Question

Is there a linear system of equation of three variables which its solutions group is as appears in the attached file. Explain your answer.

Answer 1

im not sure if i understand the question; but it appears we can create a dimension to play in

Answer 2

Can you expalin a little more?

Answer 3

since a^2 is always positive; the b axis i drew in as the positive only

but again, im not really sure what the question is asking to know if this is even along the right track

Answer 4

it should be using matrices and row reducion using Gauss elimination

Answer 5

\[\begin{pmatrix}m&n&o\\p&q&r\\s&t&v \end{pmatrix}\begin{pmatrix}x\\y\\z \end{pmatrix}=\begin{pmatrix}a\\a^2\\c \end{pmatrix}\]???

Answer 6

How did you get to this? and what does it mean?

Answer 7

too much variables how can I decide if it's true or not. I'm not sure....

Answer 8

if we input abc instead of xyz

\[\begin{pmatrix}m&n&o\\p&q&r\\s&t&v \end{pmatrix}\begin{pmatrix}a\\b\\c\end{pmatrix}=\begin{pmatrix}a\\a^2\\c \end{pmatrix}\]
\[a\begin{pmatrix}m\\p\\s\end{pmatrix}+b\begin{pmatrix}n\\q\\t \end{pmatrix}+c\begin{pmatrix}o\\r\\v\end{pmatrix}=\begin{pmatrix}a\\a^2\\c \end{pmatrix}\]

\[a\begin{pmatrix}1\\p\\0\end{pmatrix}+b\begin{pmatrix}0\\q\\0 \end{pmatrix}+c\begin{pmatrix}0\\r\\1\end{pmatrix}=\begin{pmatrix}a\\a^2\\c \end{pmatrix}\]

\[ap+bq+cr = a^2\]
\[a^2-pa-(bq+cr) = 0\]

just thinking ....

Answer 9

\[\begin{pmatrix}1&0&0&|&a\\p&q&r&|&a^2\\0&0&1&|&c \end{pmatrix}\]
\[\begin{pmatrix}1&0&0&|&a\\0&q&r&|&a^2-pa\\0&0&1&|&c \end{pmatrix}\]
\[\begin{pmatrix}1&0&0&|&a\\0&1&\frac{r}{q}&|&\frac{a^2-pa}{q}\\0&0&1&|&c \end{pmatrix}\]
\[\begin{pmatrix}1&0&0&|&a\\0&1&0&|&\frac{a^2-pa}{q}-\frac{cr}{q}\\0&0&1&|&c \end{pmatrix}\]

as long as q not= 0, it seems plausible to me

but then again, i got no idea if this idea is even correct to begin with

Answer 10

I will think about it - THANKS ALOOOOT!!!!!!

Answer 11

good luck :)