If ` A =({:( 1,1,1),( 1,omega ^(2) , omega ),( 1 ,omega , omega ^(2)) :}) ` where `omega` is a complex cube root of unity then `adj -A ` equals

Options

A. `(omega ^(2) -omega ) vec A `
B. ` (omega -omega ^(2) ) oversetto A `
C. ` -oversetto A `
D. ` oversetto A `

Answer

To find the adjugate of the matrix \( A \) given by

\[ A = \begin{pmatrix} 1 & 1 & 1 \\ 1 & \omega^2 & \omega \\ 1 & \omega & \omega^2 \end{pmatrix} \]

where \( \omega \) is a complex cube root of unity, we can follow these steps:

### Step 1: Understand the properties of \( \omega \)
Since \( \omega \) is a cube root of unity, we have:

\[ \omega^3 = 1 \quad \text{and} \quad 1 + \omega + \omega^2 = 0 \]

### Step 2: Calculate the cofactors of \( A \)
The adjugate of a matrix is the transpose of the cofactor matrix. We will calculate the cofactors for each element of the matrix.

#### Cofactor \( C_{11} \)
To find \( C_{11} \), we remove the first row and first column:

\[ C_{11} = \begin{vmatrix} \omega^2 & \omega \\ \omega & \omega^2 \end{vmatrix} = \omega^2 \cdot \omega^2 - \omega \cdot \omega = \omega^4 - \omega^2 = \omega - \omega^2 \]

#### Cofactor \( C_{12} \)
To find \( C_{12} \), we remove the first row and second column:

\[ C_{12} = -\begin{vmatrix} 1 & \omega \\ 1 & \omega^2 \end{vmatrix} = - (1 \cdot \omega^2 - 1 \cdot \omega) = -(\omega^2 - \omega) = \omega - \omega^2 \]

#### Cofactor \( C_{13} \)
To find \( C_{13} \), we remove the first row and third column:

\[ C_{13} = \begin{vmatrix} 1 & \omega^2 \\ 1 & \omega \end{vmatrix} = 1 \cdot \omega - 1 \cdot \omega^2 = \omega - \omega^2 \]

#### Cofactor \( C_{21} \)
To find \( C_{21} \), we remove the second row and first column:

\[ C_{21} = -\begin{vmatrix} 1 & 1 \\ 1 & \omega^2 \end{vmatrix} = - (1 \cdot \omega^2 - 1 \cdot 1) = -(\omega^2 - 1) = 1 - \omega^2 \]

#### Cofactor \( C_{22} \)
To find \( C_{22} \), we remove the second row and second column:

\[ C_{22} = \begin{vmatrix} 1 & 1 \\ 1 & \omega \end{vmatrix} = 1 \cdot \omega - 1 \cdot 1 = \omega - 1 \]

#### Cofactor \( C_{23} \)
To find \( C_{23} \), we remove the second row and third column:

\[ C_{23} = -\begin{vmatrix} 1 & 1 \\ 1 & \omega \end{vmatrix} = -(\omega - 1) = 1 - \omega \]

#### Cofactor \( C_{31} \)
To find \( C_{31} \), we remove the third row and first column:

\[ C_{31} = \begin{vmatrix} 1 & 1 \\ 1 & \omega^2 \end{vmatrix} = \omega^2 - 1 \]

#### Cofactor \( C_{32} \)
To find \( C_{32} \), we remove the third row and second column:

\[ C_{32} = -\begin{vmatrix} 1 & 1 \\ 1 & \omega \end{vmatrix} = -(\omega - 1) = 1 - \omega \]

#### Cofactor \( C_{33} \)
To find \( C_{33} \), we remove the third row and third column:

\[ C_{33} = \begin{vmatrix} 1 & 1 \\ 1 & \omega^2 \end{vmatrix} = \omega^2 - 1 \]

### Step 3: Assemble the cofactor matrix
The cofactor matrix \( C \) is:

\[ C = \begin{pmatrix} \omega - \omega^2 & \omega - \omega^2 & \omega - \omega^2 \\ 1 - \omega^2 & \omega - 1 & 1 - \omega \\ \omega^2 - 1 & 1 - \omega & \omega^2 - 1 \end{pmatrix} \]

### Step 4: Transpose the cofactor matrix to find the adjugate
The adjugate \( \text{adj}(A) \) is the transpose of \( C \):

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

### Final Result
Thus, the adjugate of matrix \( A \) is:

\[ \text{adj}(A) = \begin{pmatrix} \omega - \omega^2 & 1 - \omega^2 & \omega^2 - 1 \\ \omega - \omega^2 & \omega - 1 & 1 - \omega \\ \omega - \omega^2 & 1 - \omega & \omega^2 - 1 \end{pmatrix} \]