Question

 

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

 

Answer 1

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

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

Applying C₂→C₂+C₁, we get

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

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

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

Also,

${\displaystyle f\left(2x\right)=\left|\begin{array}{ccc}a& -1& 0\\ 2ax& a& -1\\ 4a{x}^2& 2ax& a\end{array}\right|}$

${\displaystyle f\left(2x\right)=a\left|\begin{array}{ccc}1& -1& 0\\ 2x& a& -1\\ 4x^2& 2ax& a\end{array}\right|}$

Applying C₂→C₂+C₁, we get

${\displaystyle f\left(2x\right)=a\left|\begin{array}{ccc}1& 0& 0\\ 2x& 2x+a& -1\\ 4x^2& 4x^2+2ax& a\end{array}\right|}$

${\displaystyle \Rightarrow f\left(2x\right)=a\{a(2x+a)+4x^2+2ax\}}$

${\displaystyle \Rightarrow f\left(2x\right)=a(4x^2+a^2+4ax)}$

${\displaystyle \therefore f\left(2x\right)-f\left(x\right)=a(4x^2+a^2+4ax-a^2-2ax-x^2) }$       

${\displaystyle =ax(3x+2a)}$