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

]

2. angles qrt and qst both measures 156 degrees.

step-by-step explanation:  

we have been given a diagram of a circle and we are told that measure of arc qvt is 156 degrees.  

we can see from our given diagram that angle qrt and angle qst subtends to arc qvt.

since we know that angles subtended by the same arc at the circumference are always equal.

by angle subtended by same arc theorem, measure of angle qrt and angle qst will be equal to the measure of arc qvt.

therefore, angles qrt and qst both measures 156 degrees and 2nd statement is the correct choice.