Question

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

• A. 0
• B. 1
• C. 2
• D. 3

Answer 1

The correct option is C 2

$f\left(x\right)=\left|\begin{array}{ccc}{1+a}^{2}x& {(1+b}^2)x& \left({1+c}^2\right)x\\ \left({1+a}^2\right)x& {1+b}^{2}x& \left({1+c}^2\right)x\\ \left({1+a}^2\right)x& \left({1+b}^2\right)x& {1+c}^{2}x\end{array}\right|$
$C_1\to C_1+C_2+C_3$
$=\left|\begin{array}{ccc}{2x+1+(a}^2+b^2+c^2)x& {(1+b}^2)x& \left({1+c}^2\right)x\\ {2x+1+(a}^2+b^2+c^2)x& {1+b}^{2}x& \left({1+c}^2\right)x\\ {2x+1+(a}^2+b^2+c^2)x& \left({1+b}^2\right)x& {1+c}^{2}x\end{array}\right|$
$=\left|\begin{array}{ccc}1& {(1+b}^2)x& \left({1+c}^2\right)x\\ 1& {1+b}^{2}x& \left({1+c}^2\right)x\\ 1& \left({1+b}^2\right)x& {1+c}^{2}x\end{array}\right|$
$R_1\to R_1-R_2,{R_2\to R}_2-R_3$
$=\left|\begin{array}{ccc}0& x-1& 0\\ 0& -(x-1)& x-1\\ 1& \left({1+b}^2\right)x& {1+c}^{2}x\end{array}\right|$
$={(x-1)}^2\left|\begin{array}{ccc}0& 1& 0\\ 0& -1& 1\\ 1& \left({1+b}^2\right)x& {1+c}^{2}x\end{array}\right|$
$f\left(x\right)={(x-1)}^2$
Hence, $f\left(x\right)$ is a polynomial of degree $2$.