If A = begin{bmatrix} 2x & 0 x & x end{bmatr | vedclass

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(D) We know that for any invertible matrix $A$,the product $A \cdot A^{-1} = I$,where $I$ is the identity matrix of the same order.Given $A = \begin{b

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$\frac{1}{2}$

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(D) We know that for any invertible matrix $A$,the product $A \cdot A^{-1} = I$,where $I$ is the identity matrix of the same order.Given $A = \begin{bmatrix} 2x & 0 \ x & x \end{bmatrix}$ and $A^{-1} = \begin{bmatrix} 1 & 0 \ -1 & 2 \end{bmatrix}$.Therefore,$A \cdot A^{-1} = \begin{bmatrix} 2x & 0 \ x & x \end{bmatrix} \begin{bmatrix} 1 & 0 \ -1 & 2 \end{bmatrix} = \begin{bmatrix} 1 & 0 \ 0 & 1 \end{bmatrix}$.Performing matrix multiplication:$\begin{bmatrix} (2x)(1) + (0)(-1) & (2x)(0) + (0)(2) \ (x)(1) + (x)(-1) & (x)(0) + (x)(2) \end{bmatrix} = \begin{bmatrix} 2x & 0 \ 0 & 2x \end{bmatrix}$.Equating this to the identity matrix $\begin{bmatrix} 1 & 0 \ 0 & 1 \end{bmatrix}$,we get:$2x = 1$.Solving for $x$,we get $x = \frac{1}{2}$.Thus,the correct option is $D$.

If $A = \begin{bmatrix} 2x & 0 \\ x & x \end{bmatrix}$ and $A^{-1} = \begin{bmatrix} 1 & 0 \\ -1 & 2 \end{bmatrix}$,then $x =$ . . . . . . .

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