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

Explanation of the correct option:

Finding the value of $ad-bc$:

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

Also $AB=B$

$$\begin{gathered} \left[\begin{array}{cc}a& b\\ c& d\end{array}\right]\left[\begin{array}{c}\alpha \\ \beta \end{array}\right]=\left[\begin{array}{c}\alpha \\ \beta \end{array}\right] \\ a\alpha +b\beta =\alpha ......\left(1\right) \\ c\alpha +d\beta =\beta ......\left(2\right) \\ \text{Simplifying equation}\left(1\right)\text{we get} \\ \alpha (a–1)=–b\beta \\ \therefore \frac{\alpha }{\beta }=\frac{–b}{(a–1)} \\ \text{and by simplifying equation}\left(2\right)\text{we get} \\ c\alpha =\beta (1–d) \\ \therefore \frac{\alpha }{\beta }=\frac{(1–d)}{c} \\ \frac{–b}{(a–1)}=\frac{(1–d)}{c} \\ –bc=(a–1)(1–d) \\ –bc=a–ad–1+d \\ ad–bc=a+d–1 \\ =2021–1[\because a+d=2021] \\ ad–bc=2020 \end{gathered}$$

Hence, the value of $ad-bc=$ $2020$.