Question

If $a^2+b^2+c^2=-2$ 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 a polynomial of degree

• A. $2$
• B. $3$
• C. $4$
• D. $5$

Answer 1

The correct option is A $2$

$\left|\begin{array}{ccc}1+a^{2}x& (1+b^2)x& (1+c^2)x\\ (1+a^2)x& 1+b^{2}x& (1+c^2)x\\ (1+a^2)x& (1+b^2)x& 1+c^{2}x\end{array}\right|$

$=\left|\begin{array}{ccc}1+2x+(a^2+b^2+c^2)x& (1+b^2)x& (1+c^2)x\\ 1+2x+(a^2+b^2+c^2)x& 1=b^{2}x& (1+c^2)x\\ 1+2x+(a^2+b^2+c^2)x& (1+b^2)x& 1+c^{2}x\end{array}\right|[C_1\to C_1+C_2+C_3]$

$=\left|\begin{array}{ccc}1& (1+b^2)x& (1+c^2)x\\ 1& 1+b^{2}x& (1+c^2)x\\ 1& (1+b^2)x& 1+c^{2}x\end{array}\right|[\because a^2+b^2+c^2=-2]$

$=\left|\begin{array}{ccc}1& (1+b^2)x& (1+c^2)x\\ 0& 1-x& 0\\ 0& 0& 1-x\end{array}\right|[R_2\to R_2–R_1;R_3\to R_3-R_1]$

$=(1-x{)}^2$

$\therefore f\left(x\right)$ is a polynomial of degree 2.