Value of `|{:(x-y-z,2x,2x),(2y,y-z-x,2y),(2z,2z,z-x-y):}|`

Options

A. `xyz(x+y+z)^2`
B. `(x+y+z)(x+y+z)^2`
C. `(x+y+z)^3`
D. `(x+y+z)^2`

Answer

To find the value of the determinant \( |{:(x-y-z, 2x, 2x),(2y, y-z-x, 2y),(2z, 2z, z-x-y):}| \), we will follow these steps:

### Step 1: Write down the determinant
We start with the given matrix:
\[ A = \begin{pmatrix} x - y - z & 2x & 2x \\ 2y & y - z - x & 2y \\ 2z & 2z & z - x - y \end{pmatrix} \]

### Step 2: Perform row operations
We will add all three rows together and replace the first row with this sum. This simplifies our calculations:
\[ R_1 \rightarrow R_1 + R_2 + R_3 \]
Calculating the new first row:
\[ R_1 = (x - y - z + 2y + 2z, 2x + (y - z - x) + 2z, 2x + 2y + (z - x - y)) \]
This simplifies to:
\[ R_1 = (x + y + z, x + y + z, x + y + z) \]
Thus, the matrix becomes:
\[ A = \begin{pmatrix} x + y + z & x + y + z & x + y + z \\ 2y & y - z - x & 2y \\ 2z & 2z & z - x - y \end{pmatrix} \]

### Step 3: Factor out common terms
Now we can factor out \( (x + y + z) \) from the first row:
\[ |A| = (x + y + z) \cdot \begin{vmatrix} 1 & 1 & 1 \\ 2y & y - z - x & 2y \\ 2z & 2z & z - x - y \end{vmatrix} \]

### Step 4: Simplify the determinant
Next, we will perform column operations to simplify the determinant:
\[ C_2 \rightarrow C_2 - C_1 \quad \text{and} \quad C_3 \rightarrow C_3 - C_1 \]
This gives us:
\[ |A| = (x + y + z) \cdot \begin{vmatrix} 1 & 0 & 0 \\ 2y & (y - z - x) - 2y & 2y \\ 2z & 2z - 2z & (z - x - y) - 2z \end{vmatrix} \]
This simplifies to:
\[ |A| = (x + y + z) \cdot \begin{vmatrix} 1 & 0 & 0 \\ 2y & -z - x - y & 2y \\ 2z & 0 & -x - y - z \end{vmatrix} \]

### Step 5: Calculate the determinant
Now we can calculate the determinant:
\[ |A| = (x + y + z) \cdot 1 \cdot \begin{vmatrix} -z - x - y & 2y \\ 0 & -x - y - z \end{vmatrix} \]
Calculating this determinant:
\[ = (x + y + z) \cdot (-z - x - y)(-x - y - z) - 0 = (x + y + z)^2 \]

### Final Result
Thus, the value of the determinant is:
\[ |A| = (x + y + z)^3 \]