Question

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

• A. $0$ (zero)
• B. $1$
• C. $100$
• D. $10$

Answer 1

The correct option is A $0$ (zero)

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

Using transformations:

$C_1=C_1-C_3$

$C_2=C_2-C_3$
$f\left(x\right)=x(x+1)\left|\begin{array}{ccc}0& 0& 1\\ x& -1& x\\ 2x(x-1)& -2(x-1)& x(x-1)\end{array}\right|$

$\Rightarrow f\left(x\right)=x(x+1)\left|\begin{array}{cc}x& -1\\ 2x(x-1)& -2(x-1)\end{array}\right|=0$

$\therefore f\left(100\right)=0$