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(C) We know that $A 	imes A^{-1} = I$,where $I$ is the identity matrix of order $3 	imes 3$.Given $A = \begin{bmatrix} 1 & 2 & 2 \ 2 & 1 & 2 \ 2 &

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$2$

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(C) We know that $A 	imes A^{-1} = I$,where $I$ is the identity matrix of order $3 	imes 3$.Given $A = \begin{bmatrix} 1 & 2 & 2 \ 2 & 1 & 2 \ 2 & 2 & 1 \end{bmatrix}$ and $A^{-1} = \frac{1}{5} \begin{bmatrix} -3 & 2 & 2 \ 2 & -3 & \alpha \ 2 & 2 & -3 \end{bmatrix}$.Multiplying the second row of $A$ with the third column of $A^{-1}$ must result in the element at position $(2, 3)$ of the identity matrix,which is $0$.So,$\frac{1}{5} [ (2 	imes 2) + (1 	imes \alpha) + (2 	imes -3) ] = 0$.$4 + \alpha - 6 = 0$.$\alpha - 2 = 0$.$\alpha = 2$.

If matrix $A = \begin{bmatrix} 1 & 2 & 2 \\ 2 & 1 & 2 \\ 2 & 2 & 1 \end{bmatrix}$,and the inverse of matrix $A$ is given by $A^{-1} = \frac{1}{5} \begin{bmatrix} -3 & 2 & 2 \\ 2 & -3 & \alpha \\ 2 & 2 & -3 \end{bmatrix}$,then find the value of $\alpha$.

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