RBSE Solutions for Class 12 Maths Chapter 4 Determinants Ex 4.4

Rajasthan Board RBSE Solutions for Class 12 Maths Chapter 4 Determinants Ex 4.4 Textbook Exercise Questions and Answers.

RBSE Class 12 Maths Solutions Chapter 4 Determinants Ex 4.4

Question 1.
(i) \(\left|\begin{array}{rr} 2 & -4 \\ 0 & 3 \end{array}\right|\)
Answer:
\(\left|\begin{array}{rr} 2 & -4 \\ 0 & 3 \end{array}\right|\)
Let Δ = \(\left|\begin{array}{ll} a_{11} & a_{12} \\ a_{21} & a_{22} \end{array}\right|=\left|\begin{array}{rr} 2 & -4 \\ 0 & 3 \end{array}\right|\)
Minor of 2(a₁₁) M₁₁ = 3,
Cofactor of 2(a₁₁) A₁₁ = (- 1)^{1 + 1} M₁₁ = 1 × 3 = 3
Minor of (- 4) a₁₂ M₁₂ = 0
Cofactor of (- 4) (a₁₂) A₁₂ = (- 1)^{1 + 2} M₁₂ = (- 1) × 0 = 0
Minor of 0(a₂₁) M₂₁ = - 4
Cofactor of 0(a₂₁) A₂₁ = (- 1)^{2 + 1} M₂₁ = (- 1) (- 4) = 4
Minor of 3(a₂₂) = 2
Cofactor of 3(a₂₂) A₂₂ = (- 1)^{2 + 2} M₂₂ = (- 1)⁴ a = a

Question 2.
(i) \(\left|\begin{array}{lll} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right|\)
Answer:
Let Δ = \(\left|\begin{array}{lll} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right|\)

(ii) \(\left|\begin{array}{rrr} 1 & 0 & 4 \\ 3 & 5 & -1 \\ 0 & 1 & 2 \end{array}\right|\)
Answer:
Let Δ = \(\left|\begin{array}{rrr} 1 & 0 & 4 \\ 3 & 5 & -1 \\ 0 & 1 & 2 \end{array}\right|\)

Question 3.
Using cofactors of elements of second row, evaluate Δ = \(\left|\begin{array}{lll} 5 & 3 & 8 \\ 2 & 0 & 1 \\ 1 & 2 & 3 \end{array}\right|\)
Answer:
Given, Δ = \(\left|\begin{array}{lll} 5 & 3 & 8 \\ 2 & 0 & 1 \\ 1 & 2 & 3 \end{array}\right|\)
Expanding along R₂
Δ = - 2(3 × 3 - 2 × 8) + 0 (5 × 3 - 8 × 1) - 1 (5 × 2 - 3 × 1)
= - 2(9 - 16) + 0 - 1 (10 - 3)
= - 2 (- 7) - 1 (7)
= 14 - 7 = 7

Question 4.
Using cofactors of elements of third column, evaluate: Δ = \(\left|\begin{array}{lll} 1 & x & y z \\ 1 & y & z x \\ 1 & z & x y \end{array}\right|\)
Answer:
Given, Δ = \(\left|\begin{array}{lll} 1 & x & y z \\ 1 & y & z x \\ 1 & z & x y \end{array}\right|\)
Expanding along third column, we get
Δ = yz × (z - y) - (z - x) + xy(y - x)
= yz² - y²z - z²x + zx² + xy² - x²y
= - (y²z - yz²) + x(y² - z²) - x²(y - z)
= - yz(y - z) + x(y - z) (y + z) - x²(y - z)
= (y - z) {- yz + xy + xz - x²}
= (y - z) {xz - x² - yz + xy}
= (y - z) {x(z - x) - y(z - x)}
= (y - z) {(z - x) (x - y)}
= (x - y) (y - z) (z - x)

Question 5.
If Δ = \(\left|\begin{array}{lll} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{array}\right|\) and A_ij is cofactor of a_ij then value of Δ is given by:
(A) a₁₁A₃₁ + a₁₂A₃₂ + a₁₃A₃₃
(B) a₁₁A₁₁ + a₁₂A₂₁ + a₁₃A₃₁
(C) a₂₁A₁₁ + a₂₂A₁₂ + a₂₃A₁₃
(D) a₁₁A₁₁ + a₂₁A₂₁ + a₃₁A₃₁
Answer:
Value of determinant = Sum of elements of any row or column and product of their corresponding cofactors
Element of first column C₁ (a₁₁, a₂₁, a₃₁)
Cofactor of a₁₁ = A₁₁
Cofactor of a₂₁ = A₂₁
Cofactor of a₃₁ = A₃₁
∴ Δ = a₁₁A₁₁ + aA₂₁A₂₁ + a₃₁A₃₁
Thus, option (D) is correct.