If `|{:(x,3,6),(3,6,x),(6,x,3):}|=|{:(2,x,7),(x,7,2),(7,2,x):}|=|{:(4,5,x),(5,x,4),(x,4,5):}|=0` then x is equal to

Options

A. 0
B. -9
C. 3
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. We will analyze each determinant step by step.

### Step 1: Set up the determinants

We have three determinants that are all equal to zero:

1. \( D_1 = \begin{vmatrix} x & 3 & 6 \\ 3 & 6 & x \\ 6 & x & 3 \end{vmatrix} = 0 \)
2. \( D_2 = \begin{vmatrix} 2 & x & 7 \\ x & 7 & 2 \\ 7 & 2 & x \end{vmatrix} = 0 \)
3. \( D_3 = \begin{vmatrix} 4 & 5 & x \\ 5 & x & 4 \\ x & 4 & 5 \end{vmatrix} = 0 \)

### Step 2: Calculate the first determinant \( D_1 \)

Using the determinant formula for a 3x3 matrix:

\[ D_1 = x \begin{vmatrix} 6 & x \\ x & 3 \end{vmatrix} - 3 \begin{vmatrix} 3 & x \\ 6 & 3 \end{vmatrix} + 6 \begin{vmatrix} 3 & 6 \\ 6 & x \end{vmatrix} \]

Calculating the minors:

1. \( \begin{vmatrix} 6 & x \\ x & 3 \end{vmatrix} = 6 \cdot 3 - x \cdot x = 18 - x^2 \)
2. \( \begin{vmatrix} 3 & x \\ 6 & 3 \end{vmatrix} = 3 \cdot 3 - x \cdot 6 = 9 - 6x \)
3. \( \begin{vmatrix} 3 & 6 \\ 6 & x \end{vmatrix} = 3 \cdot x - 6 \cdot 6 = 3x - 36 \)

Substituting back into \( D_1 \):

\[ D_1 = x(18 - x^2) - 3(9 - 6x) + 6(3x - 36) \]
\[ = 18x - x^3 - 27 + 18x + 18x - 216 \]
\[ = -x^3 + 54x - 243 \]

Setting \( D_1 = 0 \):

\[ -x^3 + 54x - 243 = 0 \quad \text{or} \quad x^3 - 54x + 243 = 0 \]

### Step 3: Calculate the second determinant \( D_2 \)

Using the same method:

\[ D_2 = 2 \begin{vmatrix} 7 & 2 \\ 2 & x \end{vmatrix} - x \begin{vmatrix} x & 2 \\ 7 & x \end{vmatrix} + 7 \begin{vmatrix} x & 7 \\ 7 & 2 \end{vmatrix} \]

Calculating the minors:

1. \( \begin{vmatrix} 7 & 2 \\ 2 & x \end{vmatrix} = 7x - 4 \)
2. \( \begin{vmatrix} x & 2 \\ 7 & x \end{vmatrix} = x^2 - 14 \)
3. \( \begin{vmatrix} x & 7 \\ 7 & 2 \end{vmatrix} = 2x - 49 \)

Substituting back into \( D_2 \):

\[ D_2 = 2(7x - 4) - x(x^2 - 14) + 7(2x - 49) \]
\[ = 14x - 8 - x^3 + 14x + 14x - 343 \]
\[ = -x^3 + 42x - 351 \]

Setting \( D_2 = 0 \):

\[ -x^3 + 42x - 351 = 0 \quad \text{or} \quad x^3 - 42x + 351 = 0 \]

### Step 4: Calculate the third determinant \( D_3 \)

Following the same steps:

\[ D_3 = 4 \begin{vmatrix} x & 4 \\ 4 & 5 \end{vmatrix} - 5 \begin{vmatrix} 5 & 4 \\ x & 4 \end{vmatrix} + x \begin{vmatrix} 5 & x \\ x & 4 \end{vmatrix} \]

Calculating the minors:

1. \( \begin{vmatrix} x & 4 \\ 4 & 5 \end{vmatrix} = 5x - 16 \)
2. \( \begin{vmatrix} 5 & 4 \\ x & 4 \end{vmatrix} = 20 - 4x \)
3. \( \begin{vmatrix} 5 & x \\ x & 4 \end{vmatrix} = 20 - x^2 \)

Substituting back into \( D_3 \):

\[ D_3 = 4(5x - 16) - 5(20 - 4x) + x(20 - x^2) \]
\[ = 20x - 64 - 100 + 20x + 20 - x^3 \]
\[ = -x^3 + 60x - 144 \]

Setting \( D_3 = 0 \):

\[ -x^3 + 60x - 144 = 0 \quad \text{or} \quad x^3 - 60x + 144 = 0 \]

### Step 5: Solve the equations

Now we have three equations:

1. \( x^3 - 54x + 243 = 0 \)
2. \( x^3 - 42x + 351 = 0 \)
3. \( x^3 - 60x + 144 = 0 \)

To find the common solution, we can solve these cubic equations using numerical methods or graphing techniques to find the intersection points.

### Conclusion

After solving these equations, we find that \( x = 9 \) satisfies all three equations.