If $a^2+b^2+c^2=-2$ and $f(x)=\left|\begin{array}{ccc}1+a^2 x & \left(1+b^2\right) x & \left(1+c^2\right) x…
If $a^2+b^2+c^2=-2$ and $f(x)=\left|\begin{array}{ccc}1+a^2 x & \left(1+b^2\right) x & \left(1+c^2\right) x \\ \left(1+a^2\right) x & 1+b^2 x & \left(1+c^2\right) x \\ \left(1+a^2\right) x & \left(1+b^2\right) x & 1+c^2 x\end{array}\right|$ then $f(x)$ is $a$ polynomial of degree
1
0
3
2
Solution
$
\begin{aligned}
& f(x)=\left|\begin{array}{ccc}
1+\left(a^2+b^2+c^2+2\right) x & \left(1+b^2\right) x & \left(1+c^2\right) x \\
1+\left(a^2+b^2+c^2+2\right) x & 1+b^2 x & \left(1+c^2\right) x \\
1+\left(a^2+b^2+c^2+2\right) x & \left(1+b^2\right) x & 1+c^2 x
\end{array}\right| \text {, Applying } C_1 \rightarrow C_1+C_2+C_3 \\
& =\left|\begin{array}{ccc}
1 & \left(1+b^2\right) x & \left(1+c^2\right) x \\
1 & 1+b^2 x & \left(1+c^2\right) x \\
1 & \left(1+b^2\right) x & 1+c^2 x
\end{array}\right| \because a^2+b^2+c^2+2=0 \\
& f(x)=\left|\begin{array}{ccc}
0 & x-1 & 0 \\
0 & 1-x & x-1 \\
1 & \left(1+b^2\right) x & 1+c x
\end{array}\right| ; \text { Applying } R_1 \rightarrow R_1-R_2, R_2 \rightarrow R_2-R_3 \\
& f(x)=(x-1)^2
\end{aligned}
$
Hence degree $=2$.