Question

If $f\left(x\right)=\left|\begin{array}{ccc}a& -1& 0\\ ax& a& -1\\ a{x}^2& ax& a\end{array}\right|$, by using properties of determinants, find the value $f\left(2x\right)-f\left(x\right)$.

Answer 1

Given,

$f\left(x\right)=\left|\begin{array}{ccc}a& -1& 0\\ ax& a& -1\\ a{x}^2& ax& a\end{array}\right|$

Taking a common from $C_1$, we get,

$f\left(x\right)=a\left|\begin{array}{ccc}1& -1& 0\\ x& a& -1\\ x^2& ax& a\end{array}\right|$

$f\left(x\right)=a\left|\begin{array}{ccc}0& -1& 0\\ x+a& a& -1\\ x^2+ax& ax& a\end{array}\right|$ $(C_1\to C_1+C_2)$

Now, on expanding through $R_1$, we get,

$f\left(x\right)=a[ax+a^2+x^2+ax]$

$f\left(x\right)=a[x^2+2ax+a^2]$

$f\left(x\right)=a(x+a{)}^2$

Therefore, $f\left(2x\right)=a(2x+a{)}^2$

Now,

$f\left(2x\right)-f\left(x\right)=a(2x+a{)}^2-a(x+a{)}^2$

$=a\left[\right(2x+a+x+a\left)\right(2x+a-x-a\left)\right]$

$=a\left[\right(3x+2a\left)\right(x\left)\right]$

$=x(3x+2a)a$