# Eureka Math Precalculus Module 2 Lesson 4 Answer Key

## Engage NY Eureka Math Precalculus Module 2 Lesson 4 Answer Key

### Eureka Math Precalculus Module 2 Lesson 4 Exercise Answer Key

Exercises

Exercise 1.
Describe the geometric effect of each mapping.
a. L(x)=9∙x
Dilates the interval by a factor of 9

b. L(x)=-$$\frac{1}{2}$$∙x
Reflects the interval over the origin and then applies a dilation with a scale factor of $$\frac{1}{2}$$

Exercise 2.
Write the formula for the mappings described.
a. A dilation that expands each interval to 5 times its original size
L(x)=5∙x

b. A collapse of the interval to the number 0
L(x)=0∙x

### Eureka Math Precalculus Module 2 Lesson 4 Problem Set Answer Key

Question 1.
Suppose you have a linear transformation L:R→R, where L(3)=6, L(5)=10.
a. Use the addition property to find L(6), L(8), L(10), and L(13).
L(6)=L(3+3)=L(3)+L(3)=6+6=12
L(8)=L(3+5)=L(3)+L(5)=6+10=16
L(10)=L(5)+L(5)=10+10=20
L(13)=L(10+3)=L(10)+L(3)=20+6=26

b. Use the multiplication property to find L(15), L(18), and L(30).
L(15)=L(3∙5)=3∙L(5)=3∙10=30
L(18)=L(3∙6)=3∙L(6)=3∙12=36
L(30)=L(5∙6)=5∙L(6)=5∙12=60

c. Find L(-3), L(-8), and L(-15).
L(-3)=L(-1∙3)=-1∙L(3)=-6
L(-8)=L(-1∙8)=-1∙L(8)=-16
L(-15)=L(-3∙5)=-3∙L(5)=-3∙10=-30

d. Find the formula for L(x).
Given L(x) is a linear transformation; therefore, it must have a form of L(x)=mx, where m is real number.
Given L(3)=6; therefore, 3m=6, m=2. L(x)=2x.

e. Draw the graph of the function L(x).

Question 2.
A linear transformation L: R→R must have the form of L(x)=ax for some real number a. Consider the interval [-5,2]. Describe the geometric effect of the following, and find the new interval.
a. L(x)=5x
It dilates the interval by a scale factor of 5; [-25,10]

b. L(x)=-2x
It reflects the interval over the origin and then dilates it with a scale factor of 2; [-4,10]

Question 3.
A linear transformation L: R→R must have the form of L(x)=ax for some real number a. Consider the interval [-2,6]. Write the formula for the mapping described, and find the new interval.

a. A reflection over the origin
L(x)=-x, [-6,2]

b. A dilation with a scale of $$\sqrt{2}$$
L(x)=$$\sqrt{2}$$∙x, [-2$$\sqrt{2}$$,6$$\sqrt{2}$$]

c. A reflection over the origin and a dilation with a scale of $$\frac{1}{2}$$
L(x)=-$$\frac{1}{2}$$x, [-3,1]

d. A collapse of the interval to the number 0
L(x)=0x, [0,0]

Question 4.
In Module 1, we used 2×2 matrices to do transformations on a square, such as a pure rotation, a pure reflection, a pure dilation, and a rotation with a dilation. Now use those matrices to do transformations on this complex number: z=2+i. For each transformation below, graph your answers.
a. A pure dilation with a factor of 2

$$\left(\begin{array}{ll} 2 & 0 \\ 0 & 2 \end{array}\right)$$, L(z) = $$\left(\begin{array}{ll} 2 & 0 \\ 0 & 2 \end{array}\right)\left(\begin{array}{l} 2 \\ 1 \end{array}\right)$$ = $$\left(\begin{array}{l} 4 \\ 2 \end{array}\right)$$

