If the system of linear equations x + 2ay + az = 0; [x + 3by + bz = 0;,,x + 4cy + cz = 0;]…

Question

If the system of linear equations

$x + 2ay + az = 0;$ $$x + 3by + bz = 0;\,\,x + 4cy + cz = 0;$$

has a non - zero solution, then $a, b, c$.

Accepted Answer

For homogeneous system of equations to have non zero solution, $$\Delta  = 0$$

$$\left| {\begin{pmatrix} 1 & {2a} & a  \
   1 & {3b} & b  \
   1 & {4c} & c  \

\end{pmatrix} } \right| = 0\,{C_2} \to {C_2} - 2{C_3}$$

$$\left| {\begin{pmatrix} 1 & 0 & a  \
   1 & b & b  \
   1 & {2c} & c  \

\end{pmatrix} } \right| = 0\,\,{R_3} \to {R_3} - {R_2},{R_2} \to {R_2} - {R_1}$$

$$\left| {\begin{pmatrix} 1 & 0 & a  \
   0 & b & {b - a}  \
   0 & {2c - b} & {c - b}  \

\end{pmatrix} } \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.

Solutions · 1