If `A=[[0, 1, 0], [1, 0, 0], [0, 0, 1]]`, then `A^(-1)=`

Options

A. `2A`
B. `A`
C. `-A`
D. `1`

Answer

To find the inverse of the matrix \( A \), we will use the formula for the inverse of a 3x3 matrix. The matrix \( A \) is given as:

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

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

The determinant of a 3x3 matrix \( A = \begin{pmatrix} a & b & c \\ d & e & f \\ g & h & i \end{pmatrix} \) is calculated as:

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

For our matrix \( A \):

- \( a = 0, b = 1, c = 0 \)
- \( d = 1, e = 0, f = 0 \)
- \( g = 0, h = 0, i = 1 \)

Now substituting these values into the determinant formula:

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

This simplifies to:

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

### Step 2: Find the Adjugate of \( A \)

The adjugate of a 3x3 matrix is found by taking the transpose of the cofactor matrix. The cofactor matrix is calculated by finding the determinant of the 2x2 matrices obtained by removing one row and one column from \( A \).

Calculating the cofactors:

- For \( a_{11} = 0 \): The minor is \( \begin{pmatrix} 0 & 0 \\ 0 & 1 \end{pmatrix} \) → cofactor = \( 0 \)
- For \( a_{12} = 1 \): The minor is \( \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} \) → cofactor = \( 1 \)
- For \( a_{13} = 0 \): The minor is \( \begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix} \) → cofactor = \( 0 \)

- For \( a_{21} = 1 \): The minor is \( \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} \) → cofactor = \( 1 \)
- For \( a_{22} = 0 \): The minor is \( \begin{pmatrix} 0 & 0 \\ 0 & 1 \end{pmatrix} \) → cofactor = \( 0 \)
- For \( a_{23} = 0 \): The minor is \( \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix} \) → cofactor = \( 0 \)

- For \( a_{31} = 0 \): The minor is \( \begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix} \) → cofactor = \( 0 \)
- For \( a_{32} = 0 \): The minor is \( \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix} \) → cofactor = \( -1 \)
- For \( a_{33} = 1 \): The minor is \( \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix} \) → cofactor = \( 1 \)

Thus, the cofactor matrix is:

\[ \text{Cof}(A) = \begin{pmatrix} 0 & 1 & 0 \\ 1 & 0 & 0 \\ 0 & -1 & 1 \end{pmatrix} \]

Now, taking the transpose gives us the adjugate:

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

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

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

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

Substituting the determinant and adjugate:

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

This simplifies to:

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

### Final Result

Thus, the inverse of matrix \( A \) is:

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