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3 and 4 .Determinants and Matrices
medium
If $\left| {\,\begin{array}{*{20}{c}}a&b&{a\alpha - b}\\b&c&{b\alpha - c}\\2&1&0\end{array}\,} \right| = 0$ and $\alpha \ne \frac{1}{2},$ then
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
None of these
Solution
(b) $\left| {\,\begin{array}{*{20}{c}}a&b&{a\alpha – b}\\b&c&{b\alpha – c}\\2&1&0\end{array}\,} \right| = 0$
==> $a[ – (b\alpha – c)] – b[ – 2(b\alpha – c)] + [a\alpha – b)(b – 2c)] = 0$
==>$ – ab\alpha + ac + 2{b^2}\alpha – 2bc + ab\alpha – 2ac\alpha – {b^2} + 2bc = 0$
==> $ac + 2{b^2}\alpha – 2ac\alpha – {b^2} = 0$
==> $(ac – {b^2}) – 2\alpha (ac – {b^2}) = 0$
==> $ac – {b^2} = 0$or $1 – 2\alpha = 0$ $ \Rightarrow $ ${b^2} = ac$ or $\alpha = \frac{1}{2}$
(As given in question)
So, ${b^2} = ac$ i.e, $a,b,c$ are in $G.P.$
Standard 12
Mathematics
