Question

Let $A=\begin{bmatrix}1&2\\ -5&1\end{bmatrix} $and$ 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

The Correct Option is A

Given,$ A=\begin{bmatrix}1&2\\ -5&1\end{bmatrix}$ we have $A=IA$ $\therefore\begin{bmatrix}1&2\\ -5&1\end{bmatrix}=\begin{bmatrix}1&0\\ 0&1\end{bmatrix}A$ Applying $R_{2}\rightarrow R_{2}+ 5R_{1}$, we get $\begin{bmatrix}1&2\\ 0&11\end{bmatrix}=\begin{bmatrix}1&0\\ 5&1\end{bmatrix}A$ Applying $R_{2}\rightarrow \frac{1}{11} R_{2}$, we get $\begin{bmatrix}1&2\\ 0&1\end{bmatrix}=\begin{bmatrix}1&0\\ \frac{5}{11}&\frac{1}{11}\end{bmatrix}A$ Applying $R_{1}\rightarrow R_{1}+ 2R_{2}$, we get $\begin{bmatrix}1&0\\ 0&1\end{bmatrix}=\begin{bmatrix}\frac{1}{11}&-\frac{2}{11}\\ \frac{5}{11}&\frac{1}{11}\end{bmatrix}A$ $\therefore A^{-1}=\frac{1}{11}\begin{bmatrix}1&-2\\ 5&1\end{bmatrix}$ Also, $A^{-1 }=xA+yI $ $\Rightarrow \frac{1}{11}\begin{bmatrix}1&-2\\ 5&1\end{bmatrix}=\begin{bmatrix}x&2x\\ -5x&x\end{bmatrix}+\begin{bmatrix}y&0\\ 0&y\end{bmatrix}$ $\Rightarrow \frac{1}{11}\begin{bmatrix}1&-2\\ 5&1\end{bmatrix}=\begin{bmatrix}x+y&2x\\ -5x&x+y\end{bmatrix}$ $\Rightarrow x+y=\frac{1}{11}, 2x = -\frac{2}{11}$ $\Rightarrow x = -\frac{1}{11}, y=\frac{2}{11}$