Answer:
The image of [tex]\left[\begin{array}{c}4&-4\end{array}\right][/tex] through T is [tex]\left[\begin{array}{c}24&-8\end{array}\right][/tex]
Step-by-step explanation:
We know that [tex]T:[/tex] [tex]IR^{2}[/tex] → [tex]IR^{2}[/tex] is a linear transformation that maps [tex]e_{1}[/tex] into [tex]y_{1}[/tex] ⇒
[tex]T(e_{1})=y_{1}[/tex]
And also maps [tex]e_{2}[/tex] into [tex]y_{2}[/tex] ⇒
[tex]T(e_{2})=y_{2}[/tex]
We need to find the image of the vector [tex]\left[\begin{array}{c}4&-4\end{array}\right][/tex]
We know that exists a matrix A from [tex]IR^{2x2}[/tex] (because of how T was defined) such that :
[tex]T(x)=Ax[/tex] for all x ∈ [tex]IR^{2}[/tex]
We can find the matrix A by applying T to a base of the domain ([tex]IR^{2}[/tex]).
Notice that we have that data :
[tex]B_{IR^{2}}=[/tex] {[tex]e_{1},e_{2}[/tex]}
Being [tex]B_{IR^{2}}[/tex] the cannonic base of [tex]IR^{2}[/tex]
The following step is to put the images from the vectors of the base into the columns of the new matrix A :
[tex]T(\left[\begin{array}{c}1&0\end{array}\right])=\left[\begin{array}{c}4&5\end{array}\right][/tex] (Data of the problem)
[tex]T(\left[\begin{array}{c}0&1\end{array}\right])=\left[\begin{array}{c}-2&7\end{array}\right][/tex] (Data of the problem)
Writing the matrix A :
[tex]A=\left[\begin{array}{cc}4&-2\\5&7\\\end{array}\right][/tex]
Now with the matrix A we can find the image of [tex]\left[\begin{array}{c}4&-4\\\end{array}\right][/tex] such as :
[tex]T(x)=Ax[/tex] ⇒
[tex]T(\left[\begin{array}{c}4&-4\end{array}\right])=\left[\begin{array}{cc}4&-2\\5&7\\\end{array}\right]\left[\begin{array}{c}4&-4\end{array}\right]=\left[\begin{array}{c}24&-8\end{array}\right][/tex]
We found out that the image of [tex]\left[\begin{array}{c}4&-4\end{array}\right][/tex] through T is the vector [tex]\left[\begin{array}{c}24&-8\end{array}\right][/tex]
