Determinants Matrices Question 8
Question: If $ a^{2}+b^{2}+c^{2}\ne -2 $ and $ f(x)= \begin{vmatrix} (1+a^{2})x & (1+b^{2})x & (1+c^{2})x \\ (1+a^{2})x & (1+b^{2})x & (1+c^{2})x \\ (1+a^{2})x & (1+b^{2})x & (1+c^{2})x \\ \end{vmatrix} $ then $ f(x) $ is a polynomial of degree
Options:
A) $ 1 $
B) 0
C) $ 3 $
D) $ 2 $
Show Answer
Answer:
Correct Answer: D
Solution:
- [d] $ f(x)= $
$ \begin{vmatrix} 1+(a^{2}+b^{2}+c^{2}+2)x & (1+b^{2})x & (1+c^{2})x \\ 1+(a^{2}+b^{2}+c^{2}+2)x & 1+b^{2}x & (1+c^{2})x \\ 1+(a^{2}+b^{2}+c^{2}+2)x & (1+b^{2})x & 1+c^{2}x \\ \end{vmatrix} $
Applying, $ C_1\to C_1+C_2+C_3 $
$ = \begin{vmatrix} 1 & (1+b^{2})x & (1+c^{2})x \\ 1 & 1+b^{2}x & (1+c^{2})x \\ 1 & (1+b^{2})x & 1+c^{2}x \\ \end{vmatrix} $
$ \therefore a^{2}+b^{2}+c^{2}+2\ne 0 $
$ f(x)= \begin{vmatrix} 0 & x-1 & 0 \\ 0 & 1-x & x-1 \\ 1 & (1+b^{2})x & 1+c^{2}x \\ \end{vmatrix} $
Applying $ R_1\to R_1-R_2, $
$ R_2\to R_2-R_3 $
$ f(x)={{(x-1)}^{2}} $ Hence degree = 2.
BETA
