## Engage NY Eureka Math Precalculus Module 1 Lesson 21 Answer Key

### Eureka Math Precalculus Module 1 Lesson 21 Exercise Answer Key

Opening Exercise

Suppose that L_{1}(x,y)=(2x-3y,3x+2y) and L_{2}(x,y)=(3x+4y,-4y+3x).

Find the result of performing L_{1}and then L_{2}on a point (p,q). That is, find L_{2}(L_{1}(p,q)).

Answer:

L_{2}(L_{1}(p,q))=L_{2}(2p-3q,3p+2q)

=(3(2p-3q)+4(3p+2q),-4(2p-3q)+3(3p+2q))

=(6p-9q+12p+8q,-8p+12q+9p+6q)

=(18p-q,p+18q)

Exercises

Exercise 1.

Calculate each of the following products.

a. \(\left(\begin{array}{cc}

3 & -2 \\

-1 & 4

\end{array}\right)\left(\begin{array}{l}

1 \\

5

\end{array}\right)\)

Answer:

\(\left(\begin{array}{c}

3-10 \\

-1+20

\end{array}\right)\) = \(\left(\begin{array}{c}

-7 \\

19

\end{array}\right)\)

b. \(\left(\begin{array}{ll}

3 & 3 \\

3 & 3

\end{array}\right)\left(\begin{array}{c}

4 \\

-4

\end{array}\right)\)

Answer:

\(\left(\begin{array}{ll}

3 & 3 \\

3 & 3

\end{array}\right)\left(\begin{array}{c}

4 \\

-4

\end{array}\right)\)\(\left(\begin{array}{l}

12-12 \\

12-12

\end{array}\right)\) = \(\left(\begin{array}{l}

\mathbf{0} \\

0

\end{array}\right)\)

c. \(\left(\begin{array}{ll}

2 & -4 \\

5 & -1

\end{array}\right)\left(\begin{array}{c}

3 \\

-2

\end{array}\right)\)

Answer:

\(\left(\begin{array}{cc}

2 & -4 \\

5 & -1

\end{array}\right)\left(\begin{array}{c}

3 \\

-2

\end{array}\right)\) = \(\) = \(\left(\begin{array}{l}

14 \\

17

\end{array}\right)\)

Exercise 2.

Find a value of k so that \(\left(\begin{array}{ll}

1 & 2 \\

k & 1

\end{array}\right)\left(\begin{array}{c}

3 \\

-1

\end{array}\right)\)=\(\left(\begin{array}{c}

1 \\

11

\end{array}\right)\).

Answer:

Multiplying this out, we have \(\left(\begin{array}{ll}

1 & 2 \\

k & 1

\end{array}\right)\left(\begin{array}{c}

3 \\

-1

\end{array}\right)\)=\(\left(\begin{array}{c}

\mathbf{1} \\

3 \boldsymbol{k}-\mathbf{1}

\end{array}\right)\)=\(\left(\begin{array}{c}

1 \\

11

\end{array}\right)\), so 3k-1=11, and thus, k=4.

Exercise 3.

Find a matrix \(\left(\begin{array}{ll}

a & b \\

c & d

\end{array}\right)\) so that we can represent the transformation L(x,y)=(2x-3y,3x+2y) by \(\).

Answer:

The matrix is \(\left(\begin{array}{cc}

2 & -3 \\

3 & 2

\end{array}\right)\).

Exercise 4.

If a transformation L\(\left(\begin{array}{l}

x \\

y

\end{array}\right)\)=\(\left(\begin{array}{ll}

a & b \\

c & d

\end{array}\right)\left(\begin{array}{l}

x \\

y

\end{array}\right)\) has the geometric effect of rotation and dilation, what do you know about the values a,b,c, and d?

Answer:

Since the transformation L(x,y)=(ax-by,bx+ay) has matrix representation L\(\left(\begin{array}{l}

x \\

y

\end{array}\right)\)=\(\left(\begin{array}{cc}

a & -b \\

b & a

\end{array}\right)\left(\begin{array}{l}

x \\

y

\end{array}\right)\), we know that a=d and c=-b.

Exercise 5.

Describe the form of a matrix \(\left(\begin{array}{ll}

a & b \\

c & d

\end{array}\right)\) so that the transformation L\(\left(\begin{array}{l}

x \\

y

\end{array}\right)\)=\(\left(\begin{array}{ll}

a & b \\

c & d

\end{array}\right)\left(\begin{array}{l}

x \\

y

\end{array}\right)\) has the geometric effect of only dilation by a scale factor r.

Answer:

The transformation that scales by factor r has the form L(x,y)=r(x,y)=(rx,ry)=(rx-0y,0x+ry), so the matrix has the form \(\left(\begin{array}{ll}

\boldsymbol{r} & \mathbf{0} \\

\mathbf{0} & \boldsymbol{r}

\end{array}\right)\).

Exercise 6.

Describe the form of a matrix \(\left(\begin{array}{ll}

a & b \\

c & d

\end{array}\right)\) so that the transformation L\(\left(\begin{array}{l}

x \\

y

\end{array}\right)\)=\(\left(\begin{array}{ll}

a & b \\

c & d

\end{array}\right)\left(\begin{array}{l}

x \\

y

\end{array}\right)\) has the geometric effect of only rotation by θ. Describe the matrix in terms of θ.

