If `|(1+a,1,1),(1+b,1+2b,1),(1+c,1+c,1+3c)|` = 0 , where , ` a ne 0 , b ne 0 , c ne 0 and a^(-1) + b^(-1) +c^(-1)` is

Options

A. 4
B. `-3`
C. `-2`
D. `-1`

Answer

To solve the determinant equation given by

\[ \begin{vmatrix} 1 + a & 1 & 1 \\ 1 + b & 1 + 2b & 1 \\ 1 + c & 1 + c & 1 + 3c \end{vmatrix} = 0 \]

we will follow these steps:

### Step 1: Simplify the Determinant
We will perform column operations to simplify the determinant. Specifically, we will subtract the third column from the first and the second columns.

\[ C_1 \rightarrow C_1 - C_3 \quad \text{and} \quad C_2 \rightarrow C_2 - C_3 \]

This gives us:

\[ \begin{vmatrix} (1 + a - 1) & (1 - 1) & 1 \\ (1 + b - 1) & (1 + 2b - 1) & 1 \\ (1 + c - (1 + 3c)) & (1 + c - (1 + 3c)) & (1 + 3c - 1) \end{vmatrix} \]

This simplifies to:

\[ \begin{vmatrix} a & 0 & 1 \\ b & 2b & 1 \\ -c & -2c & 3c \end{vmatrix} \]

### Step 2: Expand the Determinant
Now we will expand the determinant along the first column:

\[ = a \begin{vmatrix} 2b & 1 \\ -2c & 3c \end{vmatrix} - 0 + 1 \begin{vmatrix} b & 2b \\ -c & -2c \end{vmatrix} \]

Calculating the first determinant:

\[ = a (2b \cdot 3c - 1 \cdot (-2c)) = a (6bc + 2c) = a (6bc + 2c) \]

Calculating the second determinant:

\[ = 1 (b \cdot (-2c) - 2b \cdot (-c)) = -2bc + 2bc = 0 \]

Thus, we have:

\[ = a(6bc + 2c) = 0 \]

### Step 3: Set the Expression to Zero
Since \( a \neq 0 \), we must have:

\[ 6bc + 2c = 0 \]

Factoring out \( c \):

\[ c(6b + 2) = 0 \]

Since \( c \neq 0 \), we have:

\[ 6b + 2 = 0 \implies b = -\frac{1}{3} \]

### Step 4: Substitute Back to Find \( a^{-1} + b^{-1} + c^{-1} \)
Now we need to find \( a^{-1} + b^{-1} + c^{-1} \):

\[ a^{-1} + b^{-1} + c^{-1} = \frac{1}{a} + \frac{1}{-\frac{1}{3}} + \frac{1}{c} \]

This simplifies to:

\[ = \frac{1}{a} - 3 + \frac{1}{c} \]

### Step 5: Conclusion
Since we do not have specific values for \( a \) and \( c \), we cannot simplify further without additional information. However, we can conclude that:

\[ a^{-1} + b^{-1} + c^{-1} = -3 + \frac{1}{a} + \frac{1}{c} \]