If `Delta=|(-a,2b,0),(0,-a,2b),(2b,0,-a)|=0` then

Options

A. 1/b is a cube root of unit
B. a is one of the cube roots of unity
C. b is one of thecube roots of 8
D. a/b is a cube root of 8

Answer

To solve the problem where we have the determinant \( \Delta = \begin{vmatrix} -a & 2b & 0 \\ 0 & -a & 2b \\ 2b & 0 & -a \end{vmatrix} = 0 \), we will follow these steps:

### Step 1: Write the Determinant
We start with the determinant:

\[ \Delta = \begin{vmatrix} -a & 2b & 0 \\ 0 & -a & 2b \\ 2b & 0 & -a \end{vmatrix} \]

### Step 2: Calculate the Determinant
Using the formula for the determinant of a 3x3 matrix, we have:

\[ \Delta = -a \begin{vmatrix} -a & 2b \\ 0 & -a \end{vmatrix} - 2b \begin{vmatrix} 0 & 2b \\ 2b & -a \end{vmatrix} + 0 \begin{vmatrix} 0 & -a \\ 2b & 0 \end{vmatrix} \]

### Step 3: Calculate the 2x2 Determinants
Now we calculate the 2x2 determinants:

1. For the first determinant:
\[ \begin{vmatrix} -a & 2b \\ 0 & -a \end{vmatrix} = (-a)(-a) - (2b)(0) = a^2 \]

2. For the second determinant:
\[ \begin{vmatrix} 0 & 2b \\ 2b & -a \end{vmatrix} = (0)(-a) - (2b)(2b) = -4b^2 \]

### Step 4: Substitute Back into the Determinant
Now substituting these values back into the determinant calculation:

\[ \Delta = -a(a^2) - 2b(-4b^2) + 0 \]
\[ \Delta = -a^3 + 8b^3 \]

### Step 5: Set the Determinant to Zero
Since we are given that \( \Delta = 0 \):

\[ -a^3 + 8b^3 = 0 \]

### Step 6: Rearranging the Equation
Rearranging gives us:

\[ a^3 = 8b^3 \]

### Step 7: Taking the Cube Root
Taking the cube root of both sides, we find:

\[ \frac{a}{b} = \sqrt[3]{8} = 2 \]

### Final Result
Thus, the relationship between \( a \) and \( b \) is:

\[ \frac{a}{b} = 2 \]