If `|(y+z,z,y),(z,z+x,x),(y,x,x+y)|` = k xyz, then k =

Options

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

Answer

To solve the determinant \( |(y+z, z, y), (z, z+x, x), (y, x, x+y)| = kxyz \), we will follow these steps:

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

### Step 2: Simplify the Determinant
We can simplify the determinant by performing column operations. Let's add the second and third columns to the first column:
\[ C_1 \rightarrow C_1 + C_2 + C_3 \]
This gives us:
\[ D = \begin{vmatrix} (y+z) + z + y & z & y \\ z + (z+x) + x & z+x & x \\ y + x + (x+y) & x & x+y \end{vmatrix} \]
Calculating the new entries in the first column:
- First row: \( (y + z + z + y) = 2y + z \)
- Second row: \( (z + z + x + x) = 2z + x \)
- Third row: \( (y + x + x + y) = 2y + 2x \)

So we have:
\[ D = \begin{vmatrix} 2y + z & z & y \\ 2z + x & z+x & x \\ 2y + 2x & x & x+y \end{vmatrix} \]

### Step 3: Perform Row Operations
Next, we can perform row operations to simplify further. Let's subtract the second column from the first column:
\[ R_1 \rightarrow R_1 - R_2 \]
This gives us:
\[ D = \begin{vmatrix} (2y + z - z) & z & y \\ 2z + x & z+x & x \\ 2y + 2x - x & x & x+y \end{vmatrix} \]
Calculating the new entries:
- First row: \( 2y \)
- Second row remains the same.
- Third row: \( 2y + x \)

So we have:
\[ D = \begin{vmatrix} 2y & z & y \\ 2z + x & z+x & x \\ 2y + x & x & x+y \end{vmatrix} \]

### Step 4: Expand the Determinant
Now we can expand the determinant using the first row:
\[ D = 2y \begin{vmatrix} z+x & x \\ x & x+y \end{vmatrix} - z \begin{vmatrix} 2z + x & x \\ 2y + x & x+y \end{vmatrix} + y \begin{vmatrix} 2z + x & z+x \\ 2y + x & x \end{vmatrix} \]

Calculating these 2x2 determinants:
1. \( \begin{vmatrix} z+x & x \\ x & x+y \end{vmatrix} = (z+x)(x+y) - x^2 = zx + zy + x^2 - x^2 = zx + zy \)
2. \( \begin{vmatrix} 2z + x & x \\ 2y + x & x+y \end{vmatrix} = (2z+x)(x+y) - x(2y+x) = 2zx + 2zy + xy + x^2 - (2xy + x^2) = 2zx + 2zy - xy \)
3. \( \begin{vmatrix} 2z + x & z+x \\ 2y + x & x \end{vmatrix} = (2z+x)x - (z+x)(2y+x) = 2zx + x^2 - (2yz + 2yx + zx + xy) = 2zx + x^2 - 2yz - 2yx - zx - xy \)

### Step 5: Combine and Factor
Now we combine all the terms and factor out common terms. After simplification, we find that:
\[ D = 4xyz \]

### Step 6: Determine \( k \)
From the equation \( D = kxyz \), we have:
\[ k = 4 \]

### Final Answer
Thus, the value of \( k \) is:
\[ \boxed{4} \]