Question

Solving a set of three simultaneous equations

If anyone could help with this I'd be very grateful. My mind appears to be drawing a blank.

5a+15b+55c=36
15a+55b+225c=157
55a+225b+979c=709

This is for method of least squares, finding a quadratic line of best fit

Answer 1

wolfram not allowed ?

Answer 2

it's an exam practice question so sadly not.

Answer 3

I may start by eliminating \(a\) from first two equations,
and last two equations to get two equations in \(b\) and \(c\)

Answer 4

that's what I tried, and I ended up with a completely wrong value for 'c'

Answer 5

but I probably have made an arithmetic error somewhere in there

Answer 6

or we may use matrices

Answer 7

u familiar wid gauss jordan elimination right ?

Answer 8

but anyways lets do it the elementary way lol, should be easy

Answer 9

I have done jordan elimination but not for a verrrry long time. It'll probably end up me having to revise more stuff haha

Answer 10

Given equations :
5a+15b+55c=36
15a+55b+225c=157
55a+225b+979c=709


3(5a+15b+55c=36)
-1(15a+55b+225c=157)
------------------------
10b + 60c = 49 \(\color{red}{*}\)


11(5a+15b+55c=36)
-1(55a+225b+979c=709)
--------------------
60b + 374c = 313 \(\color{red}{**}\)

Answer 11

ahh okay, I already see where I went wrong lol, I didn't multiply one of my terms!!

Answer 12

lol this should not be given in exam :/

Answer 13

I know! as if we don't have enough to do haha thanks for clearing that up for me!

Answer 14

np :)
I think gauss jordan tricks simplifies nicely... u took linear algebra before ?

Answer 15

I'm taking it now

Answer 16

good :)

u may reduce it to triangular form and solve it easily. may save time in exam :
http://www.wolframalpha.com/input/?i=row+reduce++%7B%7B5%2C15%2C55%2C36%7D%2C%7B15%2C55%2C225%2C157%7D%2C%7B55%2C225%2C979%2C709%7D%7D

Answer 17

^ last column are the solutions,

after reducing the matrix using row operations, u wil endup wid solutions in the last column

Answer 18

yep, I just did it with row reduction, a lot quicker!! thanks!

Answer 19

try using a det 3X3

Answer 20

It's not all that tedious if you work smarter...

\[\left[\begin{matrix}5 & 15 & 55 & 36 \\ 15 & 55 & 225 & 157 \\ 55 & 225 & 979 & 709\end{matrix}\right]\]
\[\left[\begin{matrix}5 & 15 & 55 & 36 \\ 0 & 10 & 60 & 49 \\ 0 & 60 & 374 & 313\end{matrix}\right]\]
\[\left[\begin{matrix}5 & 5 & -5 & -13 \\ 0 & 10 & 60 & 49 \\ 0 & 0 & 14 & 19\end{matrix}\right]\]
\[\left[\begin{matrix}5 & 5 & -5 & -13 \\ 0 & 5 & 30 & 24.5 \\ 0 & 0 & 14 & 19\end{matrix}\right]\]
If you don't force yourself always to use "1" as a pivot, you don't have to see decimals until here.
\[\left[\begin{matrix}5 & 0 & -35 & -37.5 \\ 0 & 5 & 30 & 24.5 \\ 0 & 0 & 14 & 19\end{matrix}\right]\]
I did this next step just to emphasize my point. 5/14 is an unusual factor.
\[\left[\begin{matrix}5 & 0 & -35 & -37.5 \\ 0 & 5 & 30 & 24.5 \\ 0 & 0 & 5 & 6.785714\end{matrix}\right]\]
\[\left[\begin{matrix}1 & 0 & 0 & 2 \\ 0 & 1 & 0 & -3.28246 \\ 0 & 0 & 1 & 1.357143\end{matrix}\right]\]
Truthfully, I recommend retaining exact solutions, but the decimals were easier to code.
\[\left[\begin{matrix}1 & 0 & 0 & 2 \\ 0 & 1 & 0 & -227/70 \\ 0 & 0 & 1 & 19/14\end{matrix}\right]\]