Consider the system of equations :
$x + y +5z = 3$
$x + 2y + 4z = k$
$x + 2y + mz = 5$ then this system of equation is consistent if
(a) $m \ne 4$
(b) $k \ne 5$
(c) $m = 4$
(d) $k =5$
Reducing, this system by subtracting equations (2) and (3) I get :
$x + y + 5z = 3$
$x + 2y + 4z = k$
$(m -4)z = 5 -k$
Now, the given system is consistent if $m \ne 4$,so only condition I require for consistency is $m \ne 4$
but if I take $k = 5$, then again the system is consistent
It will have unique solutions if $m \ne 4$
and infinite solution if $m = 4$
So, my question is are both (a) and (d) the correct choices for this question ?
Or is only (a) correct ?
Thank you.
Best Answer
If $k=5$ the system is consistent for all values of $m$
.