Answer:

The matrix has the form \(\left(\begin{array}{cc}

a & -b \\

b & a

\end{array}\right)\), where arg(a+bi)=θ. Thus, a=cos(θ) and b=sin(θ), so the matrix has the form \(\left(\begin{array}{cc}

\cos (\theta) & -\sin (\theta) \\

\sin (\theta) & \cos (\theta)

\end{array}\right)\).

Exercise 7.

Describe the form of a matrix \(\left(\begin{array}{ll}

a & b \\

c & d

\end{array}\right)\) so that the transformation L\(\left(\begin{array}{l}

x \\

y

\end{array}\right)\)=\(\left(\begin{array}{ll}

a & b \\

c & d

\end{array}\right)\left(\begin{array}{l}

x \\

y

\end{array}\right)\) has the geometric effect of rotation by θ and dilation with scale factor r. Describe the matrix in terms of θ and r.

Answer:

The matrix has the form \(\left(\begin{array}{cc}

a & -b \\

b & a

\end{array}\right)\), where arg(a+bi)=θ and r=|a+bi|. Thus, a=r cos(θ) and b=r sin(θ), so the matrix has the form \(\left(\begin{array}{cc}

r \cos (\theta) & -r \sin (\theta) \\

r \sin (\theta) & r \cos (\theta)

\end{array}\right)\).

Exercise 8.

Suppose that we have a transformation L\(\) = \(\).

a. Does this transformation have the geometric effect of rotation and dilation?

Answer:

No, the matrix is not in the form \(\), so this transformation is not a rotation and dilation.

b. Transform each of the points A=\(\left(\begin{array}{l}

0 \\

0

\end{array}\right)\), B = \(\left(\begin{array}{l}

1 \\

0

\end{array}\right)\), C = \(\left(\begin{array}{l}

1 \\

1

\end{array}\right)\), and D = \(\left(\begin{array}{l}

0 \\

1

\end{array}\right)\) and plot the images in the plane shown.

Answer:

Exercise 9.

Describe the geometric effect of the transformation L\(\left(\begin{array}{l}

x \\

y

\end{array}\right)\)=\(\left(\begin{array}{ll}

1 & 0 \\

0 & 1

\end{array}\right)\left(\begin{array}{l}

x \\

y

\end{array}\right)\).

Answer:

This transformation does nothing to the point (x,y) in the plane; it is the identity transformation.

### Eureka Math Precalculus Module 1 Lesson 21 Problem Set Answer Key

Question 1.

Perform the indicated multiplication.

a. \(\left(\begin{array}{ll}

1 & 2 \\

4 & 8

\end{array}\right)\left(\begin{array}{c}

3 \\

-2

\end{array}\right)\)

Answer:

\(\left(\begin{array}{l}

-1 \\

-4

\end{array}\right)\)

b. \(\left(\begin{array}{cc}

3 & 5 \\

-2 & -6

\end{array}\right)\left(\begin{array}{l}

2 \\

4

\end{array}\right)\)

Answer:

\(\left(\begin{array}{c}

26 \\

-28

\end{array}\right)\)

c. \(\left(\begin{array}{cc}

1 & 1 \\

1 & -1

\end{array}\right)\left(\begin{array}{l}

6 \\

8

\end{array}\right)\)

Answer:

\(\left(\begin{array}{l}

14 \\

-2

\end{array}\right)\)

d. \(\left(\begin{array}{ll}

5 & 7 \\

4 & 9

\end{array}\right)\left(\begin{array}{c}

10 \\

100

\end{array}\right)\)

Answer:

\(\left(\begin{array}{l}

750 \\

940

\end{array}\right)\)

e. \(\left(\begin{array}{ll}

4 & 2 \\

3 & 7

\end{array}\right)\left(\begin{array}{c}

-3 \\

1

\end{array}\right)\)

Answer:

\(\left(\begin{array}{c}

-10 \\

-2

\end{array}\right)\)

f. \(\left(\begin{array}{ll}

6 & 4 \\

9 & 6

\end{array}\right)\left(\begin{array}{c}

2 \\

-3

\end{array}\right)\)

Answer:

\(\left(\begin{array}{l}

\mathbf{0} \\

0

\end{array}\right)\)

g. \(\left(\begin{array}{cc}

\cos (\theta) & -\sin (\theta) \\

\sin (\theta) & \cos (\theta)

\end{array}\right)\left(\begin{array}{l}

x \\

y

\end{array}\right)\)

Answer:

\(\left(\begin{array}{l}

x \cos (\theta)-y \sin (\theta) \\

x \sin (\theta)+y \cos (\theta)

\end{array}\right)\)

h. \(\left(\begin{array}{cc}

\pi & 1 \\

1 & -\pi

\end{array}\right)\left(\begin{array}{c}

10 \\

7

\end{array}\right)\)

Answer:

\(\left(\begin{array}{l}

10 \pi+7 \\

10-7 \pi

\end{array}\right)\)

Question 2.

