# Thread: Confusing system of simultaneous equations...

1. ## Confusing system of simultaneous equations...

Find three vectors p, v, w in R^4, such that the solutions to the system
w+3x+3y+2z=1
2w+6x+9y+5z=5
-w-3x+3y=5
have the form p + \lambda v + \mu w, with \lambda, \mu in R.

I assumed one would get rid of the z, and be left with two equations to solve three unknowns, so would write everything in terms of one vector, but simplifying the first two equations gives the third

So now I'm completely bewildered...

2. You were not taught Gauss-Jordan reduction?
Do you have a textbook? It should have some examples of the method used to solve such systems. I find it hard to believe you were asked to find a basis for the solution space without being taught how.

3. Originally Posted by PTL
Find three vectors p, v, w in R^4, such that the solutions to the system
w+3x+3y+2z=1
2w+6x+9y+5z=5
-w-3x+3y=5
have the form p + \lambda v + \mu w, with \lambda, \mu in R.

I assumed one would get rid of the z, and be left with two equations to solve three unknowns, so would write everything in terms of one vector, but simplifying the first two equations gives the third

So now I'm completely bewildered...
Let's see. You transform the system to row-echelon form. First subtracting the first equation twice from the second and adding it to the last, like this
$\begin{array}{rcrcrcrcr|lcl}
w &+& 3x &+& 3y &+& 2z &=& 1 &\cdot (-2) &|& \cdot 1\\
2w &+& 6x &+& 9y &+& 5z &=& 5 &\cdot 1 & &\\
-w &-& 3x &+& 3y & & &=& 5 & &| & \cdot 1\\\cline{1-9}
\end{array}$

which gives (if I'm not mistaken)
$\begin{array}{rcrcrcrcr|l}
w &+& 3x &+& 3y &+& 2z &=& 1 &\\
&& && 3y &+& z &=& 3 &\cdot (-2)\\
&& && 6y &+&2z&=& 6 &\cdot 1\\\cline{1-9}
\end{array}$

Then you subtract twice the second equation from the last, which gives
$\begin{array}{rcrcrcrcr|}
w &+& 3x &+& 3y &+& 2z &=& 1\\
&& && 3y &+& z &=& 3\\
&& && &&0&=& 0 \\\cline{1-9}
\end{array}$

which is now, you guessed it, in row-echelon form.
Now what...? - Now we do back-substitution from the last to the first equation, setting superfluous variables to arbitrary values and solving for the one remaining variable: The last equation is trivially true. In the second you may set $z := \mu$ to an arbitrary value and solve for y, which gives $y=1-\tfrac{1}{3}\mu$. Finally, in the first equation, you set $x :=\lambda$ to an arbitrary value, substitute $1-\tfrac{1}{3}\mu$ for y and $\mu$ for z, respectively, and then solve for w: which gives $w=-3\lambda-\mu-2$.
So we find that the general solution-vector of the given system has the form
$\begin{pmatrix}w\\x\\y\\z\end{pmatrix}=\begin{pmat rix}-3\lambda-\mu-2\\\lambda\\1-\tfrac{1}{3}\mu\\\mu\end{pmatrix}=\begin{pmatrix}-2\\0\\1\\0\end{pmatrix}+\lambda\begin{pmatrix}-3\\1\\0\\0\end{pmatrix}+\mu\begin{pmatrix}-1\\0\\-\tfrac{1}{3}\\1\end{pmatrix}$