The value of `|{:(1," "1," "1),(1,1+x," "1),(1," "1,1+y):}|`is

Options

A. x + y
B. x - y
C. xy
D. 1 + x + y

Explanation

To find the value of the determinant

\[ D = \begin{vmatrix} 1 & 1 & 1 \\ 1 & 1+x & 1 \\ 1 & 1 & 1+y \end{vmatrix} \]

we will follow these steps:

### Step 1: Write down the determinant
We start with the determinant as given:

\[ D = \begin{vmatrix} 1 & 1 & 1 \\ 1 & 1+x & 1 \\ 1 & 1 & 1+y \end{vmatrix} \]

### Step 2: Apply row operations
We can simplify the determinant by performing row operations. Let's subtract the first row from the second and third rows:

- Row 2 = Row 2 - Row 1
- Row 3 = Row 3 - Row 1

After performing these operations, the determinant becomes:

\[ D = \begin{vmatrix} 1 & 1 & 1 \\ 0 & x & 0 \\ 0 & 0 & y \end{vmatrix} \]

### Step 3: Calculate the determinant
Now we can calculate the determinant using the properties of determinants. The determinant of a triangular matrix is the product of its diagonal elements:

\[ D = 1 \cdot x \cdot y = xy \]

### Final Result
Thus, the value of the determinant is:

\[ D = xy \]