices of order n, then Proof:
𝑎𝑑𝑗(𝐴𝐵) = (𝑎𝑑𝑗 𝐵)(𝑎𝑑𝑗 𝐴) Case (i) : Suppose that A−1 exists, then 𝐴𝐴−1 = 𝐴−1 𝐴 =
 A square matrix A is called orthogonal if 𝐴𝐴𝑇 = 𝐴𝑇 𝐴 = 𝐼 By the product rule for determinants,
 A is orthogonal if and only if A is non-singular and A−1 = 𝐴𝑇 𝑑𝑒𝑡(𝐴𝐴−1 ) = 𝑑𝑒𝑡(𝐴−1 )𝑑𝑒𝑡(𝐴) = 𝑑𝑒𝑡(𝐼𝑛 ) = 1 ⟹ |𝐴| ≠
 Application of matrices is in computer graphic and cryptography. Hence A is non-singular.

,Case (ii) : Conversely, suppose A is non-singular, then |𝐴| ≠ 0. Proof:
W.k.T. A(adj A) = (adj A)A = |A|In Since A is non-singular, 𝐴−1 exists and 𝐴𝐴−1 = 𝐴−1 𝐴 = 𝐼
1 1 www.nammakalvi.in Given AB=AC. Pre multiply by 𝐴−1 on both sides,
÷ 𝑏𝑦 |𝐴| ⟹ 𝐴 ( 𝑎𝑑𝑗 𝐴) = ( 𝑎𝑑𝑗 𝐴) 𝐴 = In
|𝐴| |𝐴|
1 𝐴−1 (AB) = 𝐴−1 (AC) ⟹ (𝐴−1 𝐴)𝐵 = (𝐴−1 𝐴)𝐶 ⟹ 𝐼𝑛 𝐵 =
⟹ 𝐴𝐵 = 𝐵𝐴 = 𝐼𝑛 ; 𝐵 = |𝐴| 𝑎𝑑𝑗 𝐴 ----------------------------------------------------------------------------
1 THEOREM 8: RIGHT CANCELLATION LAW
Hence the inverse of A exists and the inverse is 𝐴−1 = |𝐴| 𝑎𝑑𝑗 𝐴
Let A,B and C be square matrices of order n. If A is non-s
PROPERTIES OF INVERSES OF MATRICES
𝟏 B=C
THEOREM 4: If A is non-singular, then |𝐀−𝟏 | = |𝐀| Proof:
Proof: Since A is non-singular, 𝐴−1 exists and 𝐴𝐴−1 = 𝐴−1 𝐴 = 𝐼
Let A be non-singular then |𝐴| ≠ 0 and 𝐴−1 exists. Given BA=CA. Post multiply by 𝐴−1 on both sides,
By definition, 𝐴𝐴−1 = 𝐴−1 𝐴 = 𝐼𝑛 ⟹ |𝐴𝐴−1 | = |𝐴−1 𝐴| = |𝐼𝑛 | (BA)𝐴−1 = (CA)𝐴−1 ⟹ 𝐵(𝐴𝐴−1 ) = 𝐶(𝐴𝐴−1 ) ⟹ 𝐵𝐼𝑛 =
By the product rule for determinants, ----------------------------------------------------------------------------
1 THEOREM 9: REVERSAL LAW FOR INVERSES
|𝐴𝐴−1 | = |𝐼𝑛 | ⟹ |𝐴||𝐴−1 | = 1 ⟹ |A−1 | =
|A|
If A and B are non-singular matrices of the same order, t
----------------------------------------------------------------------------------------------------------------------
also non-singular and (𝐀𝐁)−𝟏 = 𝐁 −𝟏 𝐀−𝟏
THEOREM 5: If A is non-singular, then (𝐀𝐓 )−𝟏 = (𝐀−𝟏 )𝐓
Proof:
Proof:
Since A and B are non-singular matrices of the same orde
By definition, 𝐴𝐴−1 = 𝐴−1 𝐴 = 𝐼𝑛
0 ⟹ 𝐴−1 𝑎𝑛𝑑 𝐵−1 of order n exists. The product of AB a
Taking transpose ,(𝐴𝐴−1 )𝑇 = (𝐴−1 𝐴)𝑇 = (𝐼𝑛 )𝑇
found.
