Question

Let $A=\left[\begin{array}{cc}a& b\\ c& d\end{array}\right]$ and $P=\left[\begin{array}{c}p\\ q\end{array}\right]\ne \left[\begin{array}{c}0\\ 0\end{array}\right]$. Such that $AP=P$ and $a+d=5050$. Let the value of $(ad-bc)=k$. Find sum of digits of $k$ ?

Answer 1

Given $A=\left[\begin{array}{cc}a& b\\ c& d\end{array}\right]$ and

$P=\left[\begin{array}{c}p\\ q\end{array}\right]\ne \left[\begin{array}{c}0\\ 0\end{array}\right]$. Such that $AP=P$ and $a+d=5050$ and $(ad-bc)=k$.

$AP=P$

$\Rightarrow \left[\begin{array}{cc}a& b\\ c& d\end{array}\right]\left[\begin{array}{c}p\\ q\end{array}\right]=\left[\begin{array}{c}p\\ q\end{array}\right]$

$\Rightarrow (a-1)p+bq=0$ and $cp+(d-1)q=0$

given $p,q\ne 0$

$\Rightarrow$ above system of equations has non-trivial solution.

$\Rightarrow \left|\begin{array}{cc}a-1& b\\ c& d-1\end{array}\right|=0$

$\Rightarrow (a-1)(d-1)-bc=0$

$\Rightarrow k=ad-bc=a+d-1=5049$

$\therefore$ Sum of digits of $k$ is $18$.