If `|(x,1,5),(1,5,x),(5,x,1)|=|(x,2,4),(2,4,x),(4,x,2)|=|(x,-1,7),(-1,7,x),(7,x,-1)|` = 0 , then x =

Options

A. 6
B. `-6`
C. 0
D. none of these

Answer

To solve the problem, we need to find the value of \( x \) such that the determinants given are equal to zero. We can start with the first determinant:

\[ D_1 = \begin{vmatrix} x & 1 & 5 \\ 1 & 5 & x \\ 5 & x & 1 \end{vmatrix} \]

### Step 1: Calculate the determinant \( D_1 \)

We can use the determinant formula for a \( 3 \times 3 \) matrix:

\[ D = a(ei - fh) - b(di - fg) + c(dh - eg) \]

For our determinant, we have:

\[ D_1 = x \begin{vmatrix} 5 & x \\ x & 1 \end{vmatrix} - 1 \begin{vmatrix} 1 & x \\ 5 & 1 \end{vmatrix} + 5 \begin{vmatrix} 1 & 5 \\ 5 & x \end{vmatrix} \]

Calculating the minors:

1. \( \begin{vmatrix} 5 & x \\ x & 1 \end{vmatrix} = 5 \cdot 1 - x \cdot x = 5 - x^2 \)
2. \( \begin{vmatrix} 1 & x \\ 5 & 1 \end{vmatrix} = 1 \cdot 1 - x \cdot 5 = 1 - 5x \)
3. \( \begin{vmatrix} 1 & 5 \\ 5 & x \end{vmatrix} = 1 \cdot x - 5 \cdot 5 = x - 25 \)

Substituting these back into the determinant:

\[ D_1 = x(5 - x^2) - 1(1 - 5x) + 5(x - 25) \]

Expanding this:

\[ D_1 = 5x - x^3 - 1 + 5x + 5x - 125 \]
\[ D_1 = -x^3 + 15x - 126 \]

### Step 2: Set \( D_1 = 0 \)

Now we set the determinant equal to zero:

\[ -x^3 + 15x - 126 = 0 \]

### Step 3: Rearranging the equation

Multiplying through by -1 gives:

\[ x^3 - 15x + 126 = 0 \]

### Step 4: Finding roots

To find the roots of the polynomial, we can use the Rational Root Theorem or synthetic division. Testing \( x = 6 \):

\[ 6^3 - 15 \cdot 6 + 126 = 216 - 90 + 126 = 252 \quad (\text{not a root}) \]

Testing \( x = -6 \):

\[ (-6)^3 - 15(-6) + 126 = -216 + 90 + 126 = 0 \quad (\text{is a root}) \]

### Step 5: Factor the polynomial

Since \( x = -6 \) is a root, we can factor the polynomial as:

\[ (x + 6)(x^2 - 6x + 21) = 0 \]

### Step 6: Solve the quadratic equation

Now we solve the quadratic \( x^2 - 6x + 21 = 0 \) using the discriminant:

\[ D = b^2 - 4ac = (-6)^2 - 4 \cdot 1 \cdot 21 = 36 - 84 = -48 \]

Since the discriminant is negative, this quadratic has no real roots.

### Conclusion

Thus, the only real solution is:

\[ \boxed{-6} \]