# Matrices

• Feb 26th 2013, 03:54 PM
JP2012
Matrices
Matrix A = 8 -1
4 -3

Matrix B= x y z
u v w

27 12 12

I need to get the answers for x,y,z,u,v and w by using Matrix A and the already given answer, anyone understand it?
• Feb 26th 2013, 03:57 PM
Esteban
Re: Matrices
that's all? I really don't know what the exercise is trying to mean XD
• Feb 26th 2013, 03:59 PM
JP2012
Re: Matrices
Yeah all it says is find x,y,z,u,v and w. I was thinking taking the inverse of A and using it but I'm not quite sure how or if that would work.
• Feb 26th 2013, 05:04 PM
HallsofIvy
Re: Matrices
Unless you are told that there is some relationship between A and B, that's impossible.

That's like saying "y= 4. Find x"!!
• Feb 26th 2013, 05:27 PM
JP2012
Re: Matrices
Ok, thank you!
• Feb 26th 2013, 06:21 PM
JP2012
Re: Matrices
Actually I figured it out, if you take the inverse of A and multiply it by the finished matrix you get the answer.
• Feb 26th 2013, 09:03 PM
Soroban
Re: Matrices
Hello, JP2012!

You forgot to give us the set-up to the problem.

Quote:

$\begin{bmatrix} 8&\text{-}1 \\ 4&\text{-}3\end{bmatrix} \cdot \begin{bmatrix} x&y&z \\ u&v&w \end{bmatrix}\;=\; \begin{bmatrix}49&4&24 \\ 27&12&12\end{bmatrix}$

$\text{Find matrix }\begin{bmatrix}x&y&z\\u&v&w\end{bmatrix}$

$\begin{bmatrix} 8&\text{-}1 \\ 4&\text{-}3\end{bmatrix} \cdot \begin{bmatrix} x&y&z \\ u&v&w \end{bmatrix}\;=\; \begin{bmatrix}49&4&24 \\ 27&12&12\end{bmatrix}$

$\begin{bmatrix}8x-u & 8y-v & 8z-w \\ 4x-3u & 4y-3v & 4z-3w\end{bmatrix} \;=\;\begin{bmatrix}49&4&24\\27&12&12\end{bmatrix}$

$\text{We have: }\:\begin{Bmatrix}8x-u &=& 49 \\ 4x-3u &=& 27 \end{Bmatrix} \quad \begin{Bmatrix}8y-v &=& 4 \\ 4y-3v &=& 12 \end{Bmatrix} \quad \begin{Bmatrix}8z-w &=& 24 \\ 4z-3w &=& 12 \end{Bmatrix}$

$\text{Solve the systems: }\:\begin{Bmatrix}x &=& 6 \\ u &=& \text{-}1 \end{Bmatrix} \quad \begin{Bmatrix}y &=& 0 \\ v &=& \text{-}4 \end{Bmatrix} \quad \begin{Bmatrix}z &=& 3 \\ w &=& 0 \end{Bmatrix}$

$\text{Therefore: }\:\begin{bmatrix}x&y&z\\u&v&w\end{bmatrix} \;=\;\begin{bmatrix}6&0&3 \\ \text{-}1&\text{-}4&0\end{bmatrix}$
• Feb 27th 2013, 06:47 AM
HallsofIvy
Re: Matrices
Quote:

Originally Posted by JP2012
Actually I figured it out, if you take the inverse of A and multiply it by the finished matrix you get the answer.

In other words, you were asked to solve AX= B for X. It would have been nice if you had told us that!