RBSE Class 12 Maths Solutions Chapter 4 Determinants Ex 4.4

RBSE Class 12 Maths Solutions Chapter 4 Determinants Ex 4.4

Question 1.

(i) $\left|\begin{array}{cc}2& -4\\ 0& 3\end{array}\right|$
Answer:
$\left|\begin{array}{cc}2& -4\\ 0& 3\end{array}\right|$
Let Δ = $\left|\begin{array}{cc}{a}_{11}& a_{12}\\ a_{21}& a_{22}\end{array}\right|=\left|\begin{array}{cc}2& -4\\ 0& 3\end{array}\right|$
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) $\left|\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right|$
Answer:
Let Δ = $\left|\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right|$

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

Question 3.
Using cofactors of elements of second row, evaluate Δ = $\left|\begin{array}{ccc}5& 3& 8\\ 2& 0& 1\\ 1& 2& 3\end{array}\right|$
Answer:
Given, Δ = $\left|\begin{array}{ccc}5& 3& 8\\ 2& 0& 1\\ 1& 2& 3\end{array}\right|$
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 = $\left|\begin{array}{ccc}1& x& yz\\ 1& y& zx\\ 1& z& xy\end{array}\right|$
Answer:
Given, Δ = $\left|\begin{array}{ccc}1& x& yz\\ 1& y& zx\\ 1& z& xy\end{array}\right|$
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 = $\left|\begin{array}{ccc}{a}_{11}& a_{12}& a_{13}\\ a_{21}& a_{22}& a_{23}\\ a_{31}& a_{32}& a_{33}\end{array}\right|$ and Aij is cofactor of aij 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.

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