Solve system of Equations I have this system:
$\pmatrix{1 & 90 & 30} = \frac{1}{1.2} \ \left( \begin{array}{c} q_u & q_m & q_d \end{array} \right)   \pmatrix{1.2 & 180 & 0 \\ 1.2 & 120 &30 \\ 1.2 &60 & 60}$
How can I solve this for $q_u$, $q_m$ and $q_d$?
ALSO - I know the answers to be
$$ q_u = .2 - \alpha \ , \ q_m = .4 + 2 \alpha \ , \ q_d = .4 - \alpha \ $$
BUT I am not sure where that alpha comes from
BTW - this is my first time posting.
Thank you!
 A: $ \frac {1}{1.2} \pmatrix {q_u q_m q_d}  * \pmatrix{1.2 & 180 & 0 \\ 1.2 & 120 &30 \\ 1.2 &60 & 60}=\frac {1}{1.2} \pmatrix{1.2q_u+1.2q_m+1.2q_d & 180q_u+120q_m+60q_d & 30q_m+60q_d}=\pmatrix{1 & 90 & 30}$, so
$q_u+q_m+q_d=1$
$150q_u+100q_m+50q_d=90$
$25q_m+50q_d=30$
Then you eliminate $q_u$ from the second equation:
$q_u+q_m+q_d=1$
$-50q_m-100q_d=-60$
$25q_m+50q_d=30$
Now notice that the second equation is the same as the third, just multiplied by $(-2)$, so it doesn't add any new information and we can forget about it. So we have
$q_u+q_m+q_d=1$
$5q_m+10q_d=6$
Now we have three unknowns, but only two equations. This means, we don't have enough information to find one specific solution. There will be infinitely many.
Let's try to find one: $q_d=.4$, $q_m=.4$, $q_u=.2$. But now notice - if I subtract a number from $q_u$ and add twice that to $q_m$, they will cancel out and $5q_m+10q_d=6$ still. Now I just have to make sure this new number won't mess up the first equation by adding appropriate multiple of it to $q_u$ and I'm done. Note this works only because the system is linear.
A: If we write the system of equations as an "augmented matrix",
$$ \left[ \begin{array}{ccc|r}
1&1&1&1 \\ 150 & 100 & 50& 90 \\ 0 & 25 & 50 & 30 \\  \end{array} \right] \ \ , $$
"row-reduction" will take this to
$$ \left[ \begin{array}{ccc|r}
1&1&1&1 \\ 0 & -50 & -100& -60 \\ 0 & 25 & 50 & 30 \\  \end{array} \right] \ \ . $$
The last row can be "zeroed-out" completely, telling us that we have a dependent system,
$$ \left[ \begin{array}{ccc|r}
1&1&1&1 \\ 0 & 1 & 2& 1.2 \\ 0 & 0 & 0 & 0 \\  \end{array} \right] \ \ . $$
We are thus "free" to give $ \ q_d \ $ any value we like, and then back-substitute to find relations between $ \ q_d \ , \ q_m \ , \ $ and $ \ q_u \ $ .  (Depending upon what you choose for $ \ q_d \ $ , you will not necessarily get the answer you show, but something related.  I made a different choice than the solver of the problem apparently did, but if we set $ \ q_d \ = \ 0.4 \ - \ \alpha \ $ , we do get the rest of the values shown.)
