The number of values of alpha for which the s | vedclass

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

Question

$x+y+z=\alpha$

$\alpha x+2 \alpha y+3 z=-1$

$x+3 \alpha y+5 z=4$

Has inconsistent solution

$D =\left|\begin{array}{ccc}1 & 1 & 1 \

Vedclass · Accepted Answer

$1$

Vedclass · Answer

$x+y+z=\alpha$

$\alpha x+2 \alpha y+3 z=-1$

$x+3 \alpha y+5 z=4$

Has inconsistent solution

$D =\left|\begin{array}{ccc}1 & 1 & 1 \ \alpha & 2 \alpha & 3 \ 1 & 3 \alpha & 5\end{array}ight|=0$

$\Rightarrow(\alpha-1)^{2}=0$

$\alpha=1$

For $\alpha=1$

$D_{1}=\left|\begin{array}{ccc}1 & 1 & 1 \ -1 & 2 & 3 \ 4 & 3 & 5\end{array}ight|$

$=(10-9)-(-5-12)+(-3-8)$

$=1+17-11 
eq 0$

For $\alpha=1$ the system of equation has Inconsistent solution

The number of values of $\alpha$ for which the system of equations: $x+y+z=\alpha$ ; $\alpha x+2 \alpha y+3 z=-1$ ; $x+3 \alpha y+5 z=4$ is inconsistent, is

Similar Questions

If $\left| {{\kern 1pt} \begin{array}{*{20}{c}}1&2&3\\2&x&3\\3&4&5\end{array}\,} \right| = 0,$ then $x =$

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

The following system of linear equations $7 x+6 y-2 z=0$ ; $3 x+4 y+2 z=0$ ; ${x}-2{y}-6{z}=0,$ has

If $A \ne O$ and $B \ne O$ are $ n × n$ matrix such that $AB = O,$ then

If the system of linear equations $x+y+3 z=0$

$x+3 y+k^{2} z=0$

$3 x+y+3 z=0$

has a non-zero solution $(x, y, z)$ for some $k \in R ,$ then $x +\left(\frac{ y }{ z }\right)$ is equal to