If `|(x-2,2x-3,3x-4),(x-4,2x-9,3x-16),(x-8,2x-27,3x-64)|` = 0 , then x =

Options

A. 1
B. 2
C. 3
D. 4

Answer

To solve the determinant equation given by

\[ \left| \begin{array}{ccc} x-2 & 2x-3 & 3x-4 \\ x-4 & 2x-9 & 3x-16 \\ x-8 & 2x-27 & 3x-64 \end{array} \right| = 0, \]

we will follow these steps:

### Step 1: Write the determinant
We start with the determinant as given:

\[ D = \left| \begin{array}{ccc} x-2 & 2x-3 & 3x-4 \\ x-4 & 2x-9 & 3x-16 \\ x-8 & 2x-27 & 3x-64 \end{array} \right|. \]

### Step 2: Apply column operations
We will perform column operations to simplify the determinant. We can subtract \(2 \times \text{Column 1}\) from Column 2 and \(3 \times \text{Column 1}\) from Column 3:

\[ C_2 \rightarrow C_2 - 2C_1 \quad \text{and} \quad C_3 \rightarrow C_3 - 3C_1. \]

This gives us:

\[ D = \left| \begin{array}{ccc} x-2 & (2x-3) - 2(x-2) & (3x-4) - 3(x-2) \\ x-4 & (2x-9) - 2(x-4) & (3x-16) - 3(x-4) \\ x-8 & (2x-27) - 2(x-8) & (3x-64) - 3(x-8) \end{array} \right|. \]

Calculating the new columns:

- For Column 2:
- Row 1: \(2x - 3 - 2(x - 2) = 2x - 3 - 2x + 4 = 1\)
- Row 2: \(2x - 9 - 2(x - 4) = 2x - 9 - 2x + 8 = -1\)
- Row 3: \(2x - 27 - 2(x - 8) = 2x - 27 - 2x + 16 = -11\)

- For Column 3:
- Row 1: \(3x - 4 - 3(x - 2) = 3x - 4 - 3x + 6 = 2\)
- Row 2: \(3x - 16 - 3(x - 4) = 3x - 16 - 3x + 12 = -4\)
- Row 3: \(3x - 64 - 3(x - 8) = 3x - 64 - 3x + 24 = -40\)

So, the determinant becomes:

\[ D = \left| \begin{array}{ccc} x-2 & 1 & 2 \\ x-4 & -1 & -4 \\ x-8 & -11 & -40 \end{array} \right|. \]

### Step 3: Expand the determinant
Now we can expand this determinant along the first row:

\[ D = (x-2) \left| \begin{array}{cc} -1 & -4 \\ -11 & -40 \end{array} \right| - 1 \left| \begin{array}{cc} x-4 & -4 \\ x-8 & -40 \end{array} \right| + 2 \left| \begin{array}{cc} x-4 & -1 \\ x-8 & -11 \end{array} \right|. \]

Calculating the 2x2 determinants:

1. \(\left| \begin{array}{cc} -1 & -4 \\ -11 & -40 \end{array} \right| = (-1)(-40) - (-4)(-11) = 40 - 44 = -4\)

2. \(\left| \begin{array}{cc} x-4 & -4 \\ x-8 & -40 \end{array} \right| = (x-4)(-40) - (-4)(x-8) = -40x + 160 + 4x - 32 = -36x + 128\)

3. \(\left| \begin{array}{cc} x-4 & -1 \\ x-8 & -11 \end{array} \right| = (x-4)(-11) - (-1)(x-8) = -11x + 44 + x - 8 = -10x + 36\)

Substituting these back into the determinant expression:

\[ D = (x-2)(-4) - 1(-36x + 128) + 2(-10x + 36). \]

Expanding this gives:

\[ D = -4x + 8 + 36x - 128 - 20x + 72. \]

Combining like terms:

\[ D = (-4x + 36x - 20x) + (8 - 128 + 72) = 12x - 48. \]

### Step 4: Set the determinant to zero
Setting \(D = 0\):

\[ 12x - 48 = 0. \]

### Step 5: Solve for x
Solving for \(x\):

\[ 12x = 48 \implies x = \frac{48}{12} = 4. \]

### Final Answer
Thus, the value of \(x\) is:

\[ \boxed{4}. \]