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Question:
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If the following three linear equations have a non-trivial solution, then
$$
\begin{array}{l}
x+4 a y+a z=0 \\
x+3 b y+b z=0 \\
x+2 c y+c z=0
\end{array}
$$
Options:
$$
\begin{array}{l}
x+4 a y+a z=0 \\
x+3 b y+b z=0 \\
x+2 c y+c z=0
\end{array}
$$
Solution:
2615 Upvotes
Verified Answer
The correct answer is:
$a, b, c$ are in $\mathrm{HP}$
For non-trivial solution, We have, $\left|\begin{array}{ccc}1 & 4 a & a \\ 1 & 3 b & b \\ 1 & 2 c & c\end{array}\right|=0$
$\Rightarrow \quad 1(3 b c-2 b c)-(4 a c-2 a c)+1(4 a b-3 a b)=0$
$\Rightarrow b c-2 a c+a b=0$
$\Rightarrow b c+a b=2 a c$
$\Rightarrow b(a+c)=2 a c$
$\Rightarrow b=\frac{2 a c}{a+c}$
$\Rightarrow a, b, c$ are in $\mathrm{HP}$
$\Rightarrow \quad 1(3 b c-2 b c)-(4 a c-2 a c)+1(4 a b-3 a b)=0$
$\Rightarrow b c-2 a c+a b=0$
$\Rightarrow b c+a b=2 a c$
$\Rightarrow b(a+c)=2 a c$
$\Rightarrow b=\frac{2 a c}{a+c}$
$\Rightarrow a, b, c$ are in $\mathrm{HP}$
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