If `[[x, y^(3)], [2, 0]]=[[1, 8], [2, 0]]`, then `[[x, y], [2, 0]]^(-1)=`

Options

A. `[[0, -2], [-2, 1]]`
B. `[[1, 0], [0, 1]]`
C. `[[0, -8], [-2, 1]]`
D. `[[0, (1)/(2)], [(1)/(2), -(1)/(4)]]`

Answer

To solve the problem, we need to find the inverse of the matrix \(\begin{bmatrix} x & y \\ 2 & 0 \end{bmatrix}\) given that \(\begin{bmatrix} x & y^3 \\ 2 & 0 \end{bmatrix} = \begin{bmatrix} 1 & 8 \\ 2 & 0 \end{bmatrix}\).

### Step 1: Compare the matrices
From the equation \(\begin{bmatrix} x & y^3 \\ 2 & 0 \end{bmatrix} = \begin{bmatrix} 1 & 8 \\ 2 & 0 \end{bmatrix}\), we can compare the corresponding elements of the matrices.

- From the first row, first column: \(x = 1\)
- From the first row, second column: \(y^3 = 8\)
- From the second row, first column: \(2 = 2\) (this is consistent)
- From the second row, second column: \(0 = 0\) (this is consistent)

### Step 2: Solve for \(y\)
We have \(y^3 = 8\). To find \(y\), we take the cube root of both sides:

\[ y = \sqrt[3]{8} = 2 \]

### Step 3: Write the matrix
Now we can substitute the values of \(x\) and \(y\) into the matrix:

\[ \begin{bmatrix} x & y \\ 2 & 0 \end{bmatrix} = \begin{bmatrix} 1 & 2 \\ 2 & 0 \end{bmatrix} \]

### Step 4: Find the determinant of the matrix
Next, we need to find the determinant of the matrix \(\begin{bmatrix} 1 & 2 \\ 2 & 0 \end{bmatrix}\):

\[ \text{det}(A) = (1)(0) - (2)(2) = 0 - 4 = -4 \]

### Step 5: Find the adjoint of the matrix
The adjoint of a \(2 \times 2\) matrix \(\begin{bmatrix} a & b \\ c & d \end{bmatrix}\) is given by:

\[ \text{adj}(A) = \begin{bmatrix} d & -b \\ -c & a \end{bmatrix} \]

For our matrix \(\begin{bmatrix} 1 & 2 \\ 2 & 0 \end{bmatrix}\), the adjoint is:

\[ \text{adj}(A) = \begin{bmatrix} 0 & -2 \\ -2 & 1 \end{bmatrix} \]

### Step 6: Calculate the inverse of the matrix
The inverse of the matrix \(A\) is given by:

\[ A^{-1} = \frac{1}{\text{det}(A)} \cdot \text{adj}(A) \]

Substituting the values we found:

\[ A^{-1} = \frac{1}{-4} \cdot \begin{bmatrix} 0 & -2 \\ -2 & 1 \end{bmatrix} = \begin{bmatrix} 0 & \frac{1}{2} \\ \frac{1}{2} & -\frac{1}{4} \end{bmatrix} \]

### Final Answer
Thus, the inverse of the matrix \(\begin{bmatrix} x & y \\ 2 & 0 \end{bmatrix}\) is:

\[ \begin{bmatrix} 0 & \frac{1}{2} \\ \frac{1}{2} & -\frac{1}{4} \end{bmatrix} \]