Quick Question on systems of eqns (more variables than eqns)

[KLS2111]

This is proabably the easiest question ever and I am simply over thinking it.

I am working with the following equations
-3w + 2x -1y +2z= 2
1w -1x + 0y -3z= 3
4w -3x + 1y -1z= 1

I worked it out and ended up with
1 0 1 8 -8
0 1 1 11 -11
0 0 0 0 0

I am trying to list the solution:
I ended up with w=0 x=-3 y=-8 and z=0

However, I thought that because there were more variables than equations, then there were multiple answers and that they should be listed in terms of one of the variables as the parameter. That row of zeros is throwing me off. Any help would be greatly appreciated.

 

[fasteddie65]

Re: Quick Question

"I worked it out and ended up with
1 0 1 8 -8
0 1 1 11 -11
0 0 0 0 0

I am trying to list the solution:
I ended up with w=0 x=-3 y=-8 and z=0"

The first line means w + y + 8 = -8
The second line means x + y + 11 = -11
The last line means z = 0.

So w = -16 - y
and x = -22 - y

y can be arbitrary.
If you let y = 1, then w = -17 and x = -23, for example.

This is called a Diophantine equation, with more variables than equations.

 

[soroban]

Re: Quick Question

Hello, KLS2111!

You are correct . . . There is an infinite number of solutions.

\(\displaystyle \begin{array}{ccc}-3w + 2x -1y +2z&= & 2 \\ \;w \;-\;x \;\qquad\;\; -3z&= &3 \\4w -3x + \;y -\;z &=& 1 \end{array}\)

\(\displaystyle \text{I changed the order of the equations: }\;\begin{array}{ccc}w \;-\;\; x \quad\;\; - 3z &=& 3 \\ 4w - 3x + \;y -\;z &=& 1 \\ -3w + 2x - \;y + 2z &=& 2 \end{array}\)

\(\displaystyle \text{We have: }\;\left[\begin{array}{cccc|c} 1 & \text{-}1 & 0 & \text{-}3 & 3 \\ 4 & \text{-}3 & 1 & \text{-}1 & 1 \\ \text{-}3 & 2 & \text{-}1 & 2 & 2 \end{array}\right]\)

\(\displaystyle \begin{array}{c} \\ R_2-4R_1 \\ R_3 + 3R_1\end{array}\left[\begin{array}{cccc|c}1 & \text{-}1 & 0 & \text{-}3 & 3 \\ 0 & 1 & 1 & 11 & \text{-}11 \\ 0 & \text{-}1 & \text{-}1 & \text{-}7 & 11 \end{array}\right]\)

\(\displaystyle \begin{array}{c}R_1+R_2 \\ \\ R_3+R_2\end{array}\left[\begin{array}{cccc|c}1&0&1&8&\text{-}8 \\ 0&1&1&11& \text{-}11 \\ 0&0&0&4&0 \end{array}\right]\)

. . . \(\displaystyle \begin{array}{c}\\ \\ \frac{1}{4}R_3\end{array} \left[\begin{array}{cccc|c}1 &0&1&8&\text{-}8 \\ 0&1&1&11&\text{-}11 \\ 0&0&0&1 & 0 \end{array}\right]\)

\(\displaystyle \begin{array}{c}R_1-8R_3 \\ R_2-11R_3 \\ \end{array} \left[\begin{array}{cccc|c}1&0&1&0& \text{-}8 \\ 0&1&1&0& \text{-}11 \\ 0&0&0&1&0 \end{array}\right]\)


\(\displaystyle \text{We have: }\;\begin{Bmatrix}w + y &=& \text{-}8\\ x+y &=& \text{-}11 \\ z &=& 0 \end{Bmatrix}\)


. . \(\displaystyle \text{That is: }\;\begin{Bmatrix} w &=& \text{-}8 - y \\ x &=& \text{-}11 - y \\ y &=& y \\ z &=& 0 \end{Bmatrix}\)


\(\displaystyle \text{On the right side, replace }y\text{ with a parameter, }t.\)

\(\displaystyle \text{Solution: }\;\boxed{\begin{array}{ccc}w &=& \text{-}8-t \\ x &=& \text{-}11-t \\ y &=& t \\ z &=&0 \end{array}}\)

 

[Deleted member 4993]

Re: Quick Question

This is called a Diophantine equation, with more variables than equations.

Diophantine problems are restricted to integer solutions only. The given problem did not state such restriction.