The system of equations begin{cases} alpha x | vedclass

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

Question

(C) For a system of linear equations to have no solution or infinitely many solutions,the determinant of the coefficient matrix $D$ must be zero.$D =

Vedclass · Accepted Answer

$-2$

Vedclass · Answer

(C) For a system of linear equations to have no solution or infinitely many solutions,the determinant of the coefficient matrix $D$ must be zero.$D = \begin{vmatrix} \alpha & 1 & 1 \ 1 & \alpha & 1 \ 1 & 1 & \alpha \end{vmatrix} = 0$Expanding the determinant:$\alpha(\alpha^2 - 1) - 1(\alpha - 1) + 1(1 - \alpha) = 0$$\alpha(\alpha - 1)(\alpha + 1) - 1(\alpha - 1) - 1(\alpha - 1) = 0$$(\alpha - 1)[\alpha(\alpha + 1) - 2] = 0$$(\alpha - 1)(\alpha^2 + \alpha - 2) = 0$$(\alpha - 1)(\alpha + 2)(\alpha - 1) = 0$$(\alpha - 1)^2(\alpha + 2) = 0$So,$\alpha = 1$ or $\alpha = -2$.If $\alpha = 1$,the equations become $x + y + z = 0$,which has infinitely many solutions.If $\alpha = -2$,the equations become:$-2x + y + z = -3$$x - 2y + z = -3$$x + y - 2z = -3$Adding these three equations gives $0 = -9$,which is a contradiction. Thus,there is no solution when $\alpha = -2$.

The system of equations $\begin{cases} \alpha x + y + z = \alpha - 1 \\ x + \alpha y + z = \alpha - 1 \\ x + y + \alpha z = \alpha - 1 \end{cases}$ has no solution,if $\alpha$ is

Similar Questions

If the system of equations $3x - 2y + z = 0$,$\lambda x - 14y + 15z = 0$,and $x + 2y + 3z = 0$ has a non-trivial solution,then $\lambda = $

If the solution for the system of equations $x+2y-z=3$,$3x-y+2z=1$ and $2x-2y+3z=2$ is $(\alpha, \beta, \gamma)$,then $\alpha^2+\beta^2+\gamma^2=$

The system of equations $(\sin\theta) x + 2z = 0$,$(\cos\theta) x + (\sin\theta) y = 0$,and $(\cos\theta) y + 2z = a$ has

If the system of linear equations : $x+y+2z=6$,$2x+3y+az=a+1$,$-x-3y+bz=2b$ where $a, b \in R$,has infinitely many solutions,then $7a+3b$ is equal to :

If the system of equations $kx + (k+1)y + (k-1)z = 0$,$(k-1)x + (k+2)y + kz = 0$,and $(k+1)x + ky + (k+2)z = 0$ has a non-trivial solution,then the sum of all possible values of $k$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module