value of `|(2,3,4),(5,6,8),(6x,9x,12x)|`

Options

A. `9x^(2)`
B. x
C. 2x
D. None of these

Answer

To find the value of the determinant \( |(2,3,4),(5,6,8),(6x,9x,12x)| \), we will follow these steps:

### Step 1: Write the Determinant
We start with the determinant:
\[ D = \begin{vmatrix} 2 & 3 & 4 \\ 5 & 6 & 8 \\ 6x & 9x & 12x \end{vmatrix} \]

### Step 2: Factor Out Common Terms
We notice that in the second column, the number 3 is common, and in the third column, the number 4 is common. We can factor these out:
\[ D = 3 \cdot 4 \cdot \begin{vmatrix} 2 & 1 & 1 \\ 5 & 2 & 2 \\ 6x & 2 & 3x \end{vmatrix} \]

### Step 3: Factor Out from the Third Row
Next, we can factor out \( 3x \) from the third row:
\[ D = 12x \cdot \begin{vmatrix} 2 & 1 & 1 \\ 5 & 2 & 2 \\ 6 & 2 & 1 \end{vmatrix} \]

### Step 4: Simplify the Determinant
Now we can calculate the determinant:
\[ D = 12x \cdot \begin{vmatrix} 2 & 1 & 1 \\ 5 & 2 & 2 \\ 6 & 2 & 1 \end{vmatrix} \]

### Step 5: Calculate the Determinant
We can calculate the determinant using the formula for 3x3 matrices:
\[ \begin{vmatrix} a & b & c \\ d & e & f \\ g & h & i \end{vmatrix} = a(ei - fh) - b(di - fg) + c(dh - eg) \]

Applying this to our determinant:
\[ = 2 \cdot (2 \cdot 1 - 2 \cdot 2) - 1 \cdot (5 \cdot 1 - 2 \cdot 6) + 1 \cdot (5 \cdot 2 - 2 \cdot 6) \]
\[ = 2 \cdot (2 - 4) - 1 \cdot (5 - 12) + 1 \cdot (10 - 12) \]
\[ = 2 \cdot (-2) - 1 \cdot (-7) + 1 \cdot (-2) \]
\[ = -4 + 7 - 2 = 1 \]

### Step 6: Final Value of the Determinant
Now substituting back, we have:
\[ D = 12x \cdot 1 = 12x \]

### Step 7: Check for Identical Columns
We notice that the second and third columns are not identical. However, we can see that the third column is a multiple of the second column, which means the determinant will still be zero.

Thus, the final value of the determinant is:
\[ \boxed{0} \]