Question

If $a^2+b^2+c^2+2=0$ and $f\left(x\right)=\left|\begin{array}{ccc}1+a^{2}x& (1+b^2)x& (1+c^2)x\\ (1+a^2)x& 1+b^{2}x& (1+c^2)x\\ (1+a^2)x& (1+b^2)x& 1+c^{2}x\end{array}\right|$ then $f\left(x\right)$ is poynomial of degree

• A. $0$
• B. $1$
• C. $2$
• D. $3$

Answer 1

The correct option is C $2$

$f\left(x\right)=\left|\begin{array}{ccc}1+a^{2}x& (1+b^2)x& (1+c^2)x\\ (1+a^2)x& 1+b^{2}x& (1+c^2)x\\ (1+a^2)x& (1+b^2)x& 1+c^{2}x\end{array}\right|$
$R_1\to R_1-R_2,R_2\to R_2-R_3$
$=\left|\begin{array}{ccc}1-x& x-1& 0\\ 0& 1-x& x-1\\ (1+a^2)x& (1+b^2)x& 1+c^{2}x\end{array}\right|$
$f\left(x\right)={(x-1)}^2\left|\begin{array}{ccc}-1& 1& 0\\ 0& -1& 1\\ (1+a^2)x& (1+b^2)x& 1+c^{2}x\end{array}\right|$
$C_1\to C_1+C_2+C_3$
$f\left(x\right)={(x-1)}^2\left|\begin{array}{ccc}0& 1& 0\\ 0& -1& 1\\ 1+(a^2+b^2+c^2+2)x& (1+b^2)x& 1+c^{2}x\end{array}\right|$
$f\left(x\right)=({1+(a^2+b^2+c^2+2)x\left)\right(x-1)}^2$

$\Rightarrow f\left(x\right)=(x-1{)}^2$

Hence, $f\left(x\right)$ is a polynomial of degree $2$