# No solutions with three planes in $\mathbb{R}^3$

Find the point in $\mathbb{R}^3$ that these three planes cross:

1. $-x+y+x=4$
2. $2x-y-4x=-2$
3. $-3x+2y+5z=0$.

I do Gauss-Jordan on this system of equation, and find $$0\ 0\ 0 \mid 6$$ So no solutions exist. The planes never cross a point.

Now, I asked myself the question: If I'd got $$0\ 0\ 0\ \mid 0$$ then, would the solution set be a line in $\mathbb{R}^3$? I think yes.

• There are typos in the eq(s). – mrs Nov 20 '13 at 12:34
• It might be a line. It could also maybe be a plane (all three equations could represent the same plane) and I think in that case in the gauss Jordan there would be two rows $0,0,0|0$. – coffeemath Nov 20 '13 at 12:34

The answer is yes. Let's consider the following system of equations: $$\left\{ \begin{array}{ll} x+2y-2z+3w=2 \\ 2x+4y-3z+4w=5 \\ 5x+10y-8z+11w=12 \end{array} \right.$$ which is equal to this one (after doing some manipulations on equations):
$$\left\{ \begin{array}{ll} x+2y-2z+3w=2 \\ ~~~~~~~~~~~~~~~~z-2w=1 \\ 0=0~~~~~~~~~~~~~~~~~~~~~ (*) \end{array} \right.$$ The system is consistent but in second system the number of variables is more than the numbers of equations so the system has infinitely many solutions. Let $y$ and $w$ be our free variables and set $w=b$ and $y=a$ to find out if we can find the general solutions. According to last system we get $z=1+2b$ and $x=4-2a+b$ so $$(4-2a+b,a,1+2b,b)$$. Indeed, if you gave us the system you had in mind in which the third equation was $0=0$, then we would find that possible line.