1. ## [SOLVED] Transofrmation Vector

Hi,

The Question:

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

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

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

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

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

Thanks

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

What is $A^{-1}$?

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

Oh and I realise that what I wrote was wrong: the $A$ 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?

$
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 $A$ matrix should be before the original matrix.
Are you saying that it should be $
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 $
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 $A (1,0)^T = (3,-2)^T$ and $A (0,-1)^T = (2,-1)^T$

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

6. Originally Posted by Solo
No. Sorry. It should be $A (1,0)^T = (3,-2)^T$ and $A (0,-1)^T = (2,-1)^T$
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 $\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]$