What is the value of the determinant ?
`|(1,1,1),(1,1+xyz,1),(1,1,1+xyz)|` ?

Options

A. A. 1 + x + y + z
B. B. 2xyz
C. C. `x^(2) y^(2) z^(2)`
D. D. `2x^(2) y^(2) z^(2)`

Answer

To find the value of the determinant of the matrix

\[ \begin{vmatrix} 1 & 1 & 1 \\ 1 & 1 + xyz & 1 \\ 1 & 1 & 1 + xyz \end{vmatrix} \]

we can follow these steps:

### Step 1: Write the matrix
We start with the given matrix:

\[ \begin{pmatrix} 1 & 1 & 1 \\ 1 & 1 + xyz & 1 \\ 1 & 1 & 1 + xyz \end{pmatrix} \]

### Step 2: Simplify the matrix
We can simplify the determinant by performing row operations. We will subtract the first row from the second and third rows:

1. R2 = R2 - R1
2. R3 = R3 - R1

After performing these operations, we have:

\[ \begin{pmatrix} 1 & 1 & 1 \\ 1 - 1 & (1 + xyz) - 1 & 1 - 1 \\ 1 - 1 & 1 - 1 & (1 + xyz) - 1 \end{pmatrix} \]

This simplifies to:

\[ \begin{pmatrix} 1 & 1 & 1 \\ 0 & xyz & 0 \\ 0 & 0 & xyz \end{pmatrix} \]

### Step 3: Calculate the determinant
Now we can calculate the determinant of the simplified matrix. The determinant of a 3x3 matrix can be calculated using the formula:

\[ \text{det}(A) = a(ei - fh) - b(di - fg) + c(dh - eg) \]

For our matrix, we can expand along the first row:

\[ \text{det} = 1 \cdot (xyz \cdot xyz - 0) - 1 \cdot (0 - 0) + 1 \cdot (0 - 0) \]

This simplifies to:

\[ \text{det} = 1 \cdot (xyz)^2 = (xyz)^2 \]

### Step 4: Final result
Thus, the value of the determinant is:

\[ \text{det} = x^2 y^2 z^2 \]

### Conclusion
The value of the determinant is \( x^2 y^2 z^2 \).

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