If `A=[(2,1),(0,x)]` and `A^(-1)=[(1/2,1/6),(0,1/x)]`, then the value of `x` is equal to

Options

A. `-3`
B. 3
C. -2
D. 6

Answer

To find the value of \( x \) given the matrix \( A = \begin{pmatrix} 2 & 1 \\ 0 & x \end{pmatrix} \) and its inverse \( A^{-1} = \begin{pmatrix} \frac{1}{2} & \frac{1}{6} \\ 0 & \frac{1}{x} \end{pmatrix} \), we will follow these steps:

### Step 1: Calculate the determinant of matrix \( A \)

The determinant of a 2x2 matrix \( \begin{pmatrix} a & b \\ c & d \end{pmatrix} \) is given by \( ad - bc \).

For matrix \( A \):
\[ \text{det}(A) = (2)(x) - (1)(0) = 2x \]

### Step 2: Use the formula for the inverse of a matrix

The inverse of a 2x2 matrix \( A \) can be calculated using the formula:
\[ A^{-1} = \frac{1}{\text{det}(A)} \begin{pmatrix} d & -b \\ -c & a \end{pmatrix} \]

Substituting the values from matrix \( A \):
\[ A^{-1} = \frac{1}{2x} \begin{pmatrix} x & -1 \\ 0 & 2 \end{pmatrix} \]

### Step 3: Simplify the expression for \( A^{-1} \)

Now, simplifying the expression:
\[ A^{-1} = \begin{pmatrix} \frac{x}{2x} & \frac{-1}{2x} \\ 0 & \frac{2}{2x} \end{pmatrix} = \begin{pmatrix} \frac{1}{2} & \frac{-1}{2x} \\ 0 & \frac{1}{x} \end{pmatrix} \]

### Step 4: Set the simplified \( A^{-1} \) equal to the given \( A^{-1} \)

We know that:
\[ A^{-1} = \begin{pmatrix} \frac{1}{2} & \frac{1}{6} \\ 0 & \frac{1}{x} \end{pmatrix} \]

Now, we can equate the corresponding elements of the two matrices:
1. From the first element: \( \frac{1}{2} = \frac{1}{2} \) (this is already satisfied)
2. From the second element: \( \frac{-1}{2x} = \frac{1}{6} \)
3. From the last element: \( \frac{1}{x} = \frac{1}{x} \) (this is also satisfied)

### Step 5: Solve for \( x \)

From the second element:
\[ \frac{-1}{2x} = \frac{1}{6} \]

Cross-multiplying gives:
\[ -6 = 2x \implies x = -3 \]

### Final Answer

Thus, the value of \( x \) is:
\[ \boxed{-3} \]