Let A = begin{bmatrix} 1 & 2 -5 & 1 end{bmat | vedclass

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(A) Given $A = \begin{bmatrix} 1 & 2 \ -5 & 1 \end{bmatrix}$.First,calculate the determinant $|A| = (1)(1) - (2)(-5) = 1 + 10 = 11$.Since $|A| 
eq 0

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$x = \frac{-1}{11}, y = \frac{2}{11}$

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(A) Given $A = \begin{bmatrix} 1 & 2 \ -5 & 1 \end{bmatrix}$.First,calculate the determinant $|A| = (1)(1) - (2)(-5) = 1 + 10 = 11$.Since $|A| 
eq 0$,$A^{-1}$ exists.The inverse is given by $A^{-1} = \frac{1}{|A|} 	ext{adj}(A) = \frac{1}{11} \begin{bmatrix} 1 & -2 \ 5 & 1 \end{bmatrix} = \begin{bmatrix} \frac{1}{11} & \frac{-2}{11} \ \frac{5}{11} & \frac{1}{11} \end{bmatrix}$.We are given $A^{-1} = xA + yI_2$.Substituting the matrices:$\begin{bmatrix} \frac{1}{11} & \frac{-2}{11} \ \frac{5}{11} & \frac{1}{11} \end{bmatrix} = x \begin{bmatrix} 1 & 2 \ -5 & 1 \end{bmatrix} + y \begin{bmatrix} 1 & 0 \ 0 & 1 \end{bmatrix} = \begin{bmatrix} x+y & 2x \ -5x & x+y \end{bmatrix}$.Comparing the corresponding elements:$2x = \frac{-2}{11} \implies x = \frac{-1}{11}$.$x + y = \frac{1}{11} \implies \frac{-1}{11} + y = \frac{1}{11} \implies y = \frac{2}{11}$.Thus,$x = \frac{-1}{11}$ and $y = \frac{2}{11}$.

Let $A = \begin{bmatrix} 1 & 2 \\ -5 & 1 \end{bmatrix}$ and $A^{-1} = xA + yI_2$,(where $I_2$ is the unit matrix of order $2$),then

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