Rajasthan Board RBSE Solutions for Class 12 Maths Chapter 4 Determinants Ex 4.2 Textbook Exercise Questions and Answers.
Rajasthan Board RBSE Solutions for Class 12 Maths in Hindi Medium & English Medium are part of RBSE Solutions for Class 12. Students can also read RBSE Class 12 Maths Important Questions for exam preparation. Students can also go through RBSE Class 12 Maths Notes to understand and remember the concepts easily.
Question 1.
\(\left|\begin{array}{lll} x & a & x+a \\ y & b & y+b \\ z & c & z+c \end{array}\right|\) = 0
Answer:
Question 2.
\(\left|\begin{array}{lll} a-b & b-c & c-a \\ b-c & c-a & a-b \\ c-a & a-b & b-c \end{array}\right|\) = 0
Answer:
Question 3.
\(\left|\begin{array}{lll} 2 & 7 & 65 \\ 3 & 8 & 75 \\ 5 & 9 & 86 \end{array}\right|\) = 0
Answer:
Question 4.
\(\left|\begin{array}{lll} 1 & b c & a(b+c) \\ 1 & c a & b(c+a) \\ 1 & a b & c(a+b) \end{array}\right|\) = 0
Answer:
Question 5.
\(\left|\begin{array}{lll} b+c & q+r & y+z \\ c+a & r+p & z+x \\ a+b & p+q & x+y \end{array}\right|=2\left|\begin{array}{lll} a & p & x \\ b & q & y \\ c & r & z \end{array}\right|\)
Answer:
Question 6.
\(\left|\begin{array}{rrr} 0 & a & -b \\ -a & 0 & -c \\ b & c & 0 \end{array}\right|\) = 0
Answer:
= - Δ
⇒ Δ + Δ = 0 ⇒ 2Δ = 0
⇒ Δ = 0
Hence proved.
Question 7.
\(\left|\begin{array}{rrr} -a^{2} & a b & a c \\ b a & -b^{2} & b c \\ c a & c b & -c^{2} \end{array}\right|\) = 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.
\(\left|\begin{array}{ccc} x & x^{2} & y z \\ y & y^{2} & z x \\ z & z^{2} & x y \end{array}\right|\) = (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) \(\left|\begin{array}{ccc} x+4 & 2 x & 2 x \\ 2 x & x+4 & 2 x \\ 2 x & 2 x & x+4 \end{array}\right|\) = (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) \(\left|\begin{array}{ccc} a-b-c & 2 a & 2 a \\ 2 b & b-c-a & 2 b \\ 2 c & 2 c & c-a-b \end{array}\right|\) = (a + b + c)3
Answer:
(ii) \(\left|\begin{array}{ccc} x+y+2 z & x & y \\ z & y+z+2 x & y \\ z & x & z+x+2 y \end{array}\right|\) = 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.
\(\left|\begin{array}{ccc} 1+a^{2}-b^{2} & 2 a b & -2 b \\ 2 a b & 1-a^{2}+b^{2} & 2 a \\ 2 b & -2 a & 1-a^{2}-b^{2} \end{array}\right|\) = (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.
\(\left|\begin{array}{ccc} a^{2}+1 & a b & a c \\ a b & b^{2}+1 & b c \\ c a & c b & c^{2}+1 \end{array}\right|\) = 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.