﻿ Class 12 NCERT Math Solution
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TOPICS
Exercise - 4.4

Question-1 :-  Write Minors and Cofactors of the elements of following determinants:

Solution :-
```(i)   Minor of the element aᵢⱼ is Mᵢⱼ
Here a₁₁ = 1.
So M₁₁ = Minor of the element a₁₁= 3
M₁₂ = Minor of the element a₁₂ = 0
M₂₁ = Minor of the element a₂₁ = –4
M₂₂ = Minor of the element a₂₂ = 2

Now, cofactor of aᵢⱼ is Aᵢⱼ.
So A₁₁ = (–1) ¹⁺¹  M₁₁ = (–1)2 (3) = 3
A₁₂ = (–1) ¹⁺²  M₁₂ = (–1)3 (0) = 0
A₂₁ = (–1) ²⁺¹  M₂₁ = (–1)3 (–4) = 4
A₂₂ = (–1) ²⁺²  M₂₂ = (–1)4 (2) = 2
```
```(ii)   Minor of the element aᵢⱼ is Mᵢⱼ
Here a₁₁ = 1.
So M₁₁ = Minor of the element a₁₁= d
M₁₂ = Minor of the element a₁₂ = b
M₂₁ = Minor of the element a₂₁ = c
M₂₂ = Minor of the element a₂₂ = a

Now, cofactor of aᵢⱼ is Aᵢⱼ.
So A₁₁ = (–1) ¹⁺¹  M₁₁ = (–1)2 (d) = d
A₁₂ = (–1) ¹⁺²  M₁₂ = (–1)3 (b) = -b
A₂₁ = (–1) ²⁺¹  M₂₁ = (–1)3 (c) = -c
A₂₂ = (–1) ²⁺²  M₂₂ = (–1)4 (a) = a
```

Question-2 :-  Write Minors and Cofactors of the elements of following determinants:

Solution :-
```(i) Minors: -	        Cofactors: -
M₁₁ = 1 – 0 = 1	    A₁₁ = (–1) ¹⁺¹ (1) = 1
M₁₂ = 0 - 0 = 0	    A₁₂ = (–1) ¹⁺² (0) = 0
M₁₃ = 0 – 0 = 0	    A₁₃ = (–1) ¹⁺³ (0) = 0
M₂₁ = 0 – 0 = 0	    A₂₁ = (–1) ²⁺¹ (0) = 0
M₂₂ = 1 – 0 = 1	    A₂₂ = (–1) ²⁺² (1) = 1
M₂₃ = 0 - 0 = 0	    A₂₃ = (–1) ²⁺³ (0) = 0
M₃₁ = 0 - 0 = 0	    A₃₁ = (–1) ³⁺¹ (0) = 0
M₃₂ = 0 – 0 = 0	    A₃₂ = (–1) ³⁺² (0) = 0
M₃₃ = 1 - 0 = 1	    A₃₃ = (–1) ³⁺³ (1) = 1
```
```(ii) Minors: -	        Cofactors: -
M₁₁ = 10 + 1 = 11	    A₁₁ = (–1) ¹⁺¹ (11) = 11
M₁₂ = 6 - 0 = 6	    A₁₂ = (–1) ¹⁺² (6) = -6
M₁₃ = 3 – 0 = 3	    A₁₃ = (–1) ¹⁺³ (3) = 3
M₂₁ = 0 – 4 = -4	    A₂₁ = (–1) ²⁺¹ (-4) = 4
M₂₂ = 2 – 0 = 2	    A₂₂ = (–1) ²⁺² (2) = 2
M₂₃ = 1 - 0 = 1	    A₂₃ = (–1) ²⁺³ (1) = -1
M₃₁ = 0 - 20 = -20        A₃₁ = (–1) ³⁺¹ (-20) = -20
M₃₂ = -1 – 12 = -13       A₃₂ = (–1) ³⁺² (-13) = 13
M₃₃ = 5 - 0 = 5	    A₃₃ = (–1) ³⁺³ (5) = 5
```

Question-3 :-  Using Cofactors of elements of second row, evaluate

Solution :-
```  Elements are : a₂₁ = 2, a₂₂ = 0, a₂₃ = 1
Minors: -	             Cofactors: -
M₂₁ = 9 – 16 = -7	    A₂₁ = (–1) ²⁺¹ (-7) = 7
M₂₂ = 15 – 8 = 7	    A₂₂ = (–1) ²⁺² (7) = 7
M₂₃ = 10 - 3 = 7	    A₂₃ = (–1) ²⁺³ (7) = -7

a₂₁ A₂₁ + a₂₂ A₂₂ + a₂₃ A₂₃
= 2 x 7 + 0 x 7 + 1 x (-7)
= 14 + 0 - 7
= 7
```

Question-4 :-  Using Cofactors of elements of third column, evaluate

Solution :-
```  Elements are : a₁₃ = yz, a₂₃ = zx, a₃₃ = xy
Minors: -	        Cofactors: -
M₁₃ = z – y	    A₁₃ = (–1) ¹⁺³ (z – y) = z – y
M₂₃ = z – x	    A₂₃ = (–1) ²⁺³ (z – x) = x - z
M₃₃ = y - x	    A₃₃ = (–1) ³⁺³ (y - x) = y - x

a₁₃ A₁₃ + a₂₃ A₂₃ + a₃₃ A₃₃
= yz . (z - y) + zx . (x - z) + xy . (y - x)
= yz² - y²z + zx² - z²x + xy² - x²y
= (x²z - y²z) + (yz² - xz²) + (xy² - yx²)
= z(x² - y²) + z²(y - x) + xy(y - x)
= z(x + y)(x - y) - z²(x - y) - xy(x - y)
= (x - y)[zx + zy - z² - xy]
= (x - y)[z(x - z) + y(z - x)]
= (x - y)(x - z)(y - z)
```

Question-5 :-  and Aᵢⱼ is Cofactors of aᵢⱼ, then value of Δ is given by
(A) a₁₁ A₃₁ + a₁₂ A₃₂ + a₁₃ A₃₃
(B) a₁₁ A₁₁ + a₁₂ A₂₁ + a₁₃ A₃₁
(C) a₂₁ A₁₁ + a₂₂ A₁₂ + a₂₃ A₁₃
(D) a₁₁ A₁₁ + a₂₁ A₂₁ + a₃₁ A₃₁

Solution :-
```  a₁₁ A₁₁ + a₂₁ A₂₁ + a₃₁ A₃₁
The correct answer is D.
```
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