If `A=[{:(2x,0),(x,x):}]and A^(-1)=[{:(1,0),(-1,2):}]`, then what is the value of x ?

Options

A. `-(1)/(2)`
B. `(1)/(2)`
C. 1
D. 2

Explanation

To find the value of \( x \) given the matrices \( A \) and \( A^{-1} \), we start by using the property that the product of a matrix and its inverse is the identity matrix.

Given:
\[ A = \begin{pmatrix} 2x & 0 \\ x & x \end{pmatrix}, \quad A^{-1} = \begin{pmatrix} 1 & 0 \\ -1 & 2 \end{pmatrix} \]

We know:
\[ A \cdot A^{-1} = I \]
where \( I \) is the identity matrix:
\[ I = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} \]

Now, we can compute the product \( A \cdot A^{-1} \):
\[ A \cdot A^{-1} = \begin{pmatrix} 2x & 0 \\ x & x \end{pmatrix} \cdot \begin{pmatrix} 1 & 0 \\ -1 & 2 \end{pmatrix} \]

Calculating the elements of the resulting matrix:
1. First row, first column:
\[ (2x \cdot 1) + (0 \cdot -1) = 2x \]
2. First row, second column:
\[ (2x \cdot 0) + (0 \cdot 2) = 0 \]
3. Second row, first column:
\[ (x \cdot 1) + (x \cdot -1) = x - x = 0 \]
4. Second row, second column:
\[ (x \cdot 0) + (x \cdot 2) = 2x \]

Thus, we have:
\[ A \cdot A^{-1} = \begin{pmatrix} 2x & 0 \\ 0 & 2x \end{pmatrix} \]

Setting this equal to the identity matrix:
\[ \begin{pmatrix} 2x & 0 \\ 0 & 2x \end{pmatrix} = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} \]

From this, we can equate the corresponding elements:
1. From the first row, first column:
\[ 2x = 1 \implies x = \frac{1}{2} \]
2. From the second row, second column:
\[ 2x = 1 \implies x = \frac{1}{2} \]

Both equations give us the same result. Therefore, the value of \( x \) is:
\[ \boxed{\frac{1}{2}} \]