getting ready for final need some help
(1) 1
Question: R squared through R cubed is a linear transformation.if T(1)= (1),
1 1
T(1 = (0) find T(2)
-1) 1 (4)
have no idea
This is completely illegible as it stands. I think you mean something like "If $\displaystyle T\begin{bmatrix}1\\1\end{bmatrix} = \begin{bmatrix}?\\?\\?\end{bmatrix}$ and $\displaystyle T\begin{bmatrix}1\\-1\end{bmatrix} = \begin{bmatrix}?\\?\\?\end{bmatrix}$, find $\displaystyle T\begin{bmatrix}2\\4\end{bmatrix}$."
I suggest you repeat the problem, writing the vectors as rows rather than columns. This will be a lot easier to read.
$\displaystyle
T\begin{bmatrix}1\\1\end{bmatrix} = \begin{bmatrix}1\\1\\1\end{bmatrix}$
$\displaystyle
T\begin{bmatrix}1\\-1\end{bmatrix} = \begin{bmatrix}1\\0\\1\end{bmatrix}$
we need to solve a(1,1)+b(1,-1)=(x,y)
This gives the system
$\displaystyle a+b=x$
$\displaystyle a-b=y$
Solving gives
$\displaystyle a=\frac{x+y}{2}, \mbox{ and } b=\frac{x-y}{2}$
Now using the linear property of transforms on a(1,1)+b(1,-1)=(x,y) we get
$\displaystyle T(x,y)=T[a(1,1)+b(1,-1)]=T[a(1,1)]+T[b(1,-1)]=aT(1,1)+bT(1,-1)$
$\displaystyle T(x,y)=a(1,1,1)+b(1,0,1)=\left(\frac{x+y}{2},\frac {x+y}{2},\frac{x+y}{2} \right)+\left( \frac{x-y}{2},0,\frac{x-y}{2}\right)$
$\displaystyle T(x,y)=\left( x, \frac{x+y}{2},x\right)$
Now we can use this to find T(2,4)
$\displaystyle T(2,4)=\left( 2, \frac{2+4}{2},2\right)=(2,3,2)$
Here is the link to the La Tex help forum
http://www.mathhelpforum.com/math-help/latex-help/
Here is a link to a few different sites with code
http://amath.colorado.edu/documentat...eX/Symbols.pdf
http://en.wikipedia.org/wiki/Help Formula
You can also look at other code by double clicking on it, or just hoover your mouse over the code.