Question

Let $A=\left[\begin{array}{cc}a& b\\ c& d\end{array}\right]$ and $B\left[\begin{array}{c}\alpha \\ \beta \end{array}\right]$$\ne \left[\begin{array}{c}0\\ 0\end{array}\right]$ such that $AB=B$ and $a+d=2021,$ then the value of $ad-bc$ is equal to

Answer 1

Given : $A=\left[\begin{array}{cc}a& b\\ c& d\end{array}\right],B=\left[\begin{array}{c}\alpha \\ \beta \end{array}\right]$
Now, we know
$AB=B$
$\left[\begin{array}{cc}a& b\\ c& d\end{array}\right]\left[\begin{array}{c}\alpha \\ \beta \end{array}\right]=\left[\begin{array}{c}\alpha \\ \beta \end{array}\right]a\alpha +b\beta =\alpha \cdots \left(1\right)c\alpha +d\beta =\beta \cdots \left(2\right)$
From equation $\left(1\right)$ and $\left(2\right)$, we get
$\frac{\alpha }{\beta }=\frac{-b}{a-1}=\frac{1-d}{c}\Rightarrow \frac{-b}{a-1}=\frac{1-d}{c}\Rightarrow -bc=a-ad-1+d\Rightarrow ad-bc=a+d-1\therefore ad-bc=2021-1=2020$