Question

The determinant$\left| \begin{matrix} a & b & a\alpha + b \\ b & c & b\alpha + c \\ a\alpha + b & b\alpha + c & 0 \end{matrix} \right| = 0$, if $a,b,c$are in.

• A. A. P.
• B. G. P.
• C. H. P.
• D. None of these

Answer 1

Answer: (B)

G. P.

Answer 2

$\Delta \equiv \left| \begin{matrix} a & b & a\alpha + b \\ b & c & b\alpha + c \\ a\alpha + b & b\alpha + c & 0 \end{matrix} \right|$

= $\left| \begin{matrix} a & b & a\alpha + b \\ b & c & b\alpha + c \\ 0 & 0 & - (a\alpha^{2} + 2b\alpha + c) \end{matrix} \right|$, by $R_{3} \rightarrow R_{3} - \alpha R_{1} - R_{2}$

= $a\{ - c(a\alpha^{2} + 2b\alpha + c) - 0\} - b\{ - b(a\alpha^{2} + 2b\alpha + c) - 0\}$

by expanding along $C_{1}$

$= (b^{2} - ac)(a\alpha^{2} + 2b\alpha + c)$

Thus, $\Delta = 0$, if either $b^{2} - ac = 0$or $a\alpha^{2} + 2b\alpha + c = 0$

i.e., $a,b,c$ in G.P. or $a\alpha^{2} + 2b\alpha + c = 0$.

Trick: Put $\alpha = 0$, then the determinant

$\left| \begin{matrix} a & b & b \\ b & c & c \\ b & c & 0 \end{matrix} \right| = \left| \begin{matrix} a & b & 0 \\ b & c & 0 \\ b & c & - c \end{matrix} \right| = - c(ac - b^{2}) = 0$. Hence the result.