The inverse of the matrix `A=[(0,1,0),(1,0,0),(0,0,1)]` is equal to

Options

A. A
B. A'
C. `[(0,1,0),(0,1,0),(0,0,1)]`
D. `[(1,0,0),(1,0,0),(0,1,0)]`

Answer

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

### Step 1: Calculate the Determinant of Matrix \( A \)

The determinant of a \( 3 \times 3 \) matrix \( A = \begin{pmatrix} a & b & c \\ d & e & f \\ g & h & i \end{pmatrix} \) can be calculated using the formula:

\[ \text{det}(A) = a(ei - fh) - b(di - fg) + c(dh - eg) \]

For our matrix \( A \):

\[ \text{det}(A) = 0 \cdot (0 \cdot 1 - 0 \cdot 0) - 1 \cdot (1 \cdot 1 - 0 \cdot 0) + 0 \cdot (1 \cdot 0 - 0 \cdot 0) \]
\[ = 0 - 1 \cdot 1 + 0 = -1 \]

### Step 2: Calculate the Adjoint of Matrix \( A \)

The adjoint of a matrix is the transpose of the cofactor matrix. We will find the cofactor matrix first.

The cofactor matrix \( C \) is calculated by taking the determinant of the \( 2 \times 2 \) submatrices formed by removing the row and column of each element.

1. For \( C_{11} \): Remove the first row and first column, we get \( \begin{pmatrix} 0 & 0 \\ 0 & 1 \end{pmatrix} \), so \( C_{11} = 0 \).
2. For \( C_{12} \): Remove the first row and second column, we get \( \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} \), so \( C_{12} = -1 \).
3. For \( C_{13} \): Remove the first row and third column, we get \( \begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix} \), so \( C_{13} = 0 \).

4. For \( C_{21} \): Remove the second row and first column, we get \( \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} \), so \( C_{21} = -1 \).
5. For \( C_{22} \): Remove the second row and second column, we get \( \begin{pmatrix} 0 & 0 \\ 0 & 1 \end{pmatrix} \), so \( C_{22} = 0 \).
6. For \( C_{23} \): Remove the second row and third column, we get \( \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix} \), so \( C_{23} = 0 \).

7. For \( C_{31} \): Remove the third row and first column, we get \( \begin{pmatrix} 1 & 0 \\ 1 & 0 \end{pmatrix} \), so \( C_{31} = 0 \).
8. For \( C_{32} \): Remove the third row and second column, we get \( \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix} \), so \( C_{32} = 1 \).
9. For \( C_{33} \): Remove the third row and third column, we get \( \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix} \), so \( C_{33} = 0 \).

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

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

Now, we take the transpose of the cofactor matrix to get the adjoint:

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

### Step 3: Calculate the Inverse of Matrix \( A \)

The inverse of matrix \( A \) is given by the formula:

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

Substituting the values we found:

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

### Final Answer

The inverse of the matrix \( A \) is:

\[ A^{-1} = \begin{pmatrix} 0 & 1 & 0 \\ 1 & 0 & -1 \\ 0 & 0 & -1 \end{pmatrix} \]