1. ## [SOLVED] Transofrmation Vector

Hi,

The Question:

A transformation $\displaystyle x \rightarrow Ax$ sends vector $\displaystyle (1,0)^T$ to $\displaystyle (3,-2)^T$ and vector $\displaystyle (0,-1)^T$ to $\displaystyle (2,-1)^T$. What is $\displaystyle A$? Find the inverse of $\displaystyle A$.

This is what I thought you're supposed to do:

Let $\displaystyle A = (x,y)^T$. Therefore:

$\displaystyle (1,2)^T (x,y)^T = (3,-2)^T$
$\displaystyle (0,-1)^T (x,y)^T = (2,-1)^T$

I thought I could solve for $\displaystyle x, y$ simultaneously but I really don't know where to go .

Thanks

2. Does this work? $\displaystyle A = \left[ {\begin{array}{rr} 3 & { - 2} \\ { - 2} & 1 \\ \end{array} } \right]$ WHY?

What is $\displaystyle A^{-1}$?

3. Yes, it does work. But I'm not too sure why, sorry .

Oh and I realise that what I wrote was wrong: the $\displaystyle A$ matrix should be before the original matrix.

I can see that if you know that it is a 2x2 matrix then you can figure it out by normal matrix multiplication.

Is the transformation matrix always a 2x2 matrix? Why is that?

$\displaystyle A^{-1} = \left[ {\begin{array}{rr} 1 & { 2} \\ { 2} & 3 \\ \end{array} } \right]$?

4. Originally Posted by Solo
Oh and I realise that what I wrote was wrong: the $\displaystyle A$ matrix should be before the original matrix.
Are you saying that it should be $\displaystyle A\left[ {\begin{array}{r} 3 \\ { - 2} \\ \end{array} } \right] = \left[ {\begin{array}{r} 1 \\ 0 \\ \end{array} } \right]\;\& \;A\left[ {\begin{array}{r} 2 \\ { - 1} \\ \end{array} } \right] = \left[ {\begin{array}{r} 0 \\ { - 1} \\ \end{array} } \right]?$

5. Originally Posted by Plato
Are you saying that it should be $\displaystyle A\left[ {\begin{array}{r} 3 \\ { - 2} \\ \end{array} } \right] = \left[ {\begin{array}{r} 1 \\ 0 \\ \end{array} } \right]\;\& \;A\left[ {\begin{array}{r} 2 \\ { - 1} \\ \end{array} } \right] = \left[ {\begin{array}{r} 0 \\ { - 1} \\ \end{array} } \right]?$
No. Sorry. It should be $\displaystyle A (1,0)^T = (3,-2)^T$ and $\displaystyle A (0,-1)^T = (2,-1)^T$

Edit: How did you know it would be a 2x2 matrix?

6. Originally Posted by Solo
No. Sorry. It should be $\displaystyle A (1,0)^T = (3,-2)^T$ and $\displaystyle A (0,-1)^T = (2,-1)^T$
Edit: How did you know it would be a 2x2 matrix?
O.K. The matrix I gave you in my O.P. is correct.
It is the transformation that $\displaystyle \left[ {\begin{array}{r} 1 \\ 0 \\ \end{array} } \right] \to \left[ {\begin{array}{r} 3 \\ { - 2} \\ \end{array} } \right]\;\& \;\left[ {\begin{array}{r} 0 \\ { - 1} \\ \end{array} } \right] \to \left[ {\begin{array}{r} 2 \\ { - 1} \\ \end{array} } \right]$