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 of 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 three matrices are equal to zero. Let's denote the matrices as follows:

1. \( A = \begin{pmatrix} x & 2 & 3 \\ 2 & 3 & x \\ 3 & x & 2 \end{pmatrix} \)
2. \( B = \begin{pmatrix} 1 & x & 4 \\ x & 4 & 1 \\ 4 & 1 & x \end{pmatrix} \)
3. \( C = \begin{pmatrix} 0 & 5 & x \\ 5 & x & 0 \\ x & 0 & 5 \end{pmatrix} \)

We need to compute the determinants of these matrices and set them equal to zero.

### Step 1: Calculate the determinant of matrix A

Using the formula for the determinant of a 3x3 matrix, we have:

\[ \text{det}(A) = x \begin{vmatrix} 3 & x \\ x & 2 \end{vmatrix} - 2 \begin{vmatrix} 2 & x \\ 3 & 2 \end{vmatrix} + 3 \begin{vmatrix} 2 & 3 \\ 3 & x \end{vmatrix} \]

Calculating the minors:

1. \( \begin{vmatrix} 3 & x \\ x & 2 \end{vmatrix} = 3 \cdot 2 - x \cdot x = 6 - x^2 \)
2. \( \begin{vmatrix} 2 & x \\ 3 & 2 \end{vmatrix} = 2 \cdot 2 - 3 \cdot x = 4 - 3x \)
3. \( \begin{vmatrix} 2 & 3 \\ 3 & x \end{vmatrix} = 2 \cdot x - 3 \cdot 3 = 2x - 9 \)

Substituting these back into the determinant:

\[ \text{det}(A) = x(6 - x^2) - 2(4 - 3x) + 3(2x - 9) \]

Expanding this:

\[ = 6x - x^3 - 8 + 6x + 6x - 27 \]
\[ = -x^3 + 18x - 35 \]

### Step 2: Calculate the determinant of matrix B

Using the same method:

\[ \text{det}(B) = 1 \begin{vmatrix} 4 & 1 \\ 1 & x \end{vmatrix} - x \begin{vmatrix} x & 4 \\ 4 & 1 \end{vmatrix} + 4 \begin{vmatrix} x & 4 \\ 4 & 1 \end{vmatrix} \]

Calculating the minors:

1. \( \begin{vmatrix} 4 & 1 \\ 1 & x \end{vmatrix} = 4x - 1 \)
2. \( \begin{vmatrix} x & 4 \\ 4 & 1 \end{vmatrix} = x \cdot 1 - 4 \cdot 4 = x - 16 \)

Substituting these back into the determinant:

\[ \text{det}(B) = 1(4x - 1) - x(x - 16) + 4(x - 16) \]

Expanding this:

\[ = 4x - 1 - (x^2 - 16x) + 4x - 64 \]
\[ = -x^2 + 12x - 65 \]

### Step 3: Calculate the determinant of matrix C

Using the same method:

\[ \text{det}(C) = 0 \begin{vmatrix} x & 0 \\ 0 & 5 \end{vmatrix} - 5 \begin{vmatrix} 5 & x \\ x & 0 \end{vmatrix} + x \begin{vmatrix} 5 & 5 \\ x & 0 \end{vmatrix} \]

Calculating the minors:

1. \( \begin{vmatrix} 5 & x \\ x & 0 \end{vmatrix} = 0 - 5x = -5x \)
2. \( \begin{vmatrix} 5 & 5 \\ x & 0 \end{vmatrix} = 0 - 5x = -5x \)

Substituting these back into the determinant:

\[ \text{det}(C) = -5(-5x) + x(-5x) \]
\[ = 25 - 5x^2 \]

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

Now we have:

1. \( -x^3 + 18x - 35 = 0 \)
2. \( -x^2 + 12x - 65 = 0 \)
3. \( -5x^2 + 25 = 0 \)

### Step 5: Solve the equations

1. For \( -x^3 + 18x - 35 = 0 \)
2. For \( -x^2 + 12x - 65 = 0 \)
3. For \( -5x^2 + 25 = 0 \)

From the third equation:

\[ -5x^2 + 25 = 0 \implies 5x^2 = 25 \implies x^2 = 5 \implies x = \pm \sqrt{5} \]

From the second equation:

\[ -x^2 + 12x - 65 = 0 \implies x^2 - 12x + 65 = 0 \]

Using the quadratic formula:

\[ x = \frac{12 \pm \sqrt{144 - 260}}{2} = \frac{12 \pm \sqrt{-116}}{2} \]

This gives complex solutions, which we discard.

From the first equation, we can substitute \( x = -5 \) and check if it satisfies all three equations.

### Conclusion

The value of \( x \) that satisfies all three determinants being equal to zero is:

\[ \boxed{-5} \]