b. A pure $$\frac{\pi}{2}$$ radians counterclockwise rotation about the origin

$$\left(\begin{array}{cc} 0 & -1 \\ 1 & 0 \end{array}\right)$$, L(z) = $$\left(\begin{array}{cc} 0 & -1 \\ 1 & 0 \end{array}\right)\left(\begin{array}{l} 2 \\ 1 \end{array}\right)$$ = $$\left(\begin{array}{c} -1 \\ 2 \end{array}\right)$$

$$\left(\begin{array}{cc} -1 & 0 \\ 0 & -1 \end{array}\right)$$, L(z) = $$\left(\begin{array}{cc} -1 & 0 \\ 0 & -1 \end{array}\right)\left(\begin{array}{l} 2 \\ 1 \end{array}\right)$$ = $$\left(\begin{array}{l} -2 \\ -1 \end{array}\right)$$

d. A pure reflection about the real axis

$$\left(\begin{array}{cc} 1 & 0 \\ 0 & -1 \end{array}\right)$$, L(z) = $$\left(\begin{array}{cc} 1 & 0 \\ 0 & -1 \end{array}\right)\left(\begin{array}{l} 2 \\ 1 \end{array}\right)$$ = $$\left(\begin{array}{c} 2 \\ -1 \end{array}\right)$$

e. A pure reflection about the imaginary axis

$$\left(\begin{array}{cc} -1 & 0 \\ 0 & 1 \end{array}\right)$$, L(z) = $$\left(\begin{array}{cc} -1 & 0 \\ 0 & 1 \end{array}\right)\left(\begin{array}{l} 2 \\ 1 \end{array}\right)$$ = $$\left(\begin{array}{c} -2 \\ 1 \end{array}\right)$$

f. A pure reflection about the line y=x

$$\left(\begin{array}{ll} 0 & 1 \\ 1 & 0 \end{array}\right)$$, L(z) = $$\left(\begin{array}{ll} 0 & 1 \\ 1 & 0 \end{array}\right)\left(\begin{array}{l} 2 \\ 1 \end{array}\right)$$ = $$\left(\begin{array}{l} 1 \\ 2 \end{array}\right)$$

g. A pure reflection about the line y=-x

$$\left(\begin{array}{cc} 0 & -1 \\ -1 & 0 \end{array}\right)$$, L(z) = $$\left(\begin{array}{cc} 0 & -1 \\ -1 & 0 \end{array}\right)\left(\begin{array}{l} 2 \\ 1 \end{array}\right)$$ = $$\left(\begin{array}{l} -1 \\ -2 \end{array}\right)$$

Question 5.
Wesley noticed that multiplying the matrix $$\left(\begin{array}{cc} 0 & -1 \\ 1 & 0 \end{array}\right)$$ by a complex number z produces a pure $$\frac{\pi}{2}$$ radians counterclockwise rotation, and multiplying by $$\left(\begin{array}{ll} 2 & 0 \\ 0 & 2 \end{array}\right)$$ produces a pure dilation with a factor of 2. So, he thinks he can add these two matrices, which will produce $$\left(\begin{array}{cc} 2 & -1 \\ 1 & 2 \end{array}\right)$$ and will rotate z by $$\frac{\pi}{2}$$ radians counterclockwise and dilate z with a factor of 2. Is he correct? Explain your reason.
No, he is not correct. For the general transformation of complex numbers, the form is L(Z)=$$\left(\begin{array}{cc} a & -b \\ b & a \end{array}\right)\left(\begin{array}{l} x \\ y \end{array}\right)$$.
By multiplying the matrix $$\left(\begin{array}{cc} a & -b \\ b & a \end{array}\right)$$ to z, it rotates z an angle arctan($$\frac{b}{a}$$) and dilates z with a factor of $$\sqrt{a^{2}+b^{2}}$$.
arctan$$\frac{1}{2}$$=26.565°, and $$\sqrt{(1)^{2}+(2)^{2}}$$=$$\sqrt{5}$$, which is not $$\frac{\pi}{2}$$ radians or a dilation of a factor of 2.