By the reversal law of transpose, (A−1 )T 𝐴𝑇 = 𝐴𝑇 (A−1 )T = |𝐼𝑛 |
|𝐴𝐵| = |𝐴||𝐵| ≠ 0 ⟹ 𝐴𝐵 is also non-singular and (AB)
Hence (AT )−1 = (A−1 )T
(AB)( B −1 A−1 ) = 𝐴(𝐵𝐵−1 )𝐴−1 = 𝐴(𝐼𝑛 )𝐴−1 = 𝐴𝐴−1 =
--------------------------------------------------------------------------------------------------------------
𝟏
( B −1 A−1 )(AB) = B −1 (𝐴−1 𝐴)𝐵 = B −1 (𝐼𝑛 )𝐵 = 𝐵−1 𝐵 =
THEOREM 6: If A is non-singular, then (𝛌𝐀)−𝟏 = 𝐀−𝟏 , where 𝛌 is a non-zero Hence (AB)−1 = B −1 A−1
𝛌
scalar. ----------------------------------------------------------------------------
Proof: THEOREM 10: LAW OF DOUBLE INVERSE
By definition, 𝐴𝐴−1 = 𝐴−1 𝐴 = 𝐼𝑛 If A is non-singular, then 𝐀−𝟏 is also non-singular and (𝐀
1 1 1
Muliply & divide by 𝛌, (𝜆𝐴) ( 𝐴−1 ) = ( 𝐴−1 ) (𝜆𝐴) = 𝐼𝑛 ⟹ (λA)−1 = A−1 Proof:
𝜆 𝜆 λ
-------------------------------------------------------------------------------------------------------------- Since A is non-singular,then |𝐴| ≠ 0 and 𝐴−1 exists.
1
THEOREM 7: LEFT CANCELLATION LAW Now |A−1 | = |A| ≠ 0 ⟹ A−1 is also non-singular and 𝐴𝐴−
Let A,B and C be square matrices of order n. If A is non-singular and AB=AC then Now 𝐴𝐴−1 = 𝐼𝑛 ⟹ (𝐴𝐴−1 )−1 = 𝐼 ⟹ (𝐴−1 )−1 𝐴−1 = 𝐼
B=C Post- multiply by A on both sides, (𝐴−1 )−1 = 𝐴

, 1
THEOREM 11: If A is non-singular square matrix of order n, then (𝐚𝐝𝐣𝐀)−𝟏 = Replace A by λA ⟹ 𝑎𝑑𝑗 (λA) = |λA|(λA)−1 = 𝜆𝑛 |𝐴| 𝐴−1
𝟏 𝜆
𝐚𝐝𝐣(𝐀−𝟏 ) = |𝐀| 𝐀 𝜆𝑛−1 adj(A)
Proof: THEOREM 15: If A is non-singular square matrix of order
1 𝟐
𝐴−1 = |𝐴| (𝑎𝑑𝑗 𝐴) ⟹ 𝑎𝑑𝑗 𝐴 = |𝐴|𝐴−1 |𝐀|(𝐧−𝟏)
1 Proof:
⟹ (𝑎𝑑𝑗 𝐴)−1 = (|𝐴|𝐴−1 )−1 = (𝐴−1 )−1 |𝐴|−1 = |A| A → (1)
W.K.T, adj (𝑎𝑑𝑗𝐴) = |A|n−2 𝐴
1 2
𝐴−1 = |𝐴| (𝑎𝑑𝑗 𝐴) ⟹ 𝑎𝑑𝑗 𝐴 = |𝐴|𝐴−1 ⟹ |adj (𝑎𝑑𝑗𝐴)| = ||A|n−2 𝐴| = (|A|n−2 )𝑛 |𝐴| = |𝐴|𝑛 −2𝑛
Replacing A by 𝐴−1 , 𝑎𝑑𝑗 (𝐴−1 ) = |𝐴−1 |(𝐴−1 )−1 = |A| A → (2)
