## Engage NY Eureka Math Precalculus Module 2 Lesson 6 Answer Key

### Eureka Math Precalculus Module 2 Lesson 6 Exercise Answer Key

Opening Exercise

Let A=\(\left(\begin{array}{ll}

7 & -2 \\

5 & -3

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

x_{1} \\

x_{2}

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

y_{1} \\

y_{2}

\end{array}\right)\). Does this represent a linear transformation? Explain how you know.

Answer:

A linear transformation satisfies the following conditions: L(x+y)=L(x)+L(y) and L(kx)=kL(x)

A(x+y)=A(x)+A(y) by the distributive property

### Eureka Math Precalculus Module 2 Lesson 6 Exploratory Challenge Answer Key

Exploratory Challenge 1:

The Geometry of 3-D Matrix Transformations

a. What matrix in R^{2} serves the role of 1 in the real number system? What is that role?

Answer:

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

1 & 0 \\

0 & 1

\end{array}\right)\) serves the role of the multiplicative identity.

b. What matrix in R^{2} serves the role of 0 in the real number system? What is that role?

Answer:

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

0 & 0 \\

0 & 0

\end{array}\right)\) serves the role of the additive identity.

c. What is the result of scalar multiplication in R^{2}?

Answer:

Multiplying by a scalar, k, dilates each point in R^{2} by a factor of k.

d. Given a complex number a+bi, what represents the transformation of that point across the real axis?

Answer:

The conjugate, a-bi

Exploratory Challenge 2: Properties of Vector Arithmetic

a. Is vector addition commutative? That is, does x+y=y+x for each pair of points in R^{2}? What about points in R^{3}?

Answer:

We want to show that x+y=y+x. First, let’s look at this problem algebraically.

When we compute x+y, we get \(\left(\begin{array}{l}

x_{1} \\

x_{2}

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

y_{1} \\

y_{2}

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

x_{1}+y_{1} \\

x_{2}+y_{2}

\end{array}\right)\).

Now let’s compute y+x. This time, we get \(\left(\begin{array}{l}

y_{1} \\

y_{2}

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

x_{1} \\

x_{2}

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

y_{1}+x_{1} \\

y_{2}+x_{2}

\end{array}\right)\).

The commutative property guarantees that the two components of these vectors are equal. For example, x_{1}+y_{1}=y_{1}+x_{1}. Since both components are equal, the vectors themselves must be equal. We can make a similar argument to show that vector addition in R^{3} is commutative.

Next, let’s see what all of this means geometrically. Let’s examine the points x=\(\left(\begin{array}{l}

5 \\

1

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

3 \\

4

\end{array}\right)\).

Thinking in terms of translations, the sum x+y=\(\left(\begin{array}{l}

5 \\

1

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

3 \\

4

\end{array}\right)\) amounts to moving 5 right and 1 up, followed by a movement that takes us 3 right and 4 up.

On the other hand, the sum y+x=\(\left(\begin{array}{l}

3 \\

4

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

5 \\

1

\end{array}\right)\) amounts to moving 3 right and 4 up, followed by a movement that takes us 5 right and 1 up.

It should be clear that, in both cases, we’ve moved a total of 8 units to the right and 5 units up. So it makes sense that, when viewed as translations, these two sums are the same.

Lastly, let’s consider vector addition in R^{3} from a geometric point of view.

Let’s consider the vectors \(\left(\begin{array}{l}

5 \\

4 \\

2

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

-3 \\

2 \\

3

\end{array}\right)\). To show that addition is commutative, let’s imagine that Jack and Jill are moving around a building. We’ll send them on two different journeys and see if they reach the same destination.

Jack’s movements will model the sum \(\left(\begin{array}{l}

5 \\

4 \\

2

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

-3 \\

2 \\

3

\end{array}\right)\). He walks 5 units east, 4 units north, and then goes up 2 flights of stairs. Next, he goes 3 units west, 2 units north, and then goes up 3 flights of stairs.

Jill starts at the same place as Jack. Her movements will model the sum \(\left(\begin{array}{l}

-3 \\

2 \\

3

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

5 \\

4 \\

2

\end{array}\right)\). She walks 3 units west, 2 units north, and then goes up 3 flights of stairs. Then, she goes 5 units east, 4 units north, and then climbs 2 flights of stairs. It should be clear from this description that Jack and Jill both end up in a location that is 2 units east, 6 units north, and 5 stories above their starting point. In particular, they end up in the same spot!

b. Is vector addition associative? That is, does (x+y)+r=x+(y+r) for any three points in R^{2}? What about points in R^{3}?

Answer:

Let’s check to see if (x+y)+r=x+(y+r) for points in R^{2}.

