If `Delta_(1) = |(7,x,2),(-5,x +1,3),(4,x,7)| and Delta_(2) = |(x,2,7),(x +1,3,-5),(x,7,4)|`, then the value of x for which `Delta_(1) + Delta_(2) = 0`, is

Options

A. 2
B. 0
C. any number
D. none of these

Answer

To solve the problem, we need to find the value of \( x \) for which \( \Delta_1 + \Delta_2 = 0 \), where:

\[ \Delta_1 = \begin{vmatrix} 7 & x & 2 \\ -5 & x + 1 & 3 \\ 4 & x & 7 \end{vmatrix} \]

\[ \Delta_2 = \begin{vmatrix} x & 2 & 7 \\ x + 1 & 3 & -5 \\ x & 7 & 4 \end{vmatrix} \]

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

To calculate \( \Delta_1 \), we can use the determinant formula:

\[ \Delta_1 = 7 \begin{vmatrix} x + 1 & 3 \\ x & 7 \end{vmatrix} - x \begin{vmatrix} -5 & 3 \\ 4 & 7 \end{vmatrix} + 2 \begin{vmatrix} -5 & x + 1 \\ 4 & x \end{vmatrix} \]

Calculating the 2x2 determinants:

1. \( \begin{vmatrix} x + 1 & 3 \\ x & 7 \end{vmatrix} = (x + 1) \cdot 7 - 3 \cdot x = 7x + 7 - 3x = 4x + 7 \)

2. \( \begin{vmatrix} -5 & 3 \\ 4 & 7 \end{vmatrix} = (-5) \cdot 7 - 3 \cdot 4 = -35 - 12 = -47 \)

3. \( \begin{vmatrix} -5 & x + 1 \\ 4 & x \end{vmatrix} = (-5) \cdot x - (x + 1) \cdot 4 = -5x - 4x - 4 = -9x - 4 \)

Now substituting back into \( \Delta_1 \):

\[ \Delta_1 = 7(4x + 7) - x(-47) + 2(-9x - 4) \]

Expanding this:

\[ \Delta_1 = 28x + 49 + 47x - 18x - 8 = (28x + 47x - 18x) + (49 - 8) = 57x + 41 \]

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

Now we calculate \( \Delta_2 \):

\[ \Delta_2 = x \begin{vmatrix} 3 & -5 \\ 7 & 4 \end{vmatrix} - 2 \begin{vmatrix} x + 1 & -5 \\ x & 4 \end{vmatrix} + 7 \begin{vmatrix} x + 1 & 3 \\ x & 7 \end{vmatrix} \]

Calculating the 2x2 determinants:

1. \( \begin{vmatrix} 3 & -5 \\ 7 & 4 \end{vmatrix} = 3 \cdot 4 - (-5) \cdot 7 = 12 + 35 = 47 \)

2. \( \begin{vmatrix} x + 1 & -5 \\ x & 4 \end{vmatrix} = (x + 1) \cdot 4 - (-5) \cdot x = 4x + 4 + 5x = 9x + 4 \)

3. We already calculated \( \begin{vmatrix} x + 1 & 3 \\ x & 7 \end{vmatrix} = 4x + 7 \)

Substituting back into \( \Delta_2 \):

\[ \Delta_2 = x(47) - 2(9x + 4) + 7(4x + 7) \]

Expanding this:

\[ \Delta_2 = 47x - 18x - 8 + 28x + 49 = (47x - 18x + 28x) + (49 - 8) = 57x + 41 \]

### Step 3: Set up the equation \( \Delta_1 + \Delta_2 = 0 \)

Now we have:

\[ \Delta_1 + \Delta_2 = (57x + 41) + (57x + 41) = 0 \]

This simplifies to:

\[ 114x + 82 = 0 \]

### Step 4: Solve for \( x \)

To solve for \( x \):

\[ 114x = -82 \]

\[ x = -\frac{82}{114} \]

Reducing the fraction:

\[ x = -\frac{41}{57} \]

### Final Answer

The value of \( x \) for which \( \Delta_1 + \Delta_2 = 0 \) is:

\[ \boxed{-\frac{41}{57}} \]