If `a != 0, b!= 0, c!= 0`, then
`|(1 +a,1,1),(1,1 +b,1),(1,1,1 +c)|` is equal to

Options

A. abc
B. `abc (1 + (1)/(a) + (1)/(b) + (1)/(c))`
C. 0
D. `1 + (1)/(a) + (1)/(b) + (1)/(c)`

Answer

To solve the determinant \( |(1 + a, 1, 1), (1, 1 + b, 1), (1, 1, 1 + c)| \), we will follow these steps:

### Step 1: Write the determinant
The determinant can be expressed as:
\[ D = \begin{vmatrix} 1 + a & 1 & 1 \\ 1 & 1 + b & 1 \\ 1 & 1 & 1 + c \end{vmatrix} \]

### Step 2: Perform column operations
We will perform column operations to simplify the determinant. Let's subtract the second column from the first column:
\[ C_1 \rightarrow C_1 - C_2 \]
This gives us:
\[ D = \begin{vmatrix} (1 + a - 1) & 1 & 1 \\ (1 - (1 + b)) & (1 + b) & 1 \\ (1 - 1) & 1 & (1 + c) \end{vmatrix} \]
Simplifying this, we have:
\[ D = \begin{vmatrix} a & 1 & 1 \\ -b & 1 + b & 1 \\ 0 & 1 & 1 + c \end{vmatrix} \]

### Step 3: Perform another column operation
Next, we will subtract the third column from the second column:
\[ C_2 \rightarrow C_2 - C_3 \]
This gives us:
\[ D = \begin{vmatrix} a & 0 & 1 \\ -b & b & 1 \\ 0 & 0 & 1 + c \end{vmatrix} \]

### Step 4: Expand the determinant
Now we can expand the determinant along the first row:
\[ D = a \begin{vmatrix} b & 1 \\ 0 & 1 + c \end{vmatrix} \]
Calculating the 2x2 determinant:
\[ D = a \cdot (b \cdot (1 + c) - 0) = ab(1 + c) \]

### Step 5: Final expression
Thus, the determinant simplifies to:
\[ D = ab(1 + c) \]

### Step 6: Factor out \(abc\)
We can express this as:
\[ D = abc \left(\frac{1}{c} + \frac{1}{b} + \frac{1}{a} + 1\right) \]

### Final Answer
The final answer for the determinant is:
\[ D = abc \left(1 + \frac{1}{a} + \frac{1}{b} + \frac{1}{c}\right) \]