Find a value of k so that .

Answer:

We have , so 4k+15=7 and 16+5k=6. Thus, 4k=-8 and 5k=-10, so

k=-2.

Question 3.

Find values of k and m so that .

Answer:

We have , so 5k+12=7 and -10+4m=-10. Therefore, k=-1 and m=0.

Question 4.

Find values of k and m so that .

Answer:

Since , we need to find values of k and m so that k+2m=0 and

-2k+5m=-9. Solving this first equation for k gives k=-2m, and substituting this expression for k into the second equation gives –9=-2(-2m)+5m=9m, so we have m=-1. Then, k=-2m gives k=2. Therefore, k=2 and m=-1.

Question 5.

Write the following transformations using matrix multiplication.

a. L(x,y)=(3x-2y,4x-5y)

Answer:

b. L(x,y)=(6x+10y,-2x+y)

Answer:

c. L(x,y)=(25x+10y,8x-64y)

Answer:

d. L(x,y)=(πx-y,-2x+3y)

Answer:

e. L(x,y)=(10x,100x)

Answer:

f. L(x,y)=(2y,7x)

Answer:

Question 6.

Identify whether or not the following transformations have the geometric effect of rotation only, dilation only, rotation and dilation only, or none of these.

a.

Answer:

The matrix cannot be written in the form , because 3≠-5, so this is neither a rotation nor a dilation. The transformation L is not one of the specified types of transformations.

b.

Answer:

This transformation has the geometric effect of dilation by a scale factor of 42.

c.

Answer:

The matrix has the form with a=-4 and b=2. Therefore, this transformation has the geometric effect of rotation and dilation.

d.

The matrix cannot be written in the form , because -1≠-(-1), so this is neither a rotation nor a dilation. The transformation L is not one of the specified types of transformations.

e.

Answer:

The matrix cannot be written in the form , because -7≠7, so this is neither a rotation nor a dilation. The transformation L is not one of the specified types of transformations.

f.

Answer:

We see that , so this transformation has the geometric effect of dilation by \(\sqrt{2}\) and rotation by \(\frac{\pi}{2}\).

Question 7.

Create a matrix representation of a linear transformation that has the specified geometric effect.

a. Dilation by a factor of 4 and no rotation

Answer:

b. Rotation by 180° and no dilation

Answer:

c. Rotation by –\(\frac{π}{2}\)rad and dilation by a scale factor of 3

Answer:

d. Rotation by 30° and dilation by a scale factor of 4

Answer:

Question 8.

Identify the geometric effect of the following transformations. Justify your answers.

a.

Answer:

Since , this transformation has the form

and, thus, represents counterclockwise rotation by \(\frac{3π}{4}\) with no dilation.

b.

Answer:

Since cos(\(\frac{π}{2}\))=0 and sin(\(\frac{π}{2}\))=1, this transformation has the form and, thus, represents counterclockwise rotation by \(\frac{π}{2}\) and dilation by a scale factor 5.

c.

Answer:

Since cos(π)=-1 and sin(π)=0, this transformation has the form

and, thus, represents counterclockwise rotation by π and dilation by a scale factor 10.

d.

Answer:

Since cos(\(\frac{5π}{3}\))=\(\frac{1}{2}\) and sin(\(\frac{5π}{3}\))=-\(\frac{\sqrt{3}}{2}\), this transformation has the form

and, thus, represents counterclockwise rotation by \(\frac{5π}{3}\) and dilation with scale factor 12.

### Eureka Math Precalculus Module 1 Lesson 21 Exit Ticket Answer Key

Question 1.

Evaluate the product \(\left(\begin{array}{cc}

10 & 2 \\

-8 & -5

\end{array}\right)\left(\begin{array}{c}

3 \\

-2

\end{array}\right)\).

Answer:

\(\left(\begin{array}{cc}

10 & 2 \\

-8 & -5

\end{array}\right)\left(\begin{array}{c}

3 \\

-2

\end{array}\right)\) = \(\left(\begin{array}{c}

30-4 \\

-24+10

\end{array}\right)\)

= \(\left(\begin{array}{c}

26 \\

-14

\end{array}\right)\)

Question 2.

Find a matrix representation of the transformation L(x,y)=(3x+4y,x-2y).

Answer:

L\(\left(\begin{array}{l}

x \\

y

\end{array}\right)\) = \(\left(\begin{array}{cc}

3 & 4 \\

1 & -2

\end{array}\right)\left(\begin{array}{l}

x \\

y

\end{array}\right)\)

Question 3.

Does the transformation L\(\left(\begin{array}{l}

x \\

y

\end{array}\right)\)= \(\left(\begin{array}{cc}

5 & 2 \\

-2 & 5

\end{array}\right)\left(\begin{array}{l}

x \\

y

\end{array}\right)\) represent a rotation and dilation in the plane? Explain how you know.

Answer:

Yes; this transformation can also be represented as L(x,y)=(5x-(-2)y,-2x+5y), which has the geometric effect of counterclockwise rotation by arg(5-2i) and dilation by |5-2i|.