Please solve this for me.

Question

Find a basis for the solution space of the system of equations.

x1 − 2x2 + 5x3 + 3x5 = 0

−2x1 + 5x2 − 7x3 − 6x5 = 0

−x1 + 3x2 − 2x3 + x4 − 3x5 = 0

−3x1 + 8x2 − 9x3 + x4 − 9x5 = 0

Answer 1

First, line up all the equations and put them into something like this table with all the like powers under the same time. Preserve all the signs.

x⁵ x⁴ x³ x² x 0

3x⁵ +5x³ -2x² +x = 0

-6x⁵ -7x³ +5x² -2x =0

First, we are going to make the +x⁴ canceled out by changing all the signs when we multiply everything on both sides of the equal sign by -1.

+3x⁵ -x⁴ +2x³ -3x² +x =0

-9x⁵ +x⁴ -9x³ +8x² -3x =0

This gives us the resulting equation below.

-6x⁵ +0 -7x³ +5x² -2x =0

-6x⁵ -7x³ +5x² -2x =0

Which is identical to the second equation. If we reverse the signs of one of these by multiplying by -1, we theoretically could make all the work disappear, but we don't want to do that. We want to isolate for just x. So, what we do is find the other things we can reduce with the first equation to isolate for just x.

These are the first two equations. Note that we can multiply the first one by 2 and then make the x⁵ variable get canceled out.

6x⁵ +10x³ -4x² +2x = 0

-6x⁵ -7x³ +5x² -2x =0

Which gives us:

0 +3x³ +x² +0 =0

Now, we can factor out some terms.

x²(3x +1) =0

Using all these things, we found that this equation can have a zero at x for x² and a zero at -1/3 for the second term. We were able to simplify everything down to this one cubic function, which can be graphed hitting the zero at y=0 and at x=-1/3.