NCERT Solutions for Chapter 4 Determinants Class 12 Maths Exercise 4.4

Class 12 Maths NCERT Solutions for Chapter 4 Determinants Exercise 4.4

Determinants Exercise 4.4 Solutions

1. Write Minors and Cofactors of the elements of following determinants : 
(i) 

(ii) 

Solution

(i) The given determinant is 
Minor of a_ij is M_ij .
M₁₁ = minor of element a₁₁ = 3
M₁₂ = minor of element a₁₂ = 0 
M₂₁ = minor of element a₂₁ = -4
M₂₂ = minor of element a₂₂ = 2 
Cofactor of a_ij is A_ij = (-1)^i+j  M_ij .
∴ 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) The given determinant is 
Minor of element a_ij is M_ij . 
∴ M₁₁ = minor of element a₁₁ = d
M₁₂ = minor of element a₁₂ = b
M₂₁ = minor of element a₂₁ = c
M₂₂ = minor of element a₂₂ = a
Cofactor of a_ij is A_ij = (-1)^i+j  M_ij .
∴ 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

2. Write Minors and Cofactors of the elements of following determinants: 
(i) 
(ii) 

Solution

(i) The given determinant is 
By the definition of minors and cofactors, we have : 

A₁₁ = cofactor of a₁₁ = (-1)¹⁺¹ M₁₁ = 1
A₁₂ = cofactor of a₁₂ = (-1)¹⁺² M₁₂ = 0
A₁₃ = cofactor of a₁₃ = (-1)¹⁺³ M₁₃ = 0
A₂₁ = cofactor of a₂₁ = (-1)²⁺¹ M₂₁ = 0
A₂₂ = cofactor of a₂₂ = (-1)²⁺² M₂₂ = 1
A₂₃ = cofactor of a₂₃ = (-1)²⁺³ M₂₃ = 0
A₃₁ = cofactor of a₃₁ = (-1)³⁺¹ M₃₁ = 0
A₃₂ = cofactor of a₃₂ = (-1)³⁺² M₃₂ = 0
A₃₃ = cofactor of a₃₃ = (-1)³⁺³ M₃₃ = 1

(ii) The given determinant is  
By definition of minors and cofactors, we have : 

A₁₁ = cofactor of a₁₁ = (-1)¹⁺¹ M₁₁ = 11
A₁₂ = cofactor of a₁₂ = (-1)¹⁺² M₁₂ = -6
A₁₃ = cofactor of a₁₃ = (-1)¹⁺³ M₁₃ = 3
A₂₁ = cofactor of a₂₁ = (-1)²⁺¹ M₂₁ = 4
A₂₂ = cofactor of a₂₂ = (-1)²⁺² M₂₂ = 2
A₂₃ = cofactor of a₂₃ = (-1)²⁺³ M₂₃ = -1
A₃₁ = cofactor of a₃₁ = (-1)³⁺¹ M₃₁ = -20
A₃₂ = cofactor of a₃₂ = (-1)³⁺² M₃₂ = 13
A₃₃ = cofactor of a₃₃ = (-1)³⁺³ M₃₃ = 5

3. Using Cofactors of elements of second row, evaluate ∆ = 

Solution

He given determinant is 
We have : 

A₂₃ = cofactor of a₂₃ = (-1)²⁺³ M₂₃ = -7
We know that ∆ is equal to the sum of the product of the elements of the second row with their corresponding cofactors.
Δ = a₂₁ A₂₁ + a₂₂ A₂₂ + a₂₃ A₂₃ = 2(7) + 0(7) + 1(-7) = 14 - 7 = 7

4. sing Cofactors of elements of third column, evaluate Δ =

Solution

The given determinant is 
We have :

A₁₃ = cofactor of a₁₃ = (-1)¹⁺³ M₁₃ = (z- y)
A₂₃ = cofactor of a₂₃ = (-1)²⁺³ M₂₃ = -(z - x) = (x - z) 
A₃₃ = cofactor of a₃₃ = (-1)³⁺³ M₃₃ = (y - x)
We know that Δ is equal to the sum of the product of the elements of the second row with their corresponding cofactors. 
∴ Δ = a₂₁ A₂₁ + a₂₂ A₂₂ + a₂₃ A₂₃ 
= yz(z - y) + zx(x - z) + xy(y - x) 
= yz² - y² z  +x² z - xz² + xy² - x² y
= (x² z - y² z) + (yz² - xz² ) + (xy² - x² y)
= z(x²  - y² ) + z² (y -x) + xy(y -x)
= z(x - y)(x + y) + z² (y -x) + xy(y - x) 
= (x - y)[zx - zy - z² - xy]
= (x - y) [z(x -z) + y(z - x)] 
= (x - y) (z - x)[-z + y]
= (x - y)(y - z)(z - x)
Hence, Δ = (x - y)(y - z)(z - x) .

5. If ∆ =  and A_ij  is Cofactors of a_ij , 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 

We know that :

∆ = Sum of the product of the elements of a column (or a row) with their corresponding cofactors
∆ = a₂₁ A₂₁ + a₂₂ A₂₂ + a₂₃ A₂₃
Hence, the value of Δ is given by the expression given in alternative D.
The correct answer is D.