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

Options

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

Answer

To solve the problem, we need to find the value of \( x \) given the matrix \( A \) and its inverse \( A^{-1} \).

### Step 1: Write down the matrices
We have:
\[ A = \begin{pmatrix} 2x & 0 \\ x & x \end{pmatrix}, \quad A^{-1} = \begin{pmatrix} 1 & 0 \\ -1 & 2 \end{pmatrix} \]

### Step 2: Use the property of inverses
The property of matrix inverses states that:
\[ A \cdot A^{-1} = I \]
where \( I \) is the identity matrix:
\[ I = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} \]

### Step 3: Multiply the matrices
Now we multiply \( A \) and \( A^{-1} \):
\[ A \cdot A^{-1} = \begin{pmatrix} 2x & 0 \\ x & x \end{pmatrix} \cdot \begin{pmatrix} 1 & 0 \\ -1 & 2 \end{pmatrix} \]

Calculating the product:
- The element at (1,1):
\[ (2x \cdot 1) + (0 \cdot -1) = 2x \]
- The element at (1,2):
\[ (2x \cdot 0) + (0 \cdot 2) = 0 \]
- The element at (2,1):
\[ (x \cdot 1) + (x \cdot -1) = x - x = 0 \]
- The element at (2,2):
\[ (x \cdot 0) + (x \cdot 2) = 2x \]

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

### Step 4: Set the product equal to the identity matrix
Now we set this equal to the identity matrix:
\[ \begin{pmatrix} 2x & 0 \\ 0 & 2x \end{pmatrix} = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} \]

### Step 5: Compare corresponding elements
From the equality of matrices, we can compare the corresponding elements:
1. From the (1,1) position:
\[ 2x = 1 \]
2. From the (2,2) position:
\[ 2x = 1 \]

### Step 6: Solve for \( x \)
Now, we can solve for \( x \):
\[ 2x = 1 \implies x = \frac{1}{2} \]

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