Question

If the following three linear equations have a non-trivial solution, then
$x+4ay+az=0x+3by+bz=0x+2cy+cz=0$

• A. $a,b,c$ are in A.P.
• B. $a,b,c$ are in G.P.
• C. $a,b,c$ are in H.P.
• D. $a+b+c=0$

Answer 1

The correct option is C $a,b,c$ are in H.P.

For non trivial solution the determinant of the matrix formed by the
co-efficient matrix of the linear equations should be equal to zero.
$\Rightarrow \left|\begin{array}{ccc}1& 4a& a\\ 1& 3b& b\\ 1& 2c& c\end{array}\right|=0$
Expanding along Column $C_1$, we get
$1(3bc-2bc)-1(4ac-2ac)+(4ab-3ab)=0\Rightarrow bc-2ac+ab=0\Rightarrow bc+ab=2ac\Rightarrow b=\frac{2ac}{a+c}\Rightarrow \frac{1}{a}+\frac{1}{c}=\frac{2}{b}$
which tells us that $a,b,c$ are in H.P.
Hence, Option c. is correct.