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 `Delta_1-Delta_2=0` for values of x equal to

Options

A. 0
B. 2
C. `x inR`
D. none

Answer

To solve the problem, we need to evaluate the determinants \( \Delta_1 \) and \( \Delta_2 \) and then find the values of \( x \) for which \( \Delta_1 - \Delta_2 = 0 \).

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

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

To calculate the determinant \( \Delta_1 \), we can use the formula for the determinant of a 3x3 matrix:
\[ \Delta = a(ei - fh) - b(di - fg) + c(dh - eg) \]
where the matrix is:
\[ \begin{vmatrix} a & b & c \\ d & e & f \\ g & h & i \end{vmatrix} \]

Applying this to \( \Delta_1 \):
\[ \Delta_1 = 7 \begin{vmatrix} x+1 & 3 \\ x & 7 \end{vmatrix} - x \begin{vmatrix} -5 & 3 \\ 4 & 7 \end{vmatrix} + 2 \begin{vmatrix} -5 & x+1 \\ 4 & x \end{vmatrix} \]

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 \( \Delta_1 \):
\[ \Delta_1 = 7(4x + 7) - x(-47) + 2(-9x - 4) \]
\[ = 28x + 49 + 47x - 18x - 8 \]
\[ = (28x + 47x - 18x) + (49 - 8) \]
\[ = 57x + 41 \]

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

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

Using the same determinant formula:
\[ \Delta_2 = x \begin{vmatrix} 3 & -5 \\ 7 & 4 \end{vmatrix} - 2 \begin{vmatrix} x+1 & -5 \\ x & 4 \end{vmatrix} + 7 \begin{vmatrix} x+1 & 3 \\ x & 7 \end{vmatrix} \]

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

Now substituting back into \( \Delta_2 \):
\[ \Delta_2 = x(47) - 2(9x + 4) + 7(4x + 7) \]
\[ = 47x - 18x - 8 + 28x + 49 \]
\[ = (47x - 18x + 28x) + (-8 + 49) \]
\[ = 57x + 41 \]

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

Now we have:
\[ \Delta_1 = 57x + 41 \]
\[ \Delta_2 = 57x + 41 \]

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

### Conclusion

The equation \( \Delta_1 - \Delta_2 = 0 \) holds true for all values of \( x \).

### Final Answer
The values of \( x \) for which \( \Delta_1 - \Delta_2 = 0 \) is:
\[ \text{for all } x \in \mathbb{R} \]