Question

If the system of linear equations

$$\begin{gathered} 2x+2ay+az=0 \\ 2x+3by+bz=0 \\ 2x+4cy+cz=0 \end{gathered}$$

where $a,b,c\in R$ are non-zero and distinct; has non-zero solution, then

• A. $a+b+c=0$
• B. $a,b,c$ are in AP
• C. $\frac{1}{a},\frac{1}{b},\frac{1}{c}$ are in AP
• D. $a,b,c$ are in GP

Answer 1

The correct option is C

$\frac{1}{a},\frac{1}{b},\frac{1}{c}$ are in AP

Explanation for the correct option.

Step 1. Form the determinant.

The given equations are

$$\begin{gathered} 2x+2ay+az=0 \\ 2x+3by+bz=0 \\ 2x+4cy+cz=0 \end{gathered}$$.

As the system has non-zero solutions, so $D=0$

$\left|\begin{array}{ccc}2& 2a& a\\ 2& 3b& b\\ 2& 4c& c\end{array}\right|=0$

Now using row operations: $R_2\to R_2-R_1$ and $R_3\to R_3-R_1$

$\left|\begin{array}{ccc}2& 2a& a\\ 0& 3b-2a& b-a\\ 0& 4c-2a& c-a\end{array}\right|=0$

Step 2. Find the relation between $a,b,c$.

Solve the determinant.

$$\begin{gathered} 2\left[\left(3b-2a\right)\left(c-a\right)-\left(b-a\right)\left(4c-2a\right)\right]=0 \\ \Rightarrow 3bc-3ab-2ac+2a^2-4bc+2ab+4ac-2a^2=0 \\ \Rightarrow -bc-ab+2ac=0 \\ \Rightarrow bc+ab=2ac \end{gathered}$$

Divide both sides by $abc$.

$$\begin{gathered} \frac{bc+ab}{abc}=\frac{2ac}{abc} \\ \Rightarrow \frac{1}{a}+\frac{1}{c}=\frac{2}{b} \end{gathered}$$

Now as $\frac{1}{a}+\frac{1}{c}=\frac{2}{b}$, so $\frac{1}{a},\frac{1}{b},\frac{1}{c}$ are in AP.

Hence, the correct option is C.