Question

If $\left[\begin{array}{cc}a& b\\ c& 1-a\end{array}\right]$ is an idempotent matrix. Find the value of f(a), Where $f\left(x\right)=x-x^2.$ when bc=1/4. Hence otherwise evaluate a.

Answer 1

Let $A=\left[\begin{array}{cc}a& b\\ c& 1-a\end{array}\right]$

$\sin a$ A is an impotent matrix.

$A^2=A$

i.e., $\left[\begin{array}{cc}a& b\\ c& 1-a\end{array}\right]\left[\begin{array}{cc}a& b\\ c& 1-a\end{array}\right]=\left[\begin{array}{cc}a& b\\ c& 1-a\end{array}\right]$

$\Rightarrow \left[\begin{array}{cc}{a}^2+bc& ab+b-ab\\ ac+c-ac& bc+(1-a{)}^2\end{array}\right]=\left[\begin{array}{cc}a& b\\ c& 1-a\end{array}\right]$

$\Rightarrow \left[\begin{array}{cc}{a}^2+bc& b\\ c& bc+(1-a{)}^2\end{array}\right]=\left[\begin{array}{cc}a& b\\ c& 1-a\end{array}\right]$

$\Rightarrow a^2+bc=a$ …………$\left(1\right)$ and $bc+(1-a{)}^2=1-a$ ………$\left(2\right)$

$\Rightarrow a^2-a+bc=0$

$\Rightarrow a^2-a+\frac{1}{4}=0$ $[\because bc=\frac{1}{4}]$

$\Rightarrow 4a^2-4a+1=0$

$\Rightarrow 4a^2-2a-2a+1=0$

$\Rightarrow a(2a-1)-1(2a-1)=0$

$\Rightarrow (2a-1)(2a-1)=0$

$\Rightarrow (2a-1{)}^2=0$

$2a-1=0$

$\Rightarrow a=\frac{1}{2}$

$\therefore f\left(a\right)=f\left(\frac{1}{4}\right)=\frac{1}{4}-{\left(\frac{1}{4}\right)}^2$

$=\frac{1}{4}-\frac{1}{16}$

$=\frac{4-1}{16}$

$=\frac{3}{16}$.