Determinants Matrices Question 23
Question: The value of the determinant $ \begin{vmatrix} kb & {k^{^{2}}}+a^{2} & 1 \\ kb & k^{2}+b^{2} & 1 \\ kc & k^{2}+c^{2} & 1 \\ \end{vmatrix} $ is
Options:
A) $ k(a+b)(b+c)(c+a) $
B) $ kabc(a^{2}+b^{2}+c^{2}) $
C) $ k(a-b)(b-c)(c-a) $
D) $ k(a+b-c)(b+c-a)(c+a-b) $
Show Answer
Answer:
Correct Answer: C
Solution:
- [c] we have, $ =k(a-b)(b-c)(c-a) $
BETA
