# Thread: Show the steps necessary to obtain the given solution set to 2 linear equations

1. ## Show the steps necessary to obtain the given solution set to 2 linear equations

I'm stuck on a homework question:

Show how the following solution set: [k1,k2,k3,k4,k5,k6,k7,k8]^T
=[1,-1,0,0,0,0,0,0]^T,
[0,-1,1,0,0,0,0,0]^T,[0,0,0,1,-1,0,0,0]^T,[0,0,0,-1,0,1,0,0]^T,[0,0,0,0,-1,0,1,0]^T,[0,0,0,0,0,-1,0,1]
is obtained from the two linear equations k1+k2+k3-k4-k5-k6-k7-k8=0 and
k4+k5+k6+k7+k8=0.

The two equations can be put into a matrix A= [1,1,1,-1,-1,-1,-1,-1,];[0,0,0,1,1,1,1,1]. Then row 2 is added to row 1 and the following solution isobtained (where v1, v2, .. v6 are arbitrary values) k1=-v1-v2; k2= v1, k3=v2, k4=-v3 -v4 -v5 -v6; k5=v3; k6=v4; k7=v5; k8=v6. But this corresponds to a solutionset different than the one provided in the question: [k1,k2,k3,k4,k5,k6,k7,k8]^T= [-1,1,0,0,0,0,0,0]^T, [-1,0,1,0,0,0,0,0]^T, [0,0,0,-1,1,0,0,0]^T,[0,0,0,-1,0,1,0,0]^T, [0,0,0,-1,0,0,1,0]^T,[0,0,0,-1,0,0,0,1]^T.

So the problem is that I don't know how to reproduce the given solution set tothe given linear equations. We were also told verbally that Excel can be usedto help us with this problem set, but I don't see how that can help if I can'teven manually reproduce this solution. I've tested the given solution byplugging it into the equations and it seems to work, but I have no clue how toreproduce it.

2. ## Re: Show the steps necessary to obtain the given solution set to 2 linear equations

Well, I would not say that is the "solution set"- rather it is a basis for the "solution space"- every vector in the solution space can be written as a linear combination of those.

Yes, one way to solve such a set of equations is to row reduce the matrix form.
$\displaystyle \begin{bmatrix}1 & 1 & 1 & -1 & -1 & -1 & -1 & -1 \\ 0 & 0 & 0 & 1 & 1 & 1 & 1 & 1\end{bmatrix}$
We can put that in reduced row echelon form by adding the second row to the first:
$\displaystyle \begin{bmatrix}1 & 1 & 1 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 & 1 & 1 & 1 & 1\end{bmatrix}$

Now, that is the same as the two equation $\displaystyle k_1+ k_2+ k_3= 0$ and $\displaystyle k_4+ k_5+ k_6+ k_7+ k_8= 0$
(We could also has simply added the second equation to the first. "Row echelon form" simply mimics the usual steps for reducing the unknowns in in a system of equations.)

First, you must recognize that such a system of equations has an infinite number of solutions and there are an infinite number of ways of writing solutions. We can solve each of those two equations for one unkown value, then take the other arbitrarily.

$\displaystyle k_1+ k_2+ k_3= 0$ is the same as $\displaystyle k_2= -k_1+ k_3$. Any such solution vector is of the form
$\displaystyle \begin{bmatrix}k_1 \\ -k_1- k_3 \\ k_3 \\ 0 \\ 0 \\ 0 \\0 \\ 0\end{bmatrix}= k_1\begin{bmatrix}1 \\ -1 \\ 0 \\ 0 \\ 0 \\ 0 \\ 0 \\ 0 \end{bmatrix}+ k_3\begin{bmatrix}0 \\ -1 \\ 1 \\ 0 \\ 0 \\ 0 \\ 0 \\ 0 \\ 0 \end{bmatrix}$
Those are the first two vectors in your set.

One way of reducing the other vector is to solve for, say, $\displaystyle k_8$
$\displaystyle k_4+ k_5+ k_6+ k_7+ k_8= 0$ is the same as, say, $\displaystyle k_8= -k_4- k_5- k_6- k_7$. Any such solution vector is of the form
$\displaystyle \begin{bmatrix}0 \\ 0 \\ 0 \\ k_4 \\ k_5 \\ k_6 \\ k_7 \\ -k_4- k_5- k_6- k_7\end{bmatrix}= k_4\begin{bmatrix}0 \\ 0 \\ 0 \\ 1 \\ 0 \\ 0 \\ 0 \\ -1\end{bmatrix}+ k_5\begin{bmatrix}0 \\ 0 \\ 0 \\ 0 \\ 1 \\ 0 \\ 0 \\ -1\end{bmatrix}+ k_6\begin{bmatrix} 0 \\ 0 \\ 0 \\ 0 \\ 0 \\ 1 \\ 0 \\ -1\end{bmatrix}+ k_7 \begin{bmatrix}0 \\ 0 \\ 0 \\ 0 \\ 0 \\ 0 \\ 1 \\ -1\end{bmatrix}$
Those vectors are not the same as you give but are an equivalent solution.

What they appear have done is to look at pairs of indices; to argue that if $\displaystyle k_6= k_7= k_8= 0$ then we must have $\displaystyle k_4+ k_5= 0$ so $\displaystyle k_5= -k_4$. That gives $\displaystyle \begin{bmatrix}0 \\ 0 \\ 0 \\ k_4 \\ -k_4 \\ 0 \\ 0 \\ 0\end{bmatrix}= k_4\begin{bmatrix} 0 \\ 0 \\ 0 \\ 1 \\ -1 \\ 0 \\ 0 \\ 0 \end{bmatrix}$.

Then, if $\displaystyle k_5= k_7= k_8= 0$ then we must have $\displaystyle k_4+ k_6= 0$ so $\displaystyle k_4= -k_6$. That gives $\displaystyle \begin{bmatrix}0 \\ 0 \\ 0 \\ -k_6 \\ 0 \\ k_6 \\ 0 \\ 0 \end{bmatrix}= k_6\begin{bmatrix}0 \\ 0 \\ 0 \\ -1 \\ 0 \\ 1 \\ 0 \\ 0 \end{bmatrix}$.

If $\displaystyle k_4= k_6= k_8= 0$ we must have $\displaystyle k_5+ k_7= 0$ so $\displaystyle k_5= -k7$. That gives $\displaystyle \begin{bmatrix}0 \\ 0 \\ 0 \\ 0 \\ -k_7 \\ 0 \\ k_7 \\ 0\end{bmatrix}= k_7\begin{bmatrix}0 \\ 0 \\ 0 \\ 0 \\ -1 \\ 0 \\ 1 \\ 0\end{bmatrix}$

Finally, if $\displaystyle k_4= k_5= k_7= 0$ we must have $\displaystyle k_6+ k_8= 0$ so $\displaystyle k_6= -k_8$. That gives $\displaystyle \begin{bmatrix} 0 \\ 0 \\ 0 \\ 0 \\ 0 \\ -k_8 \\ 0 \\ k_8\end{bmatrix}= k_8\begin{bmatrix}0 \\ 0 \\ 0 \\ 0 \\ 0 \\ -1 \\ 0 \\ 1\end{bmatrix}$.

But, again, there are many different ways to write a basis for any vector space and so many different ways to write a basis for the solution space of a set of equations. This happens to be one of them and, as I indicated earlier, not the way I personally would have arrived at a basis.

3. ## Re: Show the steps necessary to obtain the given solution set to 2 linear equations

Thank you so much. This helps a lot; I never would've thought to look at pairs of indices.