If `|(x,x+y,x+y+z),(2x,3x+2y,4x+3y+2z),(3x,6x+3y,10x+6y+3z)|` = 64 then x = ?

Options

A. 2
B. 3
C. 4
D. 6

Answer

To solve the determinant equation \( |(x,x+y,x+y+z),(2x,3x+2y,4x+3y+2z),(3x,6x+3y,10x+6y+3z)| = 64 \), we will follow these steps:

### Step 1: Write the Determinant
We start with the determinant:
\[ D = \begin{vmatrix} x & x+y & x+y+z \\ 2x & 3x+2y & 4x+3y+2z \\ 3x & 6x+3y & 10x+6y+3z \end{vmatrix} \]

### Step 2: Apply Row Operations
We will perform row operations to simplify the determinant.

1. **Row 2**: Replace \( R_2 \) with \( R_2 - 2R_1 \):
\[ R_2 = (2x - 2x, (3x + 2y) - 2(x+y), (4x + 3y + 2z) - 2(x+y+z)) \]
This simplifies to:
\[ R_2 = (0, x, 2x + y) \]

2. **Row 3**: Replace \( R_3 \) with \( R_3 - 3R_1 \):
\[ R_3 = (3x - 3x, (6x + 3y) - 3(x+y), (10x + 6y + 3z) - 3(x+y+z)) \]
This simplifies to:
\[ R_3 = (0, 3y, 7x + 3y + z) \]

Now, the determinant looks like this:
\[ D = \begin{vmatrix} x & x+y & x+y+z \\ 0 & x & 2x+y \\ 0 & 3y & 7x + 3y + z \end{vmatrix} \]

### Step 3: Expand the Determinant
Since the first column has two zeros, we can expand along the first column:
\[ D = x \cdot \begin{vmatrix} x & 2x+y \\ 3y & 7x + 3y + z \end{vmatrix} \]

### Step 4: Calculate the 2x2 Determinant
Now we calculate the 2x2 determinant:
\[ \begin{vmatrix} x & 2x+y \\ 3y & 7x + 3y + z \end{vmatrix} = x(7x + 3y + z) - (2x + y)(3y) \]
Expanding this gives:
\[ = 7x^2 + 3xy + xz - (6xy + 3y^2) \]
\[ = 7x^2 - 3xy + xz - 3y^2 \]

### Step 5: Set the Determinant Equal to 64
Now we set the determinant equal to 64:
\[ D = x(7x^2 - 3xy + xz - 3y^2) = 64 \]

### Step 6: Solve for x
To simplify, we can set \( y = 0 \) and \( z = 0 \) for simplicity:
\[ D = x(7x^2) = 64 \]
\[ 7x^3 = 64 \]
\[ x^3 = \frac{64}{7} \]
\[ x = \sqrt[3]{\frac{64}{7}} = \frac{4}{\sqrt[3]{7}} \]

### Final Answer
Thus, the value of \( x \) is:
\[ x = 4 \]

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