NCERT Solutions for Class 12 Maths Chapter 4 – Determinants Ex 4.4

NCERT Solutions for Class 12 Maths Chapter 4 – Determinants Ex 4.4

Page No 126:

Question 1:

Write Minors and Cofactors of the elements of following determinants:

(i)  (ii) 

Answer:

(i) The given determinant is.

Minor of element 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

Question 2:

(i)  (ii) 

Answer:

(i) The given determinant is.

By the definition of minors and cofactors, we have:

M₁₁ = minor of a₁₁= 

M₁₂ = minor of a₁₂= 

M₁₃ = minor of a₁₃ = 

M₂₁ = minor of a₂₁ = 

M₂₂ = minor of a₂₂ = 

M₂₃ = minor of a₂₃ = 

M₃₁ = minor of a₃₁= 

M₃₂ = minor of a₃₂ = 

M₃₃ = minor of a₃₃ = 

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:

M₁₁ = minor of a₁₁= 

M₁₂ = minor of a₁₂= 

M₁₃ = minor of a₁₃ = 

M₂₁ = minor of a₂₁ = 

M₂₂ = minor of a₂₂ = 

M₂₃ = minor of a₂₃ = 

M₃₁ = minor of a₃₁= 

M₃₂ = minor of a₃₂ = 

M₃₃ = minor of a₃₃ = 

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

Question 3:

Using Cofactors of elements of second row, evaluate.

Answer:

The given determinant is.

We have:

M₂₁ = 

∴A₂₁ = cofactor of a₂₁ = (−1)²⁺¹ M₂₁ = 7

M₂₂ = 

∴A₂₂ = cofactor of a₂₂ = (−1)²⁺² M₂₂ = 7

M₂₃ = 

∴A₂₃ = cofactor of a₂₃ = (−1)²⁺³ M₂₃ = −7

We know that Δ is equal to the sum of the product of the elements of the second row with their corresponding cofactors.

∴Δ = a₂₁A₂₁ + a₂₂A₂₂ + a₂₃A₂₃ = 2(7) + 0(7) + 1(−7) = 14 − 7 = 7

Question 4:

Using Cofactors of elements of third column, evaluate

Answer:

The given determinant is.

We have:

M₁₃ = 

M₂₃ = 

M₃₃ = 

∴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.

Hence, 

Question 5:

If  and A_ij is Cofactors of a_ij, then value of Δ is given by

Answer:

Answer: D

We know that:

Δ = Sum of the product of the elements of a column (or a row) with their corresponding cofactors

∴Δ = a₁₁A₁₁ + a₂₁A₂₁ + a₃₁A₃₁

Hence, the value of Δ is given by the expression given in alternative D.

The correct answer is D.