Question

If $f\left(x\right)=\left|\begin{array}{ccc}1& x& x+1\\ 2x& x(x-1)& (x+1)x\\ 3x(x-1)& x(x-1)(x-2)& (x+1)x(x-1)\end{array}\right|$. Then $f\left(100\right)$ is equal to

• A. $0$
• B. $1$
• C. $100$
• D. $10$

Answer 1

The correct option is A $0$

We have, $f\left(x\right)=\left|\begin{array}{ccc}1& x& x+1\\ 2x& x(x-1)& (x+1)x\\ 3x(x-1)& x(x-1)(x-2)& (x+1)x(x-1)\end{array}\right|$

Taking common $x,(x-1)$ from $r_2$ and $R_3$ respectively, we get
$f\left(x\right)=\left|\begin{array}{ccc}1& x& x+1\\ 2x& x-1& x+1\\ 3x& x(x-2)& (x+1)x\end{array}\right|$

Taking common $(x+1)$ from $C_3$, we get
$f\left(x\right)=\left|\begin{array}{ccc}1& x& 1\\ 2& x-1& 1\\ 3x& x(x-2)& x\end{array}\right|$

Taking common $x$ from $R_3$ we get
$f\left(x\right)=\left|\begin{array}{ccc}1& x& 1\\ 2& x-1& 1\\ 3& x-2& 1\end{array}\right|$

Applying $R_1\to r_1-R_2,R_2\to R_2-R_3$, we get
$f\left(x\right)=x^2(x-1)(x+1)\left|\begin{array}{ccc}-1& 1& 0\\ -1& 1& 0\\ 3& x-2& 1\end{array}\right|=0$

$[\because R_1\text{is identical with}{R}_2]$
$\Rightarrow f\left(100\right)=0$