In each part,determine whether the given vector is a solution of the linear system
\begin{align}
2x-4y-z&=1\\
x-3y+z&=1\\
3x-5y-3z&=1
\end{align}
(a) $(3,1,1)$ (b) $(3,-1,1)$ (c) $(13,5,2)$ (d) $(13/2,5/2,2)$ (e) $(17,7,5)$
It's easy to solve this question. Just plug in the given vector into 3 equations respectively and check that if the left and side = the right hand side. And it turns out that a.d.e meet the restrictions of all 3 equations.
But my question is: why is that possible? Since for a linear system with n equations and n unknowns, it has only 1 unique solution. Geometrically, this linear system is like 3 planes, and the solution is a point when these 3 planes coincide. So, I think there's only 1 point that can suit into this linear system.
Best Answer
If you subtract the first equation from the third equation, you get the same thing as when you subtract the second from the first. That means that one of the equations is redundant, so you can have more than 1 answer. It's the same thing as all three equations parallel to and intersecting at one single line.