Question

Let $A=\left[\begin{array}{cc}1& 2\\ -5& 1\end{array}\right]$and$A^1=xA+yI$, then the values of $x$ and $y$ respectively are

• A. $\frac{-1}{11},\frac{2}{11}$
• B. $\frac{-1}{11},\frac{-2}{11}$
• C. $\frac{1}{11},\frac{2}{11}$
• D. $\frac{1}{11},\frac{-2}{11}$

Answer 1

Answer: (A)

Given,$A=\left[\begin{array}{cc}1& 2\\ -5& 1\end{array}\right]$
we have $A=IA$
$\therefore \left[\begin{array}{cc}1& 2\\ -5& 1\end{array}\right]=\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]A$
Applying $R_2\to R_2+5R_1$, we get
$\left[\begin{array}{cc}1& 2\\ 0& 11\end{array}\right]=\left[\begin{array}{cc}1& 0\\ 5& 1\end{array}\right]A$
Applying $R_2\to \frac{1}{11}{R}_2$, we get
$\left[\begin{array}{cc}1& 2\\ 0& 1\end{array}\right]=\left[\begin{array}{cc}1& 0\\ \frac{5}{11}& \frac{1}{11}\end{array}\right]A$
Applying $R_1\to R_1+2R_2$, we get
$\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]=\left[\begin{array}{cc}\frac{1}{11}& -\frac{2}{11}\\ \frac{5}{11}& \frac{1}{11}\end{array}\right]A$
$\therefore A^{-1}=\frac{1}{11}\left[\begin{array}{cc}1& -2\\ 5& 1\end{array}\right]$
Also, $A^{-1}=xA+yI$
$\Rightarrow \frac{1}{11}\left[\begin{array}{cc}1& -2\\ 5& 1\end{array}\right]=\left[\begin{array}{cc}x& 2x\\ -5x& x\end{array}\right]+\left[\begin{array}{cc}y& 0\\ 0& y\end{array}\right]$
$\Rightarrow \frac{1}{11}\left[\begin{array}{cc}1& -2\\ 5& 1\end{array}\right]=\left[\begin{array}{cc}x+y& 2x\\ -5x& x+y\end{array}\right]$
$\Rightarrow x+y=\frac{1}{11},2x=-\frac{2}{11}$
$\Rightarrow x=-\frac{1}{11},y=\frac{2}{11}$