If left| begin{array}{ccc} a & b & aalpha - b | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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 $G.P.$

Vedclass · Answer

(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

Similar Questions

Three points $(p + 1, 1)$,$(2p + 1, 3)$ and $(2p + 2, 2p)$ are collinear,if $p =$

If the vertices of a triangle are $(-2, 6)$,$(3, -6)$,and $(1, 5)$,then the area of the triangle is

If $P, Q$ and $R$ are angles of $\Delta PQR$, then the value of $\left|\begin{array}{ccc}-1 & \cos R & \cos Q \\ \cos R & -1 & \cos P \\ \cos Q & \cos P & -1\end{array}\right|$ is equal to

If $x$ is a positive integer,then $\Delta = \left| \begin{array}{ccc} x! & (x+1)! & (x+2)! \\ (x+1)! & (x+2)! & (x+3)! \\ (x+2)! & (x+3)! & (x+4)! \end{array} \right|$ is equal to

If $\alpha, \beta, \gamma$ are the roots of the equation $\left|\begin{array}{ccc}x & 2 & 2 \\ 2 & x & 2 \\ 2 & 2 & x\end{array}\right|=0$ and $\min (\alpha, \beta, \gamma)=\alpha$, then $2 \alpha+3 \beta+4 \gamma=$

Vedclass Test Series

Exam Paper Generator

Online Exam Module