If `A=[(2x,0),(x,x)]` and `A^(-1)=[(1,0),(-1,2)]` then x equals to

Options

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

Explanation

To find the value of \( x \) in the given matrices \( A \) and \( A^{-1} \), we will follow these steps:

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

### Step 1: Multiply \( A \) and \( A^{-1} \)
We know that the product of a matrix and its inverse is the identity matrix \( I \):
\[ A \cdot A^{-1} = I \]

The identity matrix \( I \) for a \( 2 \times 2 \) matrix is:
\[ I = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} \]

### Step 2: Calculate the product \( A \cdot A^{-1} \)
Now we will calculate 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:
- First row, first column:
\[ (2x \cdot 1) + (0 \cdot -1) = 2x \]
- First row, second column:
\[ (2x \cdot 0) + (0 \cdot 2) = 0 \]
- Second row, first column:
\[ (x \cdot 1) + (x \cdot -1) = x - x = 0 \]
- Second row, second column:
\[ (x \cdot 0) + (x \cdot 2) = 2x \]

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

### Step 3: Set the product equal to the identity matrix
Now we set the product equal to the identity matrix:
\[ \begin{pmatrix} 2x & 0 \\ 0 & 2x \end{pmatrix} = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} \]

### Step 4: Solve for \( x \)
From the equality of the matrices, we can compare corresponding elements:
1. From the first element:
\[ 2x = 1 \implies x = \frac{1}{2} \]
2. From the second element:
\[ 2x = 1 \implies x = \frac{1}{2} \]

Both comparisons give the same result.

### Final Answer
Thus, the value of \( x \) is:
\[ \boxed{\frac{1}{2}} \]