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

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(D) Given the matrix $A = \begin{bmatrix} 2 & -1 \ -1 & 2 \end{bmatrix}$.The determinant of $A$ is $|A| = (2)(2) - (-1)(-1) = 4 - 1 = 3$.The adjoint

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$\frac{1}{3} \begin{bmatrix} 2 & 1 \ 1 & 2 \end{bmatrix}$

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(D) Given the matrix $A = \begin{bmatrix} 2 & -1 \ -1 & 2 \end{bmatrix}$.The determinant of $A$ is $|A| = (2)(2) - (-1)(-1) = 4 - 1 = 3$.The adjoint of $A$ is $	ext{adj}(A) = \begin{bmatrix} 2 & 1 \ 1 & 2 \end{bmatrix}$.The inverse of a matrix is given by $A^{-1} = \frac{1}{|A|} 	ext{adj}(A)$.Substituting the values,we get $A^{-1} = \frac{1}{3} \begin{bmatrix} 2 & 1 \ 1 & 2 \end{bmatrix}$.Alternatively,using the characteristic equation $A^{2} - 4A + 3I = 0$,multiply by $A^{-1}$:$A - 4I + 3A^{-1} = 0 \Rightarrow 3A^{-1} = 4I - A$.$3A^{-1} = 4 \begin{bmatrix} 1 & 0 \ 0 & 1 \end{bmatrix} - \begin{bmatrix} 2 & -1 \ -1 & 2 \end{bmatrix} = \begin{bmatrix} 2 & 1 \ 1 & 2 \end{bmatrix}$.Thus,$A^{-1} = \frac{1}{3} \begin{bmatrix} 2 & 1 \ 1 & 2 \end{bmatrix}$.

If $A = \begin{bmatrix} 2 & -1 \\ -1 & 2 \end{bmatrix}$,such that $A^{2} - 4A + 3I = 0$,then $A^{-1} =$

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