Finding the matrix T(x)

[sktsasus]

[FONT=MathJax_Math-italic]
[/FONT]
"Let [FONT=MathJax_Math-italic]T[/FONT] be a linear transformation on the plane with

T \begin{pmatrix}2\\ 3\end{pmatrix} = \begin{pmatrix}-3\\ 3\end{pmatrix} and T \begin{pmatrix}3\\ 4\end{pmatrix} = \begin{pmatrix}5\\ -4\end{pmatrix}

Solve for the matrix T(x)."


I tried to insert the various values of [FONT=MathJax_Math-italic]T into the matrix and managed to set up some equations, but I am not sure how you would generalize the entire thing in terms of [FONT=MathJax_Math-italic]x.[/FONT][/FONT]

Any help?

 

[barrick]

I assume you mean:

​

\(\displaystyle \displaystyle
\begin{align*}\\
T\begin{pmatrix}2\\3\end{pmatrix}&=\begin{pmatrix}-3\\3\end{pmatrix}\\
T\begin{pmatrix}3\\4\end{pmatrix}&=\begin{pmatrix}5\\-4\end{pmatrix}
\end{align*}
\)
​

Note: you should surround LaTeX code with [ t e x ]...[/ t e x ] (without the spaces) instead of $$...$$.

You should try to get the images of the unit vectors, as these correspond to the columns of the matrix.

For example, to get 0 in the second row, you would use the relation:

.
​

\(\displaystyle \displaystyle
-4\begin{pmatrix}2\\3\end{pmatrix}+3\begin{pmatrix}3\\4\end{pmatrix}=\begin{pmatrix}1\\0\end{pmatrix}
\)
​

Note that, by luck, the first component is 1; in the general case, you may have to multiply both vectors by a constant to get a unit vector.

We have therefore:

​

\(\displaystyle \displaystyle
\begin{align*}\\
T\begin{pmatrix}1\\0\end{pmatrix}&= -4\begin{pmatrix}-3\\3\end{pmatrix}+3\begin{pmatrix}5\\-4\end{pmatrix}\\
&= \begin{pmatrix}27\\-24\end{pmatrix}
\end{align*}
\)
​

and this gives you the first column of the matrix. You can then use the same technique to get the second column.