`|(1+a,1,1),(1,1+b,1),(1,1,1+c)|= `

Options

A. `1+suma`
B. `1+sum(1)/a`
C. `abc [1+sum(1)/a]`
D. none

Answer

To solve the determinant \( D = \begin{vmatrix} 1+a & 1 & 1 \\ 1 & 1+b & 1 \\ 1 & 1 & 1+c \end{vmatrix} \), we will follow these steps:

### Step 1: Rewrite the Determinant
We start with the determinant:
\[ D = \begin{vmatrix} 1+a & 1 & 1 \\ 1 & 1+b & 1 \\ 1 & 1 & 1+c \end{vmatrix} \]

### Step 2: Apply Row Operations
We can simplify the determinant by performing row operations. Specifically, we can subtract the third row from the first and second rows:
\[ R_1 \rightarrow R_1 - R_3 \quad \text{and} \quad R_2 \rightarrow R_2 - R_3 \]
This gives us:
\[ D = \begin{vmatrix} a & 0 & 0 \\ 0 & b & 0 \\ 1 & 1 & 1+c \end{vmatrix} \]

### Step 3: Expand the Determinant
Now we can expand the determinant using the first column:
\[ D = a \begin{vmatrix} b & 0 \\ 1 & 1+c \end{vmatrix} \]
Calculating the 2x2 determinant:
\[ \begin{vmatrix} b & 0 \\ 1 & 1+c \end{vmatrix} = b(1+c) - 0 = b(1+c) \]
Thus,
\[ D = a \cdot b(1+c) \]

### Step 4: Final Expression
Therefore, the final expression for the determinant is:
\[ D = ab(1+c) \]

### Summary of Steps:
1. Rewrite the determinant.
2. Apply row operations to simplify.
3. Expand the determinant using the first column.
4. Calculate the resulting 2x2 determinant.
5. Combine the results to get the final expression.