1 ----------------------------------------------------------------------------
1
THEOREM 16: If A is non-singular square matrix of order
From (1) and (2), (adjA)−1 = adj(A−1 ) = |A| A 𝐚𝐝𝐣(𝐀𝐓 )
-------------------------------------------------------------------------------------------------------------- Proof:
1
THEOREM 12: If A is non-singular square matrix of order n, then |𝐚𝐝𝐣𝐀| = W.K.T, 𝐴−1 = |𝐴| (𝑎𝑑𝑗 𝐴)
|𝐀|𝐧−𝟏 1
Proof: Replace A by 𝐴𝑇 ⟹ (𝐴𝑇 )−1 = |𝐴𝑇 | (𝑎𝑑𝑗 (𝐴𝑇 ))
A(adj A) = (adj A)A = |A|In ⟹ 𝑑𝑒𝑡[A(adj A)] = det(𝐴)det(𝑎𝑑𝑗𝐴) = det[|A|In ] ⟹ 𝑎𝑑𝑗 (𝐴𝑇 ) = |𝐴𝑇 |(𝐴𝑇 )−1 = (|𝐴|𝐴−1 )𝑇 = (adjA)T
⟹ |𝐴||𝑎𝑑𝑗 𝐴| = |𝐴|𝑛 ⟹ |adjA| = |A|n−1 ----------------------------------------------------------------------------
-------------------------------------------------------------------------------------------------------------- THEOREM 17: If A and B are any two non-singular matri
THEOREM 13: If A is non-singular square matrix of order n, then 𝐚𝐝𝐣(𝐚𝐝𝐣𝐀) = 𝐚𝐝𝐣(𝐀𝐁) = (𝐚𝐝𝐣𝐁)(𝐚𝐝𝐣𝐀)
|𝐀|𝐧−𝟐 𝐀 Proof:
Proof: W.K.T, 𝑎𝑑𝑗 𝐴 = |𝐴|𝐴−1
For any non-singular matrix B of order n, B(adj B) = (adj B)B = |B|In Replace A by AB ⟹ 𝑎𝑑𝑗 (𝐴𝐵) = |𝐴𝐵|(𝐴𝐵)−1 = |𝐴||𝐵|𝐴
Put 𝐵 = 𝑎𝑑𝑗𝐴 ⟹ 𝑎𝑑𝑗𝐴(adj (𝑎𝑑𝑗𝐴)) = |𝑎𝑑𝑗𝐴|In = (|𝐵|𝐵−1 )(|𝐴|𝐴−1 ) = (adjB)(adjA)
⟹ 𝑎𝑑𝑗𝐴(adj (𝑎𝑑𝑗𝐴)) = |A|n−1 In EXERCISE 1.1
Pre-multiply by A on both sides, 1.Find the adjoint of the following matrices.
⟹ [𝐴(𝑎𝑑𝑗𝐴)](adj (𝑎𝑑𝑗𝐴)) = 𝐴|A|n−1 In ⟹ |A|In (adj (𝑎𝑑𝑗𝐴)) = 𝐴|A|n−1 In −3 4
(i) [ ]
6 2
⟹ adj (𝑎𝑑𝑗𝐴) = |A|n−2 𝐴 −3 4 2 −4
-------------------------------------------------------------------------------------------------------------- Let 𝐴 = [ ] ⟹ 𝑎𝑑𝑗 𝐴 = [ ]
6 2 −6 −3
THEOREM 14: If A is non-singular square matrix of order n, then 𝐚𝐝𝐣(𝛌𝐀) = ----------------------------------------------------------------------------
𝛌𝐧−𝟏 𝐚𝐝𝐣(𝐀) 2 3 1
Proof: (ii) [3 4 1]
1
