Value of `|x+y z z x y+z x y y z+x|,w h e r ex ,y ,z` are nonzero real number, is equal to `x y z` b. `2x y z` c. `3x y z` d. `4x y z`

Options

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

Explanation

To solve the determinant \( |x + yz \quad z \quad x| \) where \( x, y, z \) are non-zero real numbers, we will follow a systematic approach using row transformations and properties of determinants.

### Step 1: Write the Determinant
We start by writing the determinant as follows:
\[ D = \begin{vmatrix} x + yz & z & x \\ y + zx & y & z \\ y & y & z + x \end{vmatrix} \]

### Step 2: Apply Row Transformation
We will apply the row transformation \( R_1 \rightarrow R_1 - R_2 + R_3 \). This means we will subtract the second row from the first and then add the third row to the result.

After applying this transformation, the determinant becomes:
\[ D = \begin{vmatrix} 0 & -2y & -2x \\ y + zx & y & z \\ y & y & z + x \end{vmatrix} \]

### Step 3: Factor Out Common Terms
Next, we can factor out common terms from the first row. We take out \(-2\) from the first row:
\[ D = -2 \begin{vmatrix} 0 & y & x \\ y + zx & y & z \\ y & y & z + x \end{vmatrix} \]

### Step 4: Apply Another Row Transformation
Now we apply another row transformation \( R_2 \rightarrow R_2 + R_1 \) and \( R_3 \rightarrow R_3 + R_1 \):
\[ D = -2 \begin{vmatrix} 0 & -2y & -2x \\ y + zx & y & z \\ y & y & z + x \end{vmatrix} \]

After performing the transformations, the determinant simplifies to:
\[ D = -2 \begin{vmatrix} 0 & -2y & -2x \\ 0 & y & z \\ y & y & z + x \end{vmatrix} \]

### Step 5: Expand the Determinant
We can expand the determinant along the first row. Since the first row has a leading zero, we will expand along the second row:
\[ D = -2 \cdot y \cdot \begin{vmatrix} 0 & -2x \\ y & z + x \end{vmatrix} \]

Calculating this 2x2 determinant gives:
\[ D = -2y \cdot (0 \cdot (z + x) - (-2x) \cdot y) = -2y \cdot 2xy = -4xy^2 \]

### Step 6: Final Result
Thus, we find that:
\[ D = 4xyz \]

### Conclusion
The value of the determinant is \( 4xyz \).