If `|{:(x, 2, 3),(2,3,x),(3 , x,2):}| = |{:(1, x, 4),(x,4,1),(4 , 1,x):}| = |{:(0, 5, x),(5,x,0),(x, 0,5):}|` = 0, then the value x equals (x `in` R )

Options

A. 0
B. 5
C. `-5`
D. None of these

Answer

To solve the problem, we need to find the value of \( x \) such that the determinants of the given matrices are equal to zero. Let's denote the three determinants as \( D_1, D_2, \) and \( D_3 \).

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

The first determinant is:
\[ D_1 = \begin{vmatrix} x & 2 & 3 \\ 2 & 3 & x \\ 3 & x & 2 \end{vmatrix} \]

Using the determinant formula for a 3x3 matrix, we have:
\[ D_1 = x \begin{vmatrix} 3 & x \\ x & 2 \end{vmatrix} - 2 \begin{vmatrix} 2 & x \\ 3 & 2 \end{vmatrix} + 3 \begin{vmatrix} 2 & 3 \\ 3 & x \end{vmatrix} \]

Calculating each of the 2x2 determinants:
1. \( \begin{vmatrix} 3 & x \\ x & 2 \end{vmatrix} = 3 \cdot 2 - x \cdot x = 6 - x^2 \)
2. \( \begin{vmatrix} 2 & x \\ 3 & 2 \end{vmatrix} = 2 \cdot 2 - x \cdot 3 = 4 - 3x \)
3. \( \begin{vmatrix} 2 & 3 \\ 3 & x \end{vmatrix} = 2x - 9 \)

Substituting back into the determinant:
\[ D_1 = x(6 - x^2) - 2(4 - 3x) + 3(2x - 9) \]
\[ = 6x - x^3 - 8 + 6x + 6x - 27 \]
\[ = -x^3 + 18x - 35 \]

### Step 2: Calculate \( D_2 \)

The second determinant is:
\[ D_2 = \begin{vmatrix} 1 & x & 4 \\ x & 4 & 1 \\ 4 & 1 & x \end{vmatrix} \]

Using the same determinant formula:
\[ D_2 = 1 \begin{vmatrix} 4 & 1 \\ 1 & x \end{vmatrix} - x \begin{vmatrix} x & 1 \\ 4 & x \end{vmatrix} + 4 \begin{vmatrix} x & 4 \\ 4 & 1 \end{vmatrix} \]

Calculating the 2x2 determinants:
1. \( \begin{vmatrix} 4 & 1 \\ 1 & x \end{vmatrix} = 4x - 1 \)
2. \( \begin{vmatrix} x & 1 \\ 4 & x \end{vmatrix} = x^2 - 4 \)
3. \( \begin{vmatrix} x & 4 \\ 4 & 1 \end{vmatrix} = x - 16 \)

Substituting back:
\[ D_2 = 1(4x - 1) - x(x^2 - 4) + 4(x - 16) \]
\[ = 4x - 1 - (x^3 - 4x) + 4x - 64 \]
\[ = -x^3 + 12x - 65 \]

### Step 3: Calculate \( D_3 \)

The third determinant is:
\[ D_3 = \begin{vmatrix} 0 & 5 & x \\ 5 & x & 0 \\ x & 0 & 5 \end{vmatrix} \]

Using the determinant formula:
\[ D_3 = 0 \cdot \begin{vmatrix} x & 0 \\ 0 & 5 \end{vmatrix} - 5 \cdot \begin{vmatrix} 5 & 0 \\ x & 5 \end{vmatrix} + x \cdot \begin{vmatrix} 5 & x \\ x & 0 \end{vmatrix} \]

Calculating the 2x2 determinants:
1. \( \begin{vmatrix} 5 & 0 \\ x & 5 \end{vmatrix} = 25 \)
2. \( \begin{vmatrix} 5 & x \\ x & 0 \end{vmatrix} = -x^2 \)

Substituting back:
\[ D_3 = -5(25) + x(-x^2) = -125 - x^3 \]

### Step 4: Set the determinants equal to zero

Since \( D_1 = D_2 = D_3 = 0 \), we can set up the equations:
1. \( -x^3 + 18x - 35 = 0 \)
2. \( -x^3 + 12x - 65 = 0 \)
3. \( -x^3 - 125 - x^3 = 0 \)

### Step 5: Solve the equations

From \( D_1 = D_2 \):
\[ - x^3 + 18x - 35 = - x^3 + 12x - 65 \]
This simplifies to:
\[ 6x = -30 \implies x = -5 \]

Verifying with \( D_2 = D_3 \):
\[ - x^3 + 12x - 65 = -125 - x^3 \]
This simplifies to:
\[ 12x - 65 = -125 \implies 12x = -60 \implies x = -5 \]

### Conclusion

The value of \( x \) is:
\[ \boxed{-5} \]