Question

If \[a, b\] and c  are all non-zero and 

\[\begin{vmatrix}1 + a & 1 & 1 \\ 1 & 1 + b & 1 \\ 1 & 1 & 1 + c\end{vmatrix} =\] 0, then prove that 

\[\frac{1}{a} + \frac{1}{b} + \frac{1}{c} +\]1

= 0

 

Answer 1

We have,

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

\[C_1 \to C_1 - C_2 \] 

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

\[ C_2 \to C_2 - C_3 \] 

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

\[\text{ Expanding along }R_1 , \text{ we get }\] 

\[a(b + bc + c) + 1(bc) = 0\] 

\[ \Rightarrow ab + abc + ac + bc = 0\] 

\[\text{ Dividing by abc, we get }\] 
 
\[\frac{1}{c} + 1 + \frac{1}{b} + \frac{1}{a} = 0\] 

\[ \therefore \frac{1}{a} + \frac{1}{b} + \frac{1}{c} + 1 = 0\]