Question

If T : R2 → R2 is a linear transformation such that T1
0 = 2
5 and
T0
1 = −1
6 , find T 5
−3.
(

Answer 1

oh wow that didnt work lol
It says
T(\[\left(\begin{matrix}1 \\ 0\end{matrix}\right) =\left(\begin{matrix}2 \\ 5\end{matrix}\right)\]

\[T\left(\begin{matrix}0 \\ 1\end{matrix}\right)=\left(\begin{matrix}-1 \\ 6\end{matrix}\right)\]

Find\[T\left(\begin{matrix}5 \\ -3\end{matrix}\right)\]

Answer 2

Ok, so the first equation gives you first column of matrix; the second equation the second.

\[T = \left[\begin{matrix} 2 & -1 \\ 5 & 6 \end{matrix}\right]\]

You should convince yourself of that by multiplying back out the two equations.

Now that you have T, it should be easy to calculate

\[\ T \left(\begin{matrix} 5 \\ -3 \end{matrix}\right)\]

Answer 3

\[T \left(\begin{matrix} 5 \\ -3 \end{matrix}\right)=5\cdot T \left(\begin{matrix} 1 \\ 0 \end{matrix}\right)+-3\cdot T \left(\begin{matrix} 0 \\ 1 \end{matrix}\right)\]

Answer 4

Alternatively note that

\[\left(\begin{matrix} 5 \\ -3 \end{matrix}\right) = 5 \left(\begin{matrix} 1 \\ 0 \end{matrix}\right) + -3 \left(\begin{matrix}0 \\ -1 \end{matrix}\right)\]

So .. and Zarkon just beat me too it!

Answer 5

so... the answer would be

\[\left(\begin{matrix}13\\ 7\end{matrix}\right)\]

Its that easy? lol

Answer 6

If it is that easy my professor needs to stop making it sound so complicated during class

Answer 7

it is that easy

Answer 8

Can you help me with some more "easy" ones that arent related to this kind of problem but i need some clarity?