The question I'm given is this:
Let $S$ be the subspace of $R^{4}$ consisting of the solutions to the following system of equations:
$$x_{1}+2x_{2}+2x_{3}+2x_{4}=0$$
$$x_{1}+5x_{2}+8x_{3}+5x_{4}=0$$
$$-3x_{1}-4x_{2}-2x_{3}-3x_{4}=0$$
I am asked to give a basis for $S$.
I have attempted to solve this by putting this into a matrix and solving for my $x_{1}$, $x_{2}$, $x_{3}$, and $x_{4}$, but I don't really know where to go from there. I am sure I'm missing something important.
Any help would be greatly appreciated!
Best Answer
You can also do it without matrices, simply by solving the equations. Label them (1),(2),(3).
4(1)-(2): $3x_1+3x_2+3x_4=0$ (4). Then -(2)-4(3): $11x_1+11x_2+7x_4=0$ (5). Now 11(4)-3(5): $x_4=0$ and so $x_1+x_2=0$. (2) now gives $x_2+2x_3=0$. So putting $x_3=k$ we have $x_1=2k,x_2=-2k,x_3=k,x_4=0$.
Check: (1) $2k-4k+2k=0$, (2) $2k-10k+8k=0$, (3) $-6k+8k-2k=0$.
So a basis for $S$ is $(2,-2,1,0)$.