The value of `|{:(a,a+2b,a+4b),(a+2b,a+4b,a+6b),(a+4b,a+6b,a+8b):}|` is

Options

A. `3a+6b`
B. 2b
C. 0
D. none of these

Answer

To find the value of the determinant

\[ D = \begin{vmatrix} a & a + 2b & a + 4b \\ a + 2b & a + 4b & a + 6b \\ a + 4b & a + 6b & a + 8b \end{vmatrix} \]

we will perform some column operations to simplify the determinant.

### Step 1: Perform Column Operations
We will perform two column operations:
1. Replace Column 2 with Column 2 minus Column 1.
2. Replace Column 3 with Column 3 minus Column 2.

### Step 2: Calculate the New Columns
Let's calculate the new columns after performing the operations.

- For Column 2:
\[ \text{New Column 2} = \begin{pmatrix} (a + 2b) - a \\ (a + 4b) - (a + 2b) \\ (a + 6b) - (a + 4b) \end{pmatrix} = \begin{pmatrix} 2b \\ 2b \\ 2b \end{pmatrix} \]

- For Column 3:
\[ \text{New Column 3} = \begin{pmatrix} (a + 4b) - (a + 2b) \\ (a + 6b) - (a + 4b) \\ (a + 8b) - (a + 6b) \end{pmatrix} = \begin{pmatrix} 2b \\ 2b \\ 2b \end{pmatrix} \]

### Step 3: Substitute the New Columns into the Determinant
Now, substituting the new columns into the determinant, we have:

\[ D = \begin{vmatrix} a & 2b & 2b \\ a + 2b & 2b & 2b \\ a + 4b & 2b & 2b \end{vmatrix} \]

### Step 4: Identify Duplicate Columns
Notice that Column 2 and Column 3 are now identical:

\[ \begin{pmatrix} 2b \\ 2b \\ 2b \end{pmatrix} \]

### Step 5: Conclusion
Since two columns of the determinant are the same, the value of the determinant is:

\[ D = 0 \]

Thus, the value of the determinant is **0**.