W.K.T,(adjA)−1 = A ⟹ 𝑎𝑑𝑗 𝐴 = |𝐴|𝐴−1 3 7 2
|A|

, 2 3 1 5 1 1 5 25 − 1 1 − 5 1−5
Let 𝐴 = [3 4 1] www.nammakalvi.in 1 5 1 1 ⟹ 𝑎𝑑𝑗 𝐴 = [ 1 − 5 25 − 1 1 − 5 ]
3 7 2 1 1 5 1
4 7 3 4 5 1−5 1 − 5 25 − 1
8−7 7−6 3−4 1 1 −1 5 1 1
1 2 1 1 ⟹ 𝑎𝑑𝑗 𝐴 = [ 3 − 6 24 −4 −4 6
3 4 − 3 3 − 2] = [−3 1 1] 1
𝐴−1 = |𝐴| 𝑎𝑑𝑗 𝐴 ⟹ 𝐴−1 =
1
=
1
3 3 2 [−4 24 −4 ] [ −1
4 21 − 12 9 − 14 8 − 9 9 −5 −1 112 28
4 7 3 −4 −4 24 −1
-------------------------------------------------------------------------------------------------------------- ----------------------------------------------------------------------------
2 2 1 2 3 1
1
(iii) [−2 1 2] 𝑎𝑑𝑗(𝜆𝐴) = 𝜆𝑛−1 𝑎𝑑𝑗(𝐴) (iii) [3 4 1]
3
1 −2 2 3 7 2
1 −2 2 1 2 + 4 −2 − 4 4 − 1 2 3 1
2 2 1 2 ⟹ 𝑎𝑑𝑗 𝐴 = (1)3−1 [2 + 4 4 − 1 −2 − 4] Let 𝐴 = [3 4 1]
−2 1 2 −2 3 3 7 2
4−1 2+4 2+4 2 3 1
1 −2 2 1
6 −6 3 2 −2 1 |𝐴| = |3 4 1| = 2(8 − 7) − 3(6 − 3) + 1(21 − 12)
1 1
= [6 3 −6] = [2 1 −2] 3 7 2
9 3
3 6 6 1 2 2 4 7 3 4 8−7 7−6 3−4
2. Find the inverse of the following 1 2 1 1 ⟹ 𝑎𝑑𝑗 𝐴 = [ 3 − 6
3 4 − 3 3 − 2] =
−2 4 3 3 2
(i) [ ] 21 − 12 9 − 14 8 − 9
1 −3 4 7 3 4
−2 4 −2 4 −3 −4 1 1 −1
Let 𝐴 = [ ] ; |𝐴| = | | = 6 − 4 = 2 ≠ 0 ; 𝑎𝑑𝑗 𝐴 = [ ] −1 1 −1 1
1 −3 1 −3 −1 −2 𝐴 = |𝐴| 𝑎𝑑𝑗 𝐴 ⟹ 𝐴 = [−3 1 1]
1 1 −3 −4 2
𝐴−1 = |𝐴| 𝑎𝑑𝑗 𝐴 ⟹ 𝐴−1 = [ ] 9 −5 −1
2 −1 −2 𝑎 𝑏
-------------------------------------------------------------------------------------------------------------- 𝐄𝐗𝐀𝐌𝐏𝐋𝐄 𝟏. 𝟐 Find the inverse of[ ]
𝑐 𝑑
5 1 1 2 −1 3
(ii) [1 5 1] 𝐄𝐗𝐀𝐌𝐏𝐋𝐄 𝟏. 𝟑 Find the inverse of [−5 3 1]
1 1 5 −3 2 3
5 1 1 cos 𝛼 0 sin 𝛼
Let 𝐴 = [1 5 1] 3. 𝐼𝑓 𝐹(𝛼) = [ 0 1 0 ] , 𝑠ℎ𝑜𝑤 𝑡ℎ𝑎𝑡 [𝐹(𝛼)]−1
1 1 5 − sin 𝛼 0 cos 𝛼
5 1 1 cos(−𝛼) 0 sin(−𝛼) cos 𝛼 0 −
|𝐴| = |1 5 1| = 5(25 − 1) − 1(5 − 1) + 1(1 − 5) 𝐹(−𝛼) = [ 0 1 0 ] =[ 0 1
1 1 5 − sin(−𝛼) 0 cos(−𝛼) sin 𝛼 0
= 120 − 4 − 4 cos(−𝛼) = cos 𝛼
= 112 ≠ 0 (∵ )
sin(−𝛼) = − sin 𝛼