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3 and 4 .Determinants and Matrices
hard
If the system of linear equation $x + 2ay + az = 0,$ $x + 3by + bz = 0,$ $x + 4cy + cz = 0$ has a non zero solution, then $a,b,c$
A
Are in A.P.
B
Are in G. P.
C
Are in H. P.
D
Satisfy $a + 2b + 3c = 0$
(AIEEE-2003)
Solution
(c) $\left| {\,\begin{array}{*{20}{c}}1&{2a}&a\\1&{3b}&b\\1&{4c}&c\end{array}\,} \right|\, = 0\,$, $[{C_2} \to {C_2} – 2{C_3}]$
==>$\left| {\,\begin{array}{*{20}{c}}1&0&a\\1&b&b\\1&{2c}&c\end{array}\,} \right| = 0$, $[{R_3} \to {R_3} – {R_2},\,{R_2} \to {R_2} – {R_1}]$
==> $\left| {\,\begin{array}{*{20}{c}}1&0&a\\0&b&{b – a}\\0&{2c – b}&{c – b}\end{array}\,} \right|\, = 0$ ; $b(c – b) – (b – a)\,(2c – b) = 0$
On simplification, $\frac{2}{b} = \frac{1}{a} + \frac{1}{c}$
$\therefore$ $ a, b, c$ are in Harmonic progression.
Standard 12
Mathematics
