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If a , B and C Are All Non-zero and ∣ ∣ ∣ ∣ 1 + a 1 1 1 1 + B 1 1 1 1 + C ∣ ∣ ∣ ∣ = 0, Then Prove that 1 a + 1 B + 1 C + 1 = 0

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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

 

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Solution

We have,

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

\[C_1 \to C_1 - C_2 \] 

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

\[ C_2 \to C_2 - C_3 \] 

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

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

\[a(b + bc + c) + 1(bc) = 0\] 

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

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

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

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Chapter 5: Determinants - Exercise 6.2 [Page 62]

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R.D. Sharma Mathematics Volume 1 and 2 [English] Class 12
Chapter 5 Determinants
Exercise 6.2 | Q 53 | Page 62

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