`" If " Delta_(1) =|{:(x,,b,,b),(a,,x,,b),(a,,a,,x):}|" and " Delta_(2)= |{:(x,,b),(a,,x):}|` are the given determinants then

Options

A. `Delta_(1)=3(Delta_(2))^(2)`
B. `(d)/(dx) (Delta_(1)) =3Delta_(2)`
C. `(d)/(dx) (Delta_(1)) =3(Delta_(2))^(2)`
D. `Delta_(1)=3Delta_(2)^(3//2)`

Explanation

To solve the problem, we will compute the determinants \( \Delta_1 \) and \( \Delta_2 \) step by step and find the relationship between them.

### Step 1: Calculate \( \Delta_1 \)

Given:
\[ \Delta_1 = \begin{vmatrix} x & b & b \\ a & x & b \\ a & a & x \end{vmatrix} \]

We will expand this determinant along the first column.

Using the formula for the determinant of a \( 3 \times 3 \) matrix:
\[ \Delta = a(ei - fh) - b(di - fg) + c(dh - eg) \]

For our determinant:
\[ \Delta_1 = x \begin{vmatrix} x & b \\ a & x \end{vmatrix} - b \begin{vmatrix} a & b \\ a & x \end{vmatrix} + b \begin{vmatrix} a & x \\ a & a \end{vmatrix} \]

Calculating the \( 2 \times 2 \) determinants:
1. \( \begin{vmatrix} x & b \\ a & x \end{vmatrix} = x^2 - ab \)

2. \( \begin{vmatrix} a & b \\ a & x \end{vmatrix} = ax - ab \)

3. \( \begin{vmatrix} a & x \\ a & a \end{vmatrix} = aa - ax = a(a - x) \)

Now substituting these back into the expression for \( \Delta_1 \):
\[ \Delta_1 = x(x^2 - ab) - b(ax - ab) + b[a(a - x)] \]
\[ = x^3 - abx - abx + b^2 + ab(a - x) \]
\[ = x^3 - 3abx + a^2b + b^2 \]

### Step 2: Calculate \( \Delta_2 \)

Given:
\[ \Delta_2 = \begin{vmatrix} x & b \\ a & x \end{vmatrix} \]

Calculating this determinant:
\[ \Delta_2 = x^2 - ab \]

### Step 3: Find the relationship between \( \Delta_1 \) and \( \Delta_2 \)

From the calculations, we have:
\[ \Delta_1 = x^3 - 3abx + a^2b + b^2 \]
\[ \Delta_2 = x^2 - ab \]

Now, we differentiate \( \Delta_1 \) with respect to \( x \):
\[ \frac{d(\Delta_1)}{dx} = 3x^2 - 3ab \]

We can factor this:
\[ \frac{d(\Delta_1)}{dx} = 3(x^2 - ab) \]

Notice that \( \Delta_2 = x^2 - ab \). Therefore, we can express the relationship as:
\[ \frac{d(\Delta_1)}{dx} = 3 \Delta_2 \]

Thus, we conclude:
\[ \Delta_1 = 3 \int \Delta_2 \, dx + C \]

### Final Answer

The relationship between \( \Delta_1 \) and \( \Delta_2 \) is:
\[ \frac{d(\Delta_1)}{dx} = 3 \Delta_2 \]