Class 12 NCERT Solutions – Mathematics Part I – Chapter 4 Determinants – Exercise 4.4

Write minors and cofactors of the elements of the following determinants: 

Question 1. 

(i)

(ii)

Solution: 

(i) 

Finding minors of the elements of the determinant: 

Let us assume M_ij is Minors of elements a_ij

M₁₁ = Minor of elements a₁₁ = 3

M₁₂ = Minor of elements a₁₂ = 0

M₂₁ = Minor of elements a₂₁ = −4

M₂₂ = Minor of elements a₂₂ = 2

Finding cofactor of a_ij

Let us assume cofactor of a_ij is A_ij 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) 

Finding minors of the elements of the determinant:

Let us assume M_ij is Minors of elements a_ij

M₁₁ = Minor of element a₁₁ = d

M₁₂ = Minor of elements a₁₂ = b

M₂₁ = Minor of elements a₂₁ = c

M₂₂ = Minor of elements a₂₂ = a

Finding cofactor of a_ij

Let us assume cofactor of a_ij is A_ij, which is (−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)

Solution:

(i) 

Let us find the Minors and cofactors of the elements:

Assume, M_ij is minor of element a_ij and A_ij is cofactor of a_ij

M₁₁ = Minor of elements a₁₁ == 1 − 0 = 1 and A₁₁ = 1

M₁₂ = Minor of elements a₁₂ == 0 − 0 = 0 and A₁₂ = 0

M₁₃ = Minor of elements a₁₃ == 0 − 0 = 0 and A₁₃ = 0

M₂₁ = Minor of elements a₂₁ == 0 − 0 = 0 and A₂₁ = 0

M₂₂ = Minor of elements a₂₂ == 1 − 0 = 1 and A₂₂ = 1

M₂₃ = Minor of elements a₂₃ == 0 − 0 = 0 and A₂₃ = 0

M₃₁ = Minor of elements a₃₁ == 0 − 0 = 0 and A₃₁ = 0

M₃₂ = Minor of elements a₃₂ == 0 − 0 = 0 and A₃₂ = 0

M₃₃ = Minor of elements a₃₃ == 1 − 0 = 1 and A₃₃ = 1

(ii) 

Let us find the Minors and cofactors of the elements:

Assume, M_ij is minor of element a_ij and A_ij is cofactor of a_ij

M₁₁ = Minor of elements a₁₁ == 10 − (−1) = 11 and A₁₁ = 11

M₁₂ = Minor of elements a₁₂ == 6 − 0 = 6 and A₁₂ = −6

M₁₃ = Minor of elements a₁₃ == 3 − 0 = 3 and A₁₃ = 3

M₂₁ = Minor of elements a₂₁ == 0 − 4 = −4 and A₂₁ = 4

M₂₂ = Minor of elements a₂₂ == 2 − 0 = 2 and A₂₂ = 2

M₂₃ = Minor of elements a₂₃ == 1 − 0 = 1 and A₂₃ = −1

M₃₁ = Minor of elements a₃₁ == 0 − 20 = −20 and A₃₁ = −20

M₃₂ = Minor of elements a₃₂ == −1 − 12 = −13 and A₃₂ = 13

M₃₃ = Minor of elements a₃₃ == 5 − 0 = 5 and A₃₃ = 5

Question 3. Using Cofactors of elements of second row, evaluate △?

Solution:

Finding the Cofactors of elements of second row:

A₂₁ = Cofactor of elements a₂₁ = (−1)²⁺¹ = (−1)³ (9 − 16) = 7

A₂₂ = Cofactor of elements a₂₂ = (−1)²⁺² = (−1)⁴ (15 − 8) = 7

A₂₃ = Cofactor of elements a₂₃ = (−1)²⁺³ = (−1)⁵ (10 − 3) = 7

Now, △ = a₂₁ A₂₁ + a₂₂ A₂₂ + a₂₃ A₂₃ = 14 + 0 − 7 = 7

Question 4. Using Cofactors of elements of third column, evaluate △?

Solution:

Finding the Cofactors of elements of third column:

A₁₃ = Cofactor of elements a₁₃ = (−1)¹⁺³ = (−1)⁴ (z − y) = −y

A₂₃ = Cofactor of elements a₂₃ = (−1)²⁺³ = (−1)⁵ (z − x) = x − z

A₃₃ = Cofactor of elements a₃₃ = (−1)³⁺³ = (−1)⁶ (y − x) = y − x

Now, △ = a₁₃ A₁₃ + a₂₃ A₂₃ + a₃₃ A₃₃

= yz (z − y) + zx (x − z) + xy (y − x)

= (yz² − y²z) + (xy² − xz²) + (xz² − x²y)

= (y − z)[−yz + x(y + z) − x²]

= (y − z)[−yz + x (z − x) + x (z − x)]

= (x − y)(y − x)(z − x)

Question 5. If and A_ij is cofactor of a_ij then value of △ is given by:

(A) a₁₁A₃₁ + a₁₂A₃₂ + a₁₃A₃₃

(B) a₁₁A₁₁ + a₁₂A₂₁ + a₁₃A₃₁

(C) a₂₁A₁₁ + a₂₂A₁₂ + a₂₃A₁₃

(D) a₁₁A₁₁ + a₂₁A₂₁ + a₃₁A₃₁

Solution: 

Option (D) is correct.