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VCE Stuff => VCE Mathematics => VCE Mathematics/Science/Technology => VCE Subjects + Help => VCE Mathematical Methods CAS => Topic started by: anonymous1 on June 10, 2012, 08:07:29 pm

Title: urgent matrices help
Post by: anonymous1 on June 10, 2012, 08:07:29 pm: how do i solve the following using matrices:
2x - 3y + z = 3
-3x + y - 2z =1
x- y - z = 6

thanks
Title: Re: urgent matrices help
Post by: ligands on June 10, 2012, 08:13:42 pm: Put 2 -3 1, -3 1 -2, 1 -1 -1 in 3x3 matrix call this A
Put x,y,z in a colum matrix this X
Put 3, 1, 6 in a colum matrix this B

To find x y z set up the equation X = A^-1B
Title: Re: urgent matrices help
Post by: anonymous1 on June 10, 2012, 08:21:22 pm: so B * A^-1
Title: Re: urgent matrices help
Post by: ligands on June 10, 2012, 08:25:06 pm: Yep sorry if it's messy on phone lol, the original equation will be AX = B and to find X you need to go A^-1AX = A^-1B which equals IX = A^-1B which means X = A^-1B and A will be the coefinet matrix, x will be the variable matrix and b will be the answer matrix, I think I put it the wrong way around in the other comment let me edit it
Title: Re: urgent matrices help
Post by: Flying Emu on June 11, 2012, 05:14:37 pm: This what you mean lig?

-- -- -- -- -- --
| 2 -3 1 | | X | | 3 |
| -3 1 -2 | X | Y | = | 1 |
| 1 -1 -1 | | Z | | 6 |
-- -- -- -- -- --

*Yes i know, horrible brackets.
Title: Re: urgent matrices help
Post by: Phy124 on June 11, 2012, 05:57:19 pm: Quote from: Flying Emu on June 11, 2012, 05:14:37 pm
This what you mean lig?

-- -- -- -- -- --
| 2 -3 1 | | X | | 3 |
| -3 1 -2 | X | Y | = | 1 |
| 1 -1 -1 | | Z | | 6 |
-- -- -- -- -- --

*Yes i know, horrible brackets.
Heh, nice work :P (Easier method below)

$\begin{bmatrix} 2 & -3 & 1\\ -3 & 1 & -2\\ 1 & -1 & -1 \end{bmatrix}\begin{bmatrix} x\\ y\\ z \end{bmatrix}=\begin{bmatrix} 3\\ 1\\ 6 \end{bmatrix}$

Code: [Select]
[tex]\begin{bmatrix} 2 & -3 & 1\\ -3 & 1 & -2\\ 1 & -1 & -1 \end{bmatrix}\begin{bmatrix} x\\ y\\ z \end{bmatrix}=\begin{bmatrix} 3\\ 1\\ 6 \end{bmatrix}[/tex]
Use this for help

$Let \ A = \begin{bmatrix} 2 & -3 & 1\\ -3 & 1 & -2\\ 1 & -1 & -1 \end{bmatrix}$

(http://latex.codecogs.com/gif.latex?Let%20\%20X%20=%20\begin{bmatrix}%20x\\%20y\\%20z%20\end{bmatrix})

(http://latex.codecogs.com/gif.latex?Let%20\%20B%20=%20\begin{bmatrix}%203\\%201\\%206%20\end{bmatrix})

Solve for X;

$X=A^{-1}B$

By the way, I realise I'm just reiterating what has previously been discussed, just showing you guys some latex code in the progress. (Time to try and get everyone on these boards on latex I reckon ;))

edit:

I forgot to mention you can also solve it using the solve function on your calculator.

(From what I remember regarding TI-89) The calculator has a solve function for two unknowns, you can just add a third equation to this and a third unknown to solve for. This might be quicker then putting all those values into a matrix.

IIRC it was something like;

Solve( and for {a,b})

and you could just change it to;

Solve(2x-3y+z=3 and -3x+y-2z=1 and x-y-z=6 for {x,y,z})
Title: Re: urgent matrices help
Post by: ligands on June 11, 2012, 06:41:23 pm: haha yer i was on my phone so i couldn't be bothered using latex
Title: Re: urgent matrices help
Post by: anonymous1 on June 11, 2012, 09:50:59 pm: thanks everyone :)
Title: Re: urgent matrices help
Post by: Flying Emu on June 12, 2012, 07:06:57 pm: Quote from: ~My♥Little♥Pony~ on June 11, 2012, 05:57:19 pm
Heh, nice work :P (Easier method below)

OH WOW.... T.T