# System of 4 equations (3 of them linear) - What can be said about the solution set

I have a homework at linear algebra and we have this system of linear equations:

$x+y+z+w=0$
$x+2*y+9*z+13*w=0$
$4*x+41*y+6*z+656*w = 0$

And we add this equation:

$x^3 + y^4 + 8* z^5 + 8* w^6 = 0$

What can be said about the solution set of this system?

Real solutions:

• w~~0, x~~0, y~~0, z~~0
• w~~-0.0417376, x~~-0.665814, y~~0.73708, z~~-0.0295286

Doesn't this mean that the solution set is a non empty set?
And also that it is a finite set (with 2 elements? the (0,0,0,0) and (-0.0417376, -0.665814, 0.73708, -0.0295286) ?

Also, how can I find out if the solution set of the above system is a vector space?

Thank you!

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Over $\mathbb Q$, the $3$ linear equations give you a matrix of rank $3$. So the nullspace is of dimension $1$, with basis vector $$b = (\frac{335}{21}, \frac{-2596}{147}, \frac{-2596}{147}, 1),$$ i.e., the solution space is all scalar multiples of $b$. All solutions to the system however, should be zeros of the polynomial. I'm not sure though how to use the polynomial to pick a particular solution. – user2468 Mar 12 '12 at 0:45

## 3 Answers

Your three linear equations can be satisfied by $w = 147 t, x = 2345 t, y = -2596 t, z = 104 t$ for any $t$.

Substitute that into your fourth equation and you have a sextic equation for $t$, $$80722386956232 t^6+97332232192 t^5+45417032294656 t^4+12895213625 t^3=0$$ but you can take out a factor of $t^3$ so you have in effect three roots at $t=0$ (and so $w=y=z=0$) and a cubic: $$80722386956232 t^3+97332232192 t^2+45417032294656 t+12895213625=0.$$ This particular cubic has one real root and two complex roots, which you can translate back to values for $w,x,y,z$.

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You say "This particular cubic has one real root and two complex roots, which you can translate back to values for w,x,y,z.", the one real root is the one that wolfram gave as a result? So there are two real roots? (We are looking for real roots) – Chris Mar 12 '12 at 1:21
@Chris: $t=-0.000283929\ldots$ is the same real root as you have in your question when you multiply it by $147, 2345$ etc. – Henry Mar 12 '12 at 1:36
So all in all, is the system's solution set a finite set? We are sure that 0 vector belongs to the solution set. – Chris Mar 12 '12 at 18:10
@Chris: yes - depending on how you count, there are two distinct real solutions (one of which is zero and you might count it three times) and two complex solutions – Henry Mar 12 '12 at 21:35
So could we possibly say that the solution set is finite? – Chris Mar 13 '12 at 21:38

You can solve for any three of x, y, z, and w in terms of the other by using the first three equations. For example, from the first two, you get $y+z+w = 2∗y+9∗z+13∗w$, or $y = -8z -12 w$.

From the first and third equations, you get (multiplying the first by 4), $4y+4z+4w = 41y+6z+656w$, or $37y = -2z-652w$.

This lets you get z as a multiple of w, then y as another multiple of w, and, finally, x as another multiple of w.

Substituting these into the last equation, you will get a sextic (sixth degree) equation in w. This will have 6 roots, and since the coefficients will be real, they will be real or in complex conjugate pairs.

Looking at the equation, there will be three zero roots. Since Alpha has found a non-zero real root, the other two roots are real or complex conjugate.

Since the resulting equation, after removing the zero roots is a cubic, you can solve it of divide out the root found by Alpha to get a quadratic.

I see that a solution has been entered while I was typing.

Let's look.

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As for whether or not it's a vector space: one of the axioms of vector spaces says that if $\mathbb v$ is a vector in the space, then so is $k \mathbb v$ for any $k \in \mathbb R$. This means that $k (-0.0417376, -0.665814, 0.73708, -0.0295286)$ would have to be a solution for any $k$.

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But it is not: the $x^3 + y^4 + 8 z^5 + 8 w^6 = 0$ prevents it, except for $k=0,1$ or the two complex values. – Henry Mar 12 '12 at 12:43