The associative property guarantees that each of the components is equal. Let’s look at the first coordinate. For example, we have (x_{1}+y_{1})+r_{1}, which is indeed the same as x_{1} +(y_{1} +r_{1}). We can make a similar argument for vectors in R^{3}.

Next, let’s examine the problem from a geometric point of view.

These pictures make it clear that these two sums should be the same. In both cases, the overall journey is equivalent to following each of the three paths separately. Now let’s look at the three-dimensional case.

As in the two-dimensional case, we are simply reaching the same location in two different ways, both of which are equivalent to following the three individual paths separately.

c. Does the distributive property apply to vector arithmetic? That is, does k∙(x+y)=kx+ky for each pair of points in R^{2}? What about points in R^{3}?

Answer:

We want to show that k∙(x+y)=kx+ky.

The distributive property guarantees that each of the components is equal. For example,

k∙(x-+_{1}+y_{1} )=kx_{1}+ky_{1}. This means that the vectors themselves are equal. A similar argument can be used to show that this property holds for vectors in R^{3}.

Now let’s examine this property geometrically.

Suppose that Jack walks to the point marked x+y and then walks that distance again, ending at the spot marked 2(x+y). Now suppose Jill walks to the spot marked 2x and then follows the path labeled 2y. The picture makes it clear that Jack and Jill end up at the same spot.

Next, let’s examine the three-dimensional case. Again, the picture makes it clear that the distributive property holds.

d. Is there an identity element for vector addition? That is, can you find a point a in R^{2} such that x+a=x for every point x in R^{2}? What about for R^{3}?

Answer:

We have x+a=\(\left(\begin{array}{l}

x_{1} \\

x_{2}

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

a_{1} \\

a_{2}

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

x_{1}+a_{1} \\

x_{2}+a_{2}

\end{array}\right)\)

If this sum is equal to x, then we have \(\left(\begin{array}{l}

x_{1}+a_{1} \\

x_{2}+a_{2}

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

x_{1} \\

x_{2}

\end{array}\right)\). This means that x_{1}+a_{1}=x_{1}, which implies that a_{1}=0. In the same way, we can show that a_{2}=0.

Thus, the identity element for R^{2} is \(\left(\begin{array}{l}

0 \\

0

\end{array}\right)\). Similarly, the identity element for R^{3} is \(\left(\begin{array}{l}

\mathbf{0} \\

\mathbf{0} \\

\mathbf{0}

\end{array}\right)\).

Let’s think about this geometrically. \(\left(\begin{array}{l}

2 \\

5

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

0 \\

0

\end{array}\right)\) means to go to (2,5) and translate by (0,0). That is, we don’t translate at all.

e. Does each element in R^{2} have an additive inverse? That is, if you take a point a in R^{2}, can you find a second point b such that a+b=0?

Answer:

We have a+b=\(\left(\begin{array}{l}

a_{1} \\

a_{2}

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

\boldsymbol{b}_{1} \\

\boldsymbol{b}_{2}

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

a_{1}+b_{1} \\

a_{2}+b_{2}

\end{array}\right)\). If this sum is equal to 0, then \(\left(\begin{array}{l}

a_{1}+b_{1} \\

a_{2}+b_{2}

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

0 \\

0

\end{array}\right)\), which means that a_{1}+b_{1}=0. We can conclude that b_{1}=-a_{1}. Similarly, b_{2}=-a_{2}.

Thus, the additive inverse of \(\left(\begin{array}{l}

a_{1} \\

b_{1}

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

-a_{1} \\

-b_{1}

\end{array}\right)\). For instance, the additive inverse of \(\left(\begin{array}{l}

3 \\

5

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

-3 \\

-5

\end{array}\right)\). Geometrically, we see that these vectors have the same length but point in opposite directions. Thus, their sum is (0,0). This makes sense because if Jack walks toward a spot that is 3 miles east and 5 miles north of the origin and then walks toward the spot that is 3 miles west and 5 miles south of the origin, then he’ll end up right back at the origin!

### Eureka Math Precalculus Module 2 Lesson 6 Problem Set Answer Key

Question 1.

Show that the associative property, x+(y+z)=(x+y)+z, holds for the following.

a. x=\(\left(\begin{array}{c}

3 \\

-2

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

-4 \\

2

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

-1 \\

5

\end{array}\right)\)

Answer:

b. x=\(\left(\begin{array}{c}

2 \\

-2 \\

1

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

0 \\

5 \\

-2

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

3 \\

0 \\

-3

\end{array}\right)\)

Answer:

Question 2.

Show that the distributive property, k(x+y)=kx+ky, holds for the following.

a. x=\(\left(\begin{array}{c}

5 \\

-3

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

-2 \\

4

\end{array}\right)\), k=-2

Answer:

b. x=\(\left(\begin{array}{c}

3 \\

-2 \\

5

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

-4 \\

6 \\

-7

\end{array}\right)\), k=-3

Answer:

Question 3.

