Question

Suppose $a \in R$. Let $f(x)=\begin{vmatrix} x+a & x & x \\ x & x+a & x \\ x & x & x+a \end{vmatrix}$
Then $f(2 x)-f(x)$ is equal to

• A. $3 x a^2$
• B. $3 x^2 a$
• C. $x a^2$
• D. $a^2 x$

Answer 1

Answer: (A)

Write $f(x)=\Delta_1+x \Delta_2$ where
$\Delta_1 =\begin{vmatrix} a & x & x \\ 0 & x+a & x \\ 0 & x & x+a \end{vmatrix}=a\left[(x+a)^2-x^2\right]$
$ =2 x a^2+a^3$
and $\Delta_2=\begin{vmatrix}1 & x & x \\ 1 & x+a & x \\ 1 & x & x+a\end{vmatrix}$
$=\begin{vmatrix} 1 & x & x \\ 0 & a & 0 \\ 0 & 0 & a \end{vmatrix}=a^2$
Thus $f(x)=2 x a^2+a^3+x a^2 $
$=3 x a^2+a^3 $
$\therefore f(2 x)-f(x)=3 x a^2$