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Mathematics Question on Matrices

Let A=[12 −51]A=\begin{bmatrix}1&2\\\ -5&1\end{bmatrix} andA1=xA+yI A^{1} = xA + yI, then the values of xx and yy respectively are

A

−111,211\frac{-1}{11},\frac{2}{11}

B

−111,−211\frac{-1}{11},\frac{-2}{11}

C

111,211\frac{1}{11},\frac{2}{11}

D

111,−211\frac{1}{11},\frac{-2}{11}

Answer

−111,211\frac{-1}{11},\frac{2}{11}

Explanation

Solution

Given,A=[12 −51] A=\begin{bmatrix}1&2\\\ -5&1\end{bmatrix} we have A=IAA=IA ∴[12 −51]=[10 01]A\therefore\begin{bmatrix}1&2\\\ -5&1\end{bmatrix}=\begin{bmatrix}1&0\\\ 0&1\end{bmatrix}A Applying R2→R2+5R1R_{2}\rightarrow R_{2}+ 5R_{1}, we get [12 011]=[10 51]A\begin{bmatrix}1&2\\\ 0&11\end{bmatrix}=\begin{bmatrix}1&0\\\ 5&1\end{bmatrix}A Applying R2→111R2R_{2}\rightarrow \frac{1}{11} R_{2}, we get [12 01]=[10 511111]A\begin{bmatrix}1&2\\\ 0&1\end{bmatrix}=\begin{bmatrix}1&0\\\ \frac{5}{11}&\frac{1}{11}\end{bmatrix}A Applying R1→R1+2R2R_{1}\rightarrow R_{1}+ 2R_{2}, we get [10 01]=[111−211 511111]A\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 ∴A−1=111[1−2 51]\therefore A^{-1}=\frac{1}{11}\begin{bmatrix}1&-2\\\ 5&1\end{bmatrix} Also, A−1=xA+yIA^{-1 }=xA+yI ⇒111[1−2 51]=[x2x −5xx]+[y0 0y]\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} ⇒111[1−2 51]=[x+y2x −5xx+y]\Rightarrow \frac{1}{11}\begin{bmatrix}1&-2\\\ 5&1\end{bmatrix}=\begin{bmatrix}x+y&2x\\\ -5x&x+y\end{bmatrix} ⇒x+y=111,2x=−211\Rightarrow x+y=\frac{1}{11}, 2x = -\frac{2}{11} ⇒x=−111,y=211\Rightarrow x = -\frac{1}{11}, y=\frac{2}{11}