If `abc!=0` then `|{:(1+a,1,1),(1,1+b,1),(1,1,1+c):}|` is

Options

A. `-1`
B. 1
C. abc
D. `a^(-1)b^(-1)c^(-1)`

Answer

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

### Step 1: Write down 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: Expand the determinant using the first row
Using the first row to expand the determinant, we have:
\[ D = (1+a) \begin{vmatrix} 1+b & 1 \\ 1 & 1+c \end{vmatrix} - 1 \begin{vmatrix} 1 & 1 \\ 1 & 1+c \end{vmatrix} + 1 \begin{vmatrix} 1 & 1+b \\ 1 & 1 \end{vmatrix} \]

### Step 3: Calculate the 2x2 determinants
Now, we calculate the 2x2 determinants:
1. For \( \begin{vmatrix} 1+b & 1 \\ 1 & 1+c \end{vmatrix} \):
\[ = (1+b)(1+c) - 1 \cdot 1 = 1 + b + c + bc - 1 = b + c + bc \]

2. For \( \begin{vmatrix} 1 & 1 \\ 1 & 1+c \end{vmatrix} \):
\[ = 1(1+c) - 1 \cdot 1 = 1 + c - 1 = c \]

3. For \( \begin{vmatrix} 1 & 1+b \\ 1 & 1 \end{vmatrix} \):
\[ = 1 \cdot 1 - 1(1+b) = 1 - (1+b) = -b \]

### Step 4: Substitute back into the determinant
Substituting these results back into the expression for \( D \):
\[ D = (1+a)(b+c+bc) - c - b \]

### Step 5: Expand and simplify
Now we expand \( D \):
\[ D = (1+a)(b+c+bc) - c - b = (b+c+bc) + a(b+c+bc) - c - b \]
\[ = b + c + bc + ab + ac + abc - c - b \]
The \( b \) and \( c \) terms cancel out:
\[ D = ab + ac + abc \]

### Step 6: Factor out \( abc \)
Now we can factor out \( abc \):
\[ D = abc \left( \frac{1}{a} + \frac{1}{b} + \frac{1}{c} + 1 \right) \]

### Final Result
Thus, the final result for the determinant is:
\[ D = abc \left( \frac{1}{a} + \frac{1}{b} + \frac{1}{c} + 1 \right) \]