If A=left[begin{array}{cc}1 & 2 -5 & 1end{ar | vedclass

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(B) First,find the determinant of $A$: $|A| = (1)(1) - (2)(-5) = 1 + 10 = 11$.Next,find the adjoint of $A$: $	ext{adj } A = \left[\begin{array}{cc}1

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$\frac{-1}{11}, \frac{2}{11}$

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(B) First,find the determinant of $A$: $|A| = (1)(1) - (2)(-5) = 1 + 10 = 11$.Next,find the adjoint of $A$: $	ext{adj } A = \left[\begin{array}{cc}1 & -2 \ 5 & 1\end{array}ight]$.Therefore,$A^{-1} = \frac{1}{|A|} 	ext{adj } A = \frac{1}{11} \left[\begin{array}{cc}1 & -2 \ 5 & 1\end{array}ight]$.Given the equation $A^{-1} = xA + yI$,we substitute the matrices:$\frac{1}{11} \left[\begin{array}{cc}1 & -2 \ 5 & 1\end{array}ight] = x \left[\begin{array}{cc}1 & 2 \ -5 & 1\end{array}ight] + y \left[\begin{array}{cc}1 & 0 \ 0 & 1\end{array}ight]$.$\left[\begin{array}{cc}1/11 & -2/11 \ 5/11 & 1/11\end{array}ight] = \left[\begin{array}{cc}x+y & 2x \ -5x & x+y\end{array}ight]$.Comparing the elements,we get $2x = -2/11$,which implies $x = -1/11$.Also,$x+y = 1/11$. Substituting $x = -1/11$,we get $-1/11 + y = 1/11$,so $y = 2/11$.Thus,the values are $x = -1/11$ and $y = 2/11$.

If $A=\left[\begin{array}{cc}1 & 2 \\ -5 & 1\end{array}\right]$ and $A^{-1}=x A+y I$,where $I$ is the unit matrix of order $2$,then the values of $x$ and $y$ are respectively:

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