RBSE Class 12 Maths Solutions Chapter 4 Determinants Ex 4.2

RBSE Class 12 Maths Solutions Chapter 4 Determinants Ex 4.2

Question 1.

$\left|\begin{array}{ccc}x& a& x+a\\ y& b& y+b\\ z& c& z+c\end{array}\right|$ = 0
Answer:

Question 2.

$\left|\begin{array}{ccc}a-b& b-c& c-a\\ b-c& c-a& a-b\\ c-a& a-b& b-c\end{array}\right|$ = 0
Answer:

Question 3.
$\left|\begin{array}{ccc}2& 7& 65\\ 3& 8& 75\\ 5& 9& 86\end{array}\right|$ = 0
Answer:

Question 4.
$\left|\begin{array}{ccc}1& bc& a(b+c)\\ 1& ca& b(c+a)\\ 1& ab& c(a+b)\end{array}\right|$ = 0
Answer:

Question 5.

$\left|\begin{array}{ccc}b+c& q+r& y+z\\ c+a& r+p& z+x\\ a+b& p+q& x+y\end{array}\right|=2\left|\begin{array}{ccc}a& p& x\\ b& q& y\\ c& r& z\end{array}\right|$
Answer:

Question 6.
$\left|\begin{array}{ccc}0& a& -b\\ -a& 0& -c\\ b& c& 0\end{array}\right|$ = 0
Answer:

= - Δ
⇒ Δ + Δ = 0 ⇒ 2Δ = 0
⇒ Δ = 0
Hence proved.

Question 7.

$\left|\begin{array}{ccc}-a^2& ab& ac\\ ba& -b^2& bc\\ ca& cb& -c^2\end{array}\right|$ = 4a2b2c2
Answer:

Question 8.
(i) $\left|\begin{array}{ccc}1& a& a^2\\ 1& b& b^2\\ 1& c& c^2\end{array}\right|$ = (a - b) (b - c) (c - a)
Answer:

Taking (b - a) and (c - a) common from R2 and R3 respectively
Δ = (b - a) (c - a) $\left|\begin{array}{ccc}1& a& a^2\\ 0& 1& b+a\\ 0& 1& c+a\end{array}\right|$
Expanding along C1
Δ = (b - a) (c - a) $\left\{1\left|\begin{array}{cc}1& b+a\\ 1& c+a\end{array}\right|\right\}$
= (b - a) (c - a) {c + a - b - a}
= (b - a) (c - a) (c - b)
= (a - b) (b - c) (c - a)
Hence proved

(ii) $\left|\begin{array}{ccc}1& 1& 1\\ a& b& c\\ a^3& b^3& c^3\end{array}\right|$ = (a - b) (b - c) (c - a) (a + b + c)
Answer:

= (a - b) (b - c) × {b2 + bc + c2 - a2 - ab - b2}
= (a - b) (b - c) {c2 - a2 + bc - ab}
= (a - b) (b - c) × {(c - a) (c + a) + b(c - a)}
= (a - b) (b - c) (c - a) (a + b + c)
Hence Proved.

Question 9.

$\left|\begin{array}{ccc}x& x^2& yz\\ y& y^2& zx\\ z& z^2& xy\end{array}\right|$ = (x - y) (y - z) (z - x) (xy + yz + zx)
Answer:

Taking (x - y) and (y - z) common from R1 and R2 respectively

= (x - y) (y - z) (z - x) × {xy - z2 + z(x + y + z)}
= (x - y) (y - z) (z - x) × (xy - z2 + xz + yz + z2)
= (x - y) (y - z) (z - x) (xy + yz + zx)
Hence Proved.

Question 10.
(i) $\left|\begin{array}{ccc}x+4& 2x& 2x\\ 2x& x+4& 2x\\ 2x& 2x& x+4\end{array}\right|$ = (5x + 4) (4 - x)2
Answer:

= (5x - 4) {(x - 4) (x - 4) - 0 × (4 - x)}
= (5x + 4) (x - 4)2
= (5x + 4) (4 - x)2
Hence Proved.

(ii) $\left|\begin{array}{ccc}y+k& y& y\\ y& y+k& y\\ y& y& y+k\end{array}\right|$ = k2(3y + k)
Answer:

Question 11.

(i) $\left|\begin{array}{ccc}a-b-c& 2a& 2a\\ 2b& b-c-a& 2b\\ 2c& 2c& c-a-b\end{array}\right|$ = (a + b + c)3
Answer:

(ii) $\left|\begin{array}{ccc}x+y+2z& x& y\\ z& y+z+2x& y\\ z& x& z+x+2y\end{array}\right|$ = 2(x + y + z)3
Answer:

Question 12.

$\left|\begin{array}{ccc}1& x& x^2\\ x^2& 1& x\\ x& x^2& 1\end{array}\right|$ = (1 - x3)2
Answer:

= (1 + x + x2) (1 - x2) (1 + x + x2)
= (1 + x + x2)2 (1 - x)2 = {(1 + x + x2) (1 - x)}2
= (1 - x3)2
Hence proved.

Question 13.
$\left|\begin{array}{ccc}1+a^2-b^2& 2ab& -2b\\ 2ab& 1-a^2+b^2& 2a\\ 2b& -2a& 1-a^2-b^2\end{array}\right|$ = (1 + a2 + b2)3
Answer:

= (1 + a2 + b2)2 {1 - a2 - b2 + 2a2 + 2b2}
= (1 + a2 + b2)2 {1 + a2 + b2}
= (1 + a2 + b2)3
Hence Proved.

Question 14.

$\left|\begin{array}{ccc}{a}^2+1& ab& ac\\ ab& b^2+1& bc\\ ca& cb& c^2+1\end{array}\right|$ = 1 + a2 + b2 + c2
Answer:

= (1 + a2 + b2 + c2) {1 (1 × 1 - 0)}
= (1 + a2 + b2 + c2).1
= 1 + a2 + b2 + c2
Hence Proved.

Question 15.
Let A be a square matrix of order 3 × 3,then | kA | is equal to:
(A) k|A|
(B) k2|A|
(C) k3|A|
(D) 3k |A|
Answer:

Thus, option (C) is correct.

Question 16.
Which of the following is correct:
(A) Determinant is a square matrix.
(B) Determinant is a number associated to a matrix.
(C) Determinant is a number associated to a square matrix.
(D) None of these
Answer:
Option (C) is correct.

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