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

If the matrix $\begin{bmatrix} 0 & 1 & -2 \\ -1 & 0 & 3 \\ \lambda & -3 & 0 \end{bmatrix}$ is singular,then $\lambda = $

$\left| {\begin{array}{*{20}{c}}1&1&1\\1&{1 + x}&1\\1&1&{1 + y}\end{array}} \right| = $

Let $S = \left\{ A = \begin{bmatrix} 0 & 1 & c \\ 1 & a & d \\ 1 & b & e \end{bmatrix} : a, b, c, d, e \in \{0, 1\} \text{ and } |A| \in \{-1, 1\} \right\}$,where $|A|$ denotes the determinant of $A$. Then the number of elements in $S$ is:

If the area of a triangle is $35$ sq. units with vertices $(2, -6)$,$(5, 4)$,and $(k, 4)$,then $k$ is . . . . . . .

If $\left|\begin{array}{cc}\sin ^2 \theta & \cos ^2 \theta \\ -\cos ^2 \theta & \sin ^2 \theta\end{array}\right| = $ . . . . . . .

Vedclass Test Series

Exam Paper Generator

Online Exam Module