The inverse of the matrix `[[2, 0, 0], [0, 1, 0], [0, 0, -1]]` is

Options

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

Answer

To find the inverse of the matrix \( A = \begin{bmatrix} 2 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & -1 \end{bmatrix} \), we will follow these steps:

### Step 1: Calculate the Determinant of Matrix A

The determinant of a diagonal matrix can be calculated by multiplying its diagonal elements.

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

**Hint**: For a diagonal matrix, the determinant is simply the product of the diagonal elements.

### Step 2: Check if the Determinant is Non-Zero

Since \(\text{det}(A) = -2 \neq 0\), the inverse of matrix \( A \) exists.

**Hint**: An inverse exists only if the determinant is not equal to zero.

### Step 3: Calculate the Adjoint of Matrix A

The adjoint of a matrix is the transpose of its cofactor matrix. For a diagonal matrix, the cofactor matrix is obtained by taking the determinant of the submatrices formed by removing one row and one column at a time, and applying the appropriate sign based on the position.

The cofactor matrix \( C \) for matrix \( A \) is:

\[ C = \begin{bmatrix} C_{11} & C_{12} & C_{13} \\ C_{21} & C_{22} & C_{23} \\ C_{31} & C_{32} & C_{33} \end{bmatrix} \]

Calculating the cofactors:
- \( C_{11} = \text{det} \begin{bmatrix} 1 & 0 \\ 0 & -1 \end{bmatrix} = -1 \)
- \( C_{12} = -\text{det} \begin{bmatrix} 0 & 0 \\ 0 & -1 \end{bmatrix} = 0 \)
- \( C_{13} = \text{det} \begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix} = 0 \)

- \( C_{21} = -\text{det} \begin{bmatrix} 0 & 0 \\ 0 & -1 \end{bmatrix} = 0 \)
- \( C_{22} = \text{det} \begin{bmatrix} 2 & 0 \\ 0 & -1 \end{bmatrix} = -2 \)
- \( C_{23} = -\text{det} \begin{bmatrix} 2 & 0 \\ 0 & 0 \end{bmatrix} = 0 \)

- \( C_{31} = \text{det} \begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix} = 0 \)
- \( C_{32} = -\text{det} \begin{bmatrix} 2 & 0 \\ 0 & 0 \end{bmatrix} = 0 \)
- \( C_{33} = \text{det} \begin{bmatrix} 2 & 1 \\ 0 & 0 \end{bmatrix} = 2 \)

Thus, the cofactor matrix \( C \) is:

\[ C = \begin{bmatrix} -1 & 0 & 0 \\ 0 & -2 & 0 \\ 0 & 0 & 2 \end{bmatrix} \]

Now, taking the transpose to get the adjoint:

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

**Hint**: The adjoint is the transpose of the cofactor matrix. For diagonal matrices, the cofactor matrix will also be diagonal.

### Step 4: Calculate the Inverse of Matrix A

Using the formula for the inverse:

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

Substituting the values we found:

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

This simplifies to:

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

**Final Answer**:

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