Question

Let $A$ be a square matrix of order $2$ such that $A+adjA=O.$ If $det\left(A\right)=r$ and $f\left(r\right)=det(A+det(A)\cdot adjA)$ and local maximum value of $f\left(r\right)$ is $\frac{a}{b},$ where $a,b\in N,$ then the value of $81\left(\frac{a}{b}\right)$ is

Answer 1

Let $A=\left[\begin{array}{cc}a& b\\ c& d\end{array}\right]$
Then, $adjA=\left[\begin{array}{cc}d& -b\\ -c& a\end{array}\right]$
Given, $A+adjA=O$
$\Rightarrow \left[\begin{array}{cc}a& b\\ c& d\end{array}\right]+\left[\begin{array}{cc}d& -b\\ -c& a\end{array}\right]=\left[\begin{array}{cc}0& 0\\ 0& 0\end{array}\right]$
$\Rightarrow \left[\begin{array}{cc}a+d& 0\\ 0& a+d\end{array}\right]=\left[\begin{array}{cc}0& 0\\ 0& 0\end{array}\right]$
$\Rightarrow a+d=0\Rightarrow d=-a$
$\therefore A=\left[\begin{array}{cc}a& b\\ c& -a\end{array}\right]$
$\Rightarrow det\left(A\right)=-a^2-bc=r$

Also, $f\left(r\right)=det(A+det(A)\cdot adjA)$
$\Rightarrow f\left(r\right)=det(A+r\cdot adjA)$
We have
$A+r\cdot adjA=\left[\begin{array}{cc}a& b\\ c& -a\end{array}\right]+r\left[\begin{array}{cc}-a& -b\\ -c& a\end{array}\right]$
$\Rightarrow A+r\cdot adjA=\left[\begin{array}{cc}a(1-r)& b(1-r)\\ c(1-r)& -a(1-r)\end{array}\right]$

Now, $f\left(r\right)=det(A+r\cdot adjA)$
$\Rightarrow f\left(r\right)=\left|\begin{array}{cc}a(1-r)& b(1-r)\\ c(1-r)& -a(1-r)\end{array}\right|$
$\Rightarrow f\left(r\right)=(1-r{)}^2(-a^2-bc)$
$\therefore f\left(r\right)=r(1-r{)}^2$
Differentiation w.r.t. $x,$ we get
$f^{\prime }\left(r\right)=(1-r{)}^2\cdot 1-r\cdot 2(1-r)$
$\Rightarrow f^{\prime }\left(r\right)=(r-1)(3r-1)$
Sign of $f^{\prime }\left(r\right):$
$\therefore r=\frac{1}{3}$ is point of local maximum
and local maximum value of $f\left(r\right)$ is $\frac{1}{3}\times \frac{4}{9}=\frac{4}{27}=\frac{a}{b}$