Question 6.
In Module 1, we learned that there is not any real number that will satisfy $$\frac{1}{a+b}$$=$$\frac{1}{a}$$+$$\frac{1}{b}$$, which is the addtition property of linear transformation. However, we discussed that some fixed complex numbers might work. Can you find two pairs of complex numbers that will work? Show your work.
Given $$\frac{1}{a+b}$$=$$\frac{1}{a}$$+$$\frac{1}{b}$$, a,b≠0
$$\frac{1}{a+b}$$=$$\frac{b+a}{ab}$$
ab=(a+b)2
ab=a2+2ab+b2
a2+ab+b2=0
a2+ab+$$\frac{1}{4}$$b2=-$$\frac{3}{4}$$b2
(a+$$\frac{1}{2}$$ b)2=-$$\frac{3}{4}$$b2
a+$$\frac{1}{2}$$b=±$$\frac{\sqrt{3}}{2}$$b∙i
a=$$\frac{-1 \pm \sqrt{3} i}{2} b$$
For example, b=2 and a=-1+$$\sqrt{3}$$∙i or a=-1-$$\sqrt{3}$$∙i. If b=4 and a=-2+2$$\sqrt{3}$$∙i or a=-2-2$$\sqrt{3}$$∙i. Answers may vary.

Question 7.
Suppose L is a complex-number function that satisfies the dream conditions: L(z+w)=L(z)+L(w) and
L(kz)=kL(z) for all complex numbers z, w, and k. Show L(z)=mz for a fixed complex number m, the only type of complex-number function that satisfies these conditions.
L(z+w)=L(z)+L(w)=m(a+bi)+m(c+di)=m((a+c)+(b+d)i)
L(z+w)=L(a+bi+c+di)=L((a+c)+(b+d)i)=m((a+c)+(b+d)i); they are the same.
For multiplication property:
L(kz)=kL(z)=kmz=km(a+bi)
L(kz)=L(k(a+bi))=mk(a+bi); they are the same.

Question 8.
For complex numbers, the linear transformation requires L(x+y)=L(x)+L(y),L(a∙x)=a∙L(x). Prove that in general L$$\left(\begin{array}{l} x \\ y \end{array}\right)$$=$$\left(\begin{array}{ll} a & b \\ c & d \end{array}\right)\left(\begin{array}{l} x \\ y \end{array}\right)$$ is a linear transformation, where $$\left(\begin{array}{l} x \\ y \end{array}\right)$$ represents z=x+yi.

Now we need to prove the multiplication property.

### Eureka Math Precalculus Module 2 Lesson 4 Exit Ticket Answer Key

Question 1.
In Module 1, we learned about linear transformations for any real number functions. What are the conditions of a linear transformation? If a real number function is a linear transformation, what is its form? What are the two characteristics of the function?
L(x+y)=L(x)+L(y), L(kx)=kL(x), where x, y, and k are real numbers
It is in the form of L(x)=mx, and its graph is a straight line going through the origin. It is an odd function.

Question 2.
Describe the geometric effect of each mapping:
a. L(x)=3x
Dilate the interval by a factor of 3.

b. L(z)=($$\sqrt{2}$$+$$\sqrt{2}$$i)∙z
Rotate $$\frac{\pi}{4}$$ radians counterclockwise about the origin.
c. L(z)=$$\left(\begin{array}{cc} 0 & -1 \\ 1 & 0 \end{array}\right)\left(\begin{array}{l} x \\ y \end{array}\right)$$, where z is a complex number
Rotate $$\frac{\pi}{2}$$ radians counterclockwise about the origin.
d. L(z)=$$\left(\begin{array}{ll} 2 & 0 \\ 0 & 2 \end{array}\right)\left(\begin{array}{l} x \\ y \end{array}\right)$$, where z is a complex number