JEE MAIN - Mathematics (2003 - No. 17)
If the system of linear equations
$$x + 2ay + az = 0;$$ $$x + 3by + bz = 0;\,\,x + 4cy + cz = 0;$$
has a non - zero solution, then $$a, b, c$$.
$$x + 2ay + az = 0;$$ $$x + 3by + bz = 0;\,\,x + 4cy + cz = 0;$$
has a non - zero solution, then $$a, b, c$$.
satisfy $$a+2b+3c=0$$
are in A.P
are in G.P
are in H.P.
Explanation
For homogeneous system of equations to have non zero solution, $$\Delta = 0$$
$$\left| {\matrix{ 1 & {2a} & a \cr 1 & {3b} & b \cr 1 & {4c} & c \cr } } \right| = 0\,{C_2} \to {C_2} - 2{C_3}$$
$$\left| {\matrix{ 1 & 0 & a \cr 1 & b & b \cr 1 & {2c} & c \cr } } \right| = 0\,\,{R_3} \to {R_3} - {R_2},{R_2} \to {R_2} - {R_1}$$
$$\left| {\matrix{ 1 & 0 & a \cr 0 & b & {b - a} \cr 0 & {2c - b} & {c - b} \cr } } \right| = 0$$
$$b\left( {c - b} \right) - \left( {b - a} \right)\left( {2c - b} \right) = 0$$
On simplification, $${2 \over b} = {1 \over a} + {1 \over c}$$
$$\therefore$$ $$a,b,c$$ are in Harmonic Progression.
$$\left| {\matrix{ 1 & {2a} & a \cr 1 & {3b} & b \cr 1 & {4c} & c \cr } } \right| = 0\,{C_2} \to {C_2} - 2{C_3}$$
$$\left| {\matrix{ 1 & 0 & a \cr 1 & b & b \cr 1 & {2c} & c \cr } } \right| = 0\,\,{R_3} \to {R_3} - {R_2},{R_2} \to {R_2} - {R_1}$$
$$\left| {\matrix{ 1 & 0 & a \cr 0 & b & {b - a} \cr 0 & {2c - b} & {c - b} \cr } } \right| = 0$$
$$b\left( {c - b} \right) - \left( {b - a} \right)\left( {2c - b} \right) = 0$$
On simplification, $${2 \over b} = {1 \over a} + {1 \over c}$$
$$\therefore$$ $$a,b,c$$ are in Harmonic Progression.
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