Let e1 = 1
0

and e2 =

0
1

and let T : R2 ! R2 be a linear transformation that
maps e1 into z1 and maps e2 into z2 where z1 =

6
(d + 5)

, and z2 =

(d + 3)
(d + 4)

.
Find image of the vector

4
(d + 7)

:

The image of \begin{gathered}\left[\begin{array}{c}4\\-4\end{array}\right]\end{gathered}

[

4

−4

]

through T is \begin{gathered}\left[\begin{array}{c}24\\-8\end{array}\right]\end{gathered}

[

24

−8

]

Step-by-step explanation:

We know that T:T: IR^{2}IR

2

→ IR^{2}IR

2

is a linear transformation that maps e_{1}e

1

into y_{1}y

1

⇒

T(e_{1})=y_{1}T(e

1

)=y

1

And also maps e_{2}e

2

into y_{2}y

2

⇒

T(e_{2})=y_{2}T(e

2

)=y

2

We need to find the image of the vector \begin{gathered}\left[\begin{array}{c}4\\-4\end{array}\right]\end{gathered}

[

4

−4

]

We know that exists a matrix A from IR^{2x2}IR

2x2

(because of how T was defined) such that :

T(x)=AxT(x)=Ax for all x ∈ IR^{2}IR

2

We can find the matrix A by applying T to a base of the domain (IR^{2}IR

2

).

Notice that we have that data :

B_{IR^{2}}=B

IR

2

= {e_{1},e_{2}e

1

,e

2

}

Being B_{IR^{2}}B

IR

2

the cannonic base of IR^{2}IR

2

The following step is to put the images from the vectors of the base into the columns of the new matrix A :

\begin{gathered}T(\left[\begin{array}{c}1\\0\end{array}\right])=\left[\begin{array}{c}4\\5\end{array}\right]\end{gathered}

T([

1

0

])=[

4

5

]

(Data of the problem)

\begin{gathered}T(\left[\begin{array}{c}0\\1\end{array}\right])=\left[\begin{array}{c}-2\\7\end{array}\right]\end{gathered}

T([

0

1

])=[

−2

7

]

(Data of the problem)

Writing the matrix A :

\begin{gathered}A=\left[\begin{array}{cc}4&-2\\5&7\\\end{array}\right]\end{gathered}

A=[

4

5

−2

7

]

Now with the matrix A we can find the image of \begin{gathered}\left[\begin{array}{c}4\\-4\\\end{array}\right]\end{gathered}

[

4

−4

]

such as :

T(x)=AxT(x)=Ax ⇒

\begin{gathered}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]\end{gathered}

T([

4

−4

])=[

4

5

−2

7

][

4

−4

]=[

24

−8

]

We found out that the image of \begin{gathered}\left[\begin{array}{c}4\\-4\end{array}\right]\end{gathered}

[

4

−4

]

through T is the vector \begin{gathered}\left[\begin{array}{c}24\\-8\end{array}\right]\end{gathered}

[

24

−8

]

kbmflmklbhml; bmlzb; bdhzz

step-by-step explanation: