If A = begin{bmatrix} x & 1 1 & 0 end{bmatri | vedclass

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(A) Given $A = \begin{bmatrix} x & 1 \ 1 & 0 \end{bmatrix}$.We know that $A^{-1} = \frac{1}{|A|} 	ext{adj}(A)$.First,calculate the determinant $|A|

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(A) Given $A = \begin{bmatrix} x & 1 \ 1 & 0 \end{bmatrix}$.We know that $A^{-1} = \frac{1}{|A|} 	ext{adj}(A)$.First,calculate the determinant $|A| = (x 	imes 0) - (1 	imes 1) = -1$.Next,find the adjoint of $A$ by swapping the diagonal elements and changing the signs of the off-diagonal elements:$	ext{adj}(A) = \begin{bmatrix} 0 & -1 \ -1 & x \end{bmatrix}$.Therefore,$A^{-1} = \frac{1}{-1} \begin{bmatrix} 0 & -1 \ -1 & x \end{bmatrix} = \begin{bmatrix} 0 & 1 \ 1 & -x \end{bmatrix}$.Given $A = A^{-1}$,we equate the matrices:$\begin{bmatrix} x & 1 \ 1 & 0 \end{bmatrix} = \begin{bmatrix} 0 & 1 \ 1 & -x \end{bmatrix}$.Comparing the corresponding elements,we get $x = 0$ and $0 = -x$,which implies $x = 0$.

If $A = \begin{bmatrix} x & 1 \\ 1 & 0 \end{bmatrix}$ and $A = A^{-1}$,then $x = \dots$

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