Question

Let $A=\left[\begin{array}{cc}a& b\\ c& d\end{array}\right]$ and $B=\left[\begin{array}{c}p\\ q\end{array}\right]\ne \left[\begin{array}{c}0\\ 0\end{array}\right]$. If $AB=B$ and $a+d=2$, then the value of $ad-bc$ is

Answer 1

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

$\Rightarrow \left[\begin{array}{c}ap+bq\\ cp+dq\end{array}\right]=\left[\begin{array}{c}p\\ q\end{array}\right]$

$\Rightarrow ap+bq=p$
and $cp+dq=q$
i.e., $(a-1)p+bq=0$
and $cp+(d-1)q=0$
Since $\left[\begin{array}{c}p\\ q\end{array}\right]\ne \left[\begin{array}{c}0\\ 0\end{array}\right]$, the above system of equations have non-trivial solution.
So, $\left|\begin{array}{cc}a-1& b\\ c& d-1\end{array}\right|=0$

$\Rightarrow (a-1)(d-1)-bc=0$
$\Rightarrow ad-(a+d)+1-bc=0$
$\Rightarrow ad-bc=a+d-1$
$\Rightarrow ad-bc=2-1=1$