If `{:[(a,b^3),(2,0)]=[(1,8),(2,0)]," then " [(a,b),(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)]:}`

Explanation

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

### Step 1: Equate the matrices
Since the two matrices are equal, we can equate the corresponding elements:
\[ a = 1 \quad \text{and} \quad b^3 = 8 \]

### Step 2: Solve for \(b\)
From the equation \(b^3 = 8\), we can find \(b\):
\[ b = \sqrt[3]{8} = 2 \]

### Step 3: Substitute \(a\) and \(b\) into the matrix
Now that we have \(a\) and \(b\), we can substitute these values into the matrix:
\[ \begin{pmatrix} a & b \\ 2 & 0 \end{pmatrix} = \begin{pmatrix} 1 & 2 \\ 2 & 0 \end{pmatrix} \]

### Step 4: Find the inverse of the matrix
To find the inverse of the matrix \(\begin{pmatrix} 1 & 2 \\ 2 & 0 \end{pmatrix}\), we use the formula for the inverse of a 2x2 matrix:
\[ \text{If } A = \begin{pmatrix} x & y \\ w & z \end{pmatrix}, \text{ then } A^{-1} = \frac{1}{(xz - wy)} \begin{pmatrix} z & -y \\ -w & x \end{pmatrix} \]
Here, \(x = 1\), \(y = 2\), \(w = 2\), and \(z = 0\).

### Step 5: Calculate the determinant
First, we calculate the determinant:
\[ \text{det}(A) = (1)(0) - (2)(2) = 0 - 4 = -4 \]

### Step 6: Apply the inverse formula
Now we can apply the inverse formula:
\[ A^{-1} = \frac{1}{-4} \begin{pmatrix} 0 & -2 \\ -2 & 1 \end{pmatrix} = \begin{pmatrix} 0 & \frac{1}{2} \\ \frac{1}{2} & -\frac{1}{4} \end{pmatrix} \]

### Final Answer
Thus, the inverse of the matrix \(\begin{pmatrix} a & b \\ 2 & 0 \end{pmatrix}\) is:
\[ \begin{pmatrix} 0 & \frac{1}{2} \\ \frac{1}{2} & -\frac{1}{4} \end{pmatrix} \]