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(B) Given the determinant equation: $\left| \begin{array}{ccc} a & b & a\alpha - b \ b & c & b\alpha - c \ 2 & 1 & 0 \end{array} ight| = 0$Expandi

Vedclass · Accepted Answer

$a, b, c$ are in $G.P.$

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(B) Given the determinant equation: $\left| \begin{array}{ccc} a & b & a\alpha - b \ b & c & b\alpha - c \ 2 & 1 & 0 \end{array} ight| = 0$Expanding along the third row:$2(b(b\alpha - c) - c(a\alpha - b)) - 1(a(b\alpha - c) - b(a\alpha - b)) + 0 = 0$$2(b^2\alpha - bc - ac\alpha + bc) - (ab\alpha - ac - ab\alpha + b^2) = 0$$2(b^2\alpha - ac\alpha) - (b^2 - ac) = 0$$2\alpha(b^2 - ac) - (b^2 - ac) = 0$$(2\alpha - 1)(b^2 - ac) = 0$Since it is given that $\alpha 
eq \frac{1}{2}$,then $2\alpha - 1 
eq 0$.Therefore,we must have $b^2 - ac = 0$,which implies $b^2 = ac$.This condition indicates that $a, b, c$ are in $G.P.$

If $\left| \begin{array}{ccc} a & b & a\alpha - b \\ b & c & b\alpha - c \\ 2 & 1 & 0 \end{array} \right| = 0$ and $\alpha \neq \frac{1}{2}$,then

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