If `Delta_(1)=|{:(Ax,x^(2),1),(By,y^(2),1),(Cz,z^(2),1):}|" and "Delta_(2)=|{:(A,B,C),(x,y,z),(yz,zx,xy):}|`, then

Options

A. `Delta_(1)+Delta_(2)=0`
B. `Delta_(1)-Delta_(2)=0`
C. `Delta_(1)!=Delta_(2)`
D. none of these

Answer

To solve the problem, we need to analyze the determinants \( \Delta_1 \) and \( \Delta_2 \) given in the question.

### Step 1: Write down the determinants

The determinants are given as follows:

\[ \Delta_1 = \begin{vmatrix} Ax & x^2 & 1 \\ By & y^2 & 1 \\ Cz & z^2 & 1 \end{vmatrix} \]

\[ \Delta_2 = \begin{vmatrix} A & B & C \\ x & y & z \\ yz & zx & xy \end{vmatrix} \]

### Step 2: Transform \( \Delta_1 \)

We can apply the property of determinants that allows us to interchange rows and columns. By interchanging rows and columns, we can express \( \Delta_1 \) in a different form:

\[ \Delta_1 = \begin{vmatrix} Ax & By & Cz \\ x^2 & y^2 & z^2 \\ 1 & 1 & 1 \end{vmatrix} \]

### Step 3: Factor out common elements

Next, we can factor out \( x, y, z \) from the first column:

\[ \Delta_1 = xyz \begin{vmatrix} A & B & C \\ x & y & z \\ 1 & 1 & 1 \end{vmatrix} \]

### Step 4: Simplify \( \Delta_1 \)

Now we can simplify the determinant:

\[ \Delta_1 = xyz \cdot \Delta_2 \]

### Step 5: Compare \( \Delta_1 \) and \( \Delta_2 \)

From the above step, we have:

\[ \Delta_1 = \Delta_2 \]

This implies:

\[ \Delta_1 - \Delta_2 = 0 \]

### Conclusion

Thus, we conclude that:

\[ \Delta_1 - \Delta_2 = 0 \]

So the correct option is:

**Option B: \( \Delta_1 - \Delta_2 = 0 \)**

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