Question

Let A =$\left(\begin{array}{cc}p& q\\ r& s\end{array}\right]$ and B =$\left(\begin{array}{c}a\\ b\end{array}\right]$$\ne \left(\begin{array}{c}0\\ 0\end{array}\right]$. Such that AB = B and p+s = 2, then find the value of (ps-qr).

• A. 5
• B. 1
• C. 2
• D. 3

Answer 1

The correct option is B 1

Let A=$\left(\begin{array}{cc}p& q\\ r& s\end{array}\right]\mathrm{and}B=\left(\begin{array}{c}a\\ b\end{array}\right]$
Given AB = B. Thus,
$\left(\begin{array}{cc}p& q\\ r& s\end{array}\right]\left(\begin{array}{c}a\\ b\end{array}\right]=\left(\begin{array}{c}a\\ b\end{array}\right]\mathrm{or}\left(\begin{array}{c}\mathrm{pa}+\mathrm{qb}\\ \mathrm{ra}+\mathrm{sb}\end{array}\right]=\left(\begin{array}{c}a\\ b\end{array}\right]\mathrm{or}\mathrm{pa}+\mathrm{qb}=a-\left(1\right)\mathrm{ra}+\mathrm{sb}=b-\left(2\right)\mathrm{Eliminating}a,b\mathrm{from}\left(1\right)\mathrm{and}\left(2\right),\mathrm{we}\mathrm{have}-\left(p-1\right)a+\mathrm{qb}=0\mathrm{ra}+\left(s-1\right)b=0\mathrm{or}\left(\begin{array}{cc}p-1& q\\ r& s-1\end{array}\right]\left(\begin{array}{c}a\\ b\end{array}\right]=\left(\begin{array}{c}0\\ 0\end{array}\right]\left(\begin{array}{cc}p-1& q\\ r& s-1\end{array}\right|=0\mathrm{or}\left(p-1\right)\left(s-1\right)-\mathrm{qr}=0\mathrm{or}\mathrm{ps}-\left(p+s\right)+1-\mathrm{qr}=0\mathrm{or}\mathrm{ps}-\mathrm{qr}=\left(p+s\right)-1=2-1=1$