Question

Solve the folowing matrix for c3,c4 and c1
\[\huge \left[\matrix{n&&c_3&&c_4&&c_1\\-\frac{20pa^3}{3}&&2a&&1&&-2a\\-6pa^3&&3a&&1&&0\\-2pa^2&&1&&0&&-1} \right]\]

Answer 1

sorry, I have no idea what that even is! I can toss the link around to people that night tho

Answer 2

Ok if you could just solve this I would appreciate it
\[\huge \left[\matrix{-\frac{20pa^3}{3}&&2a&&1&&-2a\\-6pa^3&&3a&&1&&0\\-2pa^2&&1&&0&&-1} \right]\]

Answer 3

This is what I did
\[\huge \left[\matrix{-\frac{20pa^3}{3}&&2a&&1&&-2a\\-6pa^3&&3a&&1&&0\\-2pa^2&&1&&0&&-1} \right]\]
Multiplied top row by 1/(2a), second row by 1/(3a) and third row by -1 to get this new matrix
\[\huge \left[\matrix{-\frac{10pa^2}{3}&&1&&\frac{1}{2a}&&-1\\-2pa^2&&1&&\frac{1}{3a}&&0\\2pa^2&&-1&&0&&+1} \right]\]

I then added the first row to the last row and replaced the first row with the result to obtain this new matrix
\[\huge \left[\matrix{-\frac{10pa^2}{3}+2pa^2&&0&&\frac{1}{2a}&&0\\-2pa^2&&1&&\frac{1}{3a}&&0\\2pa^2&&-1&&0&&+1} \right]\]
\[\huge \frac{-10pa^2+6pa^2}{3} = \frac{-4pa^2}{3}(2a)=c_4\] Then I get
\[\huge c_4 = -\frac{8pa^3}{3}\] WHICH IS WRONG
THE RIGHT ANSWER IS \[\huge c_4 = \frac{4}{3}Pa^3\]

Answer 4

what do you mean by "solve"?

Answer 5

its a system of equations

Answer 6

can you show the equations? and whats the top row represent?

Answer 7

in a second

Answer 8

P and a are constants

Answer 9

an augment is usually, as far as ive seen; written with the coeffs on the left and the =stuff to the right

Answer 10

Same thing

Answer 11

You get the picture

Answer 12

i see it ....

Answer 13

\[\begin{vmatrix}-\frac{20pa^3}{3}&|&2a&1&-2a\\-6pa^3&| &3a&1&0\\-2pa^2&| &1&0&-1\end{vmatrix}\]

\[\begin{vmatrix}-2pa^2&| &1&0&-1\\ -\frac{20pa^3}{-2a.3}&|&-1&1/-2a&1\\-6pa^3/-3a&| &-1&1/-3a&0\end{vmatrix}\]

\[\begin{vmatrix}-2pa^2&| &1&0&-1\\ \frac{10pa^2}{3}-2pa^2&|&0&-1/2a&0\\2pa^2-2pa^2&| &0&-1/3a&-1\end{vmatrix}\]

\[\begin{vmatrix}-2pa^2&| &1&0&-1\\ \frac{-4pa^2.2a}{3}&|&0&1&0\\0&| &0&-1&-3a\end{vmatrix}\]

\[\begin{vmatrix}-2pa^2&| &1&0&-1\\ \frac{-8pa^3}{3}&|&0&1&0\\\frac{-8pa^3}{3}&| &0&0&-3a\end{vmatrix}\]

\[\begin{vmatrix}-2pa^2&| &1&0&-1\\ \frac{-8pa^3}{3}&|&0&1&0\\\frac{8pa^2}{9}&| &0&0&1\end{vmatrix}\]

\[\begin{vmatrix}\frac{-10pa^2}{9}&| &1&0&0\\ \frac{-8pa^3}{3}&|&0&1&0\\\frac{8pa^2}{9}&| &0&0&1\end{vmatrix}\]

maybe .....

Answer 14

http://www.wolframalpha.com/input/?i=rref%7B%7B2a%2C1%2C-2a%2C-20xa%5E3%2F3%7D%2C%7B3a%2C1%2C0%2C-6xa%5E3%7D%2C%7B1%2C0%2C-1%2C-2xa%5E2%7D%7D

same same

Answer 15

you sure you got your eqs correct?

Answer 16

No I had one wrong equation all along
http://www.wolframalpha.com/input/?i=rref%7B%7B2a%2C-2a%2C-1%2C8a%5E3z%2F3%7D%2C%7B1%2C-1%2C0%2C2za%5E2%7D%2C%7B0%2C3a%2C1%2C-6za%5E3%7D%7D

I apologize for wasting your time.
Really really sorry

Answer 17

c1 c3 c4 in that order

Answer 18

yeah, it helps to have the right inputs :)