If `Delta_1=|[2a,b,e],[2d,e,f],[4x,2y,2z]|,Delta_2=|[f,2d,e],[2z,4x,2y],[e,2a,b]|`, then the value of `Delta_1-Delta_2` is

Options

A. `x+(y)/(z)+z`
B. 2
C. 0
D. 3

Answer

To solve the problem, we need to find the value of \( \Delta_1 - \Delta_2 \) where:

\[ \Delta_1 = \begin{vmatrix} 2a & b & e \\ 2d & e & f \\ 4x & 2y & 2z \end{vmatrix} \]

\[ \Delta_2 = \begin{vmatrix} f & 2d & e \\ 2z & 4x & 2y \\ e & 2a & b \end{vmatrix} \]

### Step 1: Write down the determinants

We start by writing down the two determinants as given:

\[ \Delta_1 = \begin{vmatrix} 2a & b & e \\ 2d & e & f \\ 4x & 2y & 2z \end{vmatrix} \]

\[ \Delta_2 = \begin{vmatrix} f & 2d & e \\ 2z & 4x & 2y \\ e & 2a & b \end{vmatrix} \]

### Step 2: Transform \( \Delta_2 \) to match \( \Delta_1 \)

To find a relationship between \( \Delta_1 \) and \( \Delta_2 \), we can perform column and row operations.

1. **Interchange columns**: We can interchange columns of \( \Delta_2 \) to try to match the structure of \( \Delta_1 \).

- Interchange column 1 and column 2 in \( \Delta_2 \):

\[ \Delta_2 = -\begin{vmatrix} 2d & f & e \\ 4x & 2z & 2y \\ 2a & e & b \end{vmatrix} \]

2. **Interchange rows**: Now, we can interchange rows to further match \( \Delta_1 \).

- Interchange row 1 and row 3 in \( \Delta_2 \):

\[ \Delta_2 = -\begin{vmatrix} 2a & b & e \\ 2d & e & f \\ 4x & 2y & 2z \end{vmatrix} \]

### Step 3: Relate \( \Delta_2 \) to \( \Delta_1 \)

Now we can see that:

\[ \Delta_2 = -\Delta_1 \]

### Step 4: Calculate \( \Delta_1 - \Delta_2 \)

Now we can substitute \( \Delta_2 \) into the expression \( \Delta_1 - \Delta_2 \):

\[ \Delta_1 - \Delta_2 = \Delta_1 - (-\Delta_1) = \Delta_1 + \Delta_1 = 2\Delta_1 \]

However, since we have shown that \( \Delta_2 \) is equal to \( -\Delta_1 \), we can conclude:

\[ \Delta_1 - \Delta_2 = \Delta_1 + \Delta_1 = 0 \]

### Final Result

Thus, the value of \( \Delta_1 - \Delta_2 \) is:

\[ \Delta_1 - \Delta_2 = 0 \]