If the system of equations (alpha + 1)^3 x + | vedclass

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

Question

(D) For the system of equations to be consistent,the determinant of the coefficient matrix must be zero.The system is:$(\alpha + 1)^3 x + (\alpha + 2)

Vedclass · Accepted Answer

$-2$

Vedclass · Answer

(D) For the system of equations to be consistent,the determinant of the coefficient matrix must be zero.The system is:$(\alpha + 1)^3 x + (\alpha + 2)^3 y = (\alpha + 3)^3$$(\alpha + 1) x + (\alpha + 2) y = (\alpha + 3)$$x + y = 1$The determinant of the augmented matrix must be zero:$\begin{vmatrix} (\alpha + 1)^3 & (\alpha + 2)^3 & -(\alpha + 3)^3 \ \alpha + 1 & \alpha + 2 & -(\alpha + 3) \ 1 & 1 & -1 \end{vmatrix} = 0$Applying column operations $C_1 	o C_1 + C_2$ and $C_2 	o C_2 + C_3$ (or evaluating directly),we find:$6\alpha + 12 = 0$$6\alpha = -12$$\alpha = -2$Thus,the correct option is $D$.

If the system of equations $(\alpha + 1)^3 x + (\alpha + 2)^3 y - (\alpha + 3)^3 = 0$,$(\alpha + 1)x + (\alpha + 2)y - (\alpha + 3) = 0$,and $x + y - 1 = 0$ is consistent,what is the value of $\alpha$?

Similar Questions

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

Find the equation of the line joining $(1, 2)$ and $(3, 6)$ using determinants.

$\left| {\begin{array}{*{20}{c}}a&b&c\\b&c&a\\c&a&b\end{array}} \right| = $

The area of a triangle with vertices $(K, 0), (4, 0), (0, 2)$ is $4$ square units. Find the value of $K$.

The area of $\triangle PQR$ with the vertices $P(k, 1)$,$Q(2, 4)$,and $R(1, 1)$ is $3$ sq. units. Then,$k = $ . . . . . . .

Vedclass Test Series

Exam Paper Generator

Online Exam Module