Question

If $ a^2 +b^2 +c^2 = -2 $ and $ f\left(x\right)=\begin{vmatrix}1+a^{2}x&\left(1+b^{2}\right)x&\left(1+c^{2}\right)x\\ \left(1+a^{2}\right)x&1+b^{2}x&\left(1+c^{2}\right)x\\ \left(1+a^{2}\right)x&\left(1+b^{2}\right)x&1+c^{2}x\end{vmatrix} $ , then $ f(x) $ is a polynomial of degree

• A. $ 3 $
• B. $ 2 $
• C. $ 1 $
• D. $ 0 $

Answer 1

Answer: (B)

Applying $C_{1} \rightarrow C_{1}+C_{2}+C_{3}$, we get
$f(x)=\begin{vmatrix}1 & \left(1+b^{2}\right) x & \left(1+c^{2}\right) x \\ 1 & \left(1+b^{2} x\right) & \left(1+c^{2}\right) x \\ 1 & \left(1+b^{2}\right) x & 1+c^{2} x\end{vmatrix}$
$\left(\because a^{2}+b^{2}+c^{2}+2=0\right)$
Again, applying $R_{2} \rightarrow R_{2}-R_{1}, R_{3} \rightarrow R_{3}-R_{1}$,
we get
$=\begin{vmatrix}1 & \left(1+b^{2}\right) x & \left(1+c^{2}\right) x \\ 0 & 1-x & 0 \\ 0 & 0 & 1-x\end{vmatrix}=(1-x)^{2}$
Hence, degree of $f(x)=2$