Compute the following.

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

1 & 0 & 2 \\

0 & 1 & 2 \\

2 & 2 & 3

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

2 \\

1 \\

3

\end{array}\right)\)

Answer:

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

8 \\

7 \\

15

\end{array}\right)\)

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

-1 & 2 & 3 \\

3 & 1 & -2 \\

1 & -2 & 3

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

1 \\

-2 \\

1

\end{array}\right)\)

Answer:

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

-2 \\

-1 \\

8

\end{array}\right)\)

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

1 & 0 & 2 \\

0 & 1 & 0 \\

2 & 0 & 3

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

2 \\

1 \\

2

\end{array}\right)\)

Answer:

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

6 \\

1 \\

10

\end{array}\right)\)

Question 4.

Let x=\(\left(\begin{array}{l}

3 \\

1 \\

2

\end{array}\right)\). Compute L(x)=\(\left[\begin{array}{lll}

a & d & g \\

b & e & h \\

c & f & i

\end{array}\right]\)∙x, plot the points, and describe the geometric effect to x.

a. \(\left[\begin{array}{ccc}

-1 & 0 & 0 \\

0 & 1 & 0 \\

0 & 0 & 1

\end{array}\right]\)

Answer:

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

-3 \\

1 \\

2

\end{array}\right)\) It is reflected about the yz-plane.

b. \(\left[\begin{array}{lll}

2 & 0 & 0 \\

0 & 2 & 0 \\

0 & 0 & 2

\end{array}\right]\)

Answer:

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

6 \\

2 \\

4

\end{array}\right)\) It is dilated by a factor of 2.

c. \(\left[\begin{array}{lll}

0 & 1 & 0 \\

1 & 0 & 0 \\

0 & 0 & 1

\end{array}\right]\)

Answer:

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

1 \\

3 \\

2

\end{array}\right)\) It is reflected about the vertical plane through the line y=x on the xy-plane.

d. \(\left[\begin{array}{lll}

1 & 0 & 0 \\

0 & 0 & 1 \\

0 & 1 & 0

\end{array}\right]\)

Answer:

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

3 \\

2 \\

1

\end{array}\right)\) It is reflected about the vertical plane through the line y=z on the zy-plane.

Question 5.

Let x=\(\left(\begin{array}{l}

3 \\

1 \\

2

\end{array}\right)\). Compute L(x)=\(\left[\begin{array}{lll}

a & d & g \\

b & e & h \\

c & f & i

\end{array}\right]\)∙x. Describe the geometric effect to x.

a. \(\left[\begin{array}{lll}

0 & 0 & 0 \\

0 & 0 & 0 \\

0 & 0 & 0

\end{array}\right]\)

Answer:

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

\mathbf{0} \\

\mathbf{0} \\

\mathbf{0}

\end{array}\right)\) It is mapped to the origin.

b. \(\left[\begin{array}{lll}

3 & 0 & 0 \\

0 & 3 & 0 \\

0 & 0 & 3

\end{array}\right]\)

Answer:

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

9 \\

3 \\

6

\end{array}\right)\) It is dilated by a factor of 3.

c. \(\left[\begin{array}{ccc}

-1 & 0 & 0 \\

0 & -1 & 0 \\

0 & 0 & -1

\end{array}\right]\)

Answer:

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

-3 \\

-1 \\

-2

\end{array}\right)\) It is mapped to the opposite side of the origin on the same line that is equal distance from the origin.

d. \(\left[\begin{array}{ccc}

-1 & 0 & 0 \\

0 & 1 & 0 \\

0 & 0 & 1

\end{array}\right]\)

Answer:

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

-3 \\

1 \\

2

\end{array}\right)\) It is reflected about the yz-plane.

e. \(\left[\begin{array}{ccc}

1 & 0 & 0 \\

0 & -1 & 0 \\

0 & 0 & 1

\end{array}\right]\)

Answer:

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

3 \\

-1 \\

2

\end{array}\right)\) It is reflected about the xz-plane.

f. \(\left[\begin{array}{ccc}

1 & 0 & 0 \\

0 & 1 & 0 \\

0 & 0 & -1

\end{array}\right]\)

Answer:

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

3 \\

\mathbf{1} \\

-2

\end{array}\right)\) It is reflected about the xy-plane.

g. \(\left[\begin{array}{lll}

0 & 1 & 0 \\

1 & 0 & 0 \\

0 & 0 & 1

\end{array}\right]\)

Answer:

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

1 \\

3 \\

2

\end{array}\right)\) It is reflected about the vertical plane through the line y=x on the xy-plane.

h. \(\left[\begin{array}{lll}

1 & 0 & 0 \\

0 & 0 & 1 \\

0 & 1 & 0

\end{array}\right]\)

Answer:

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

3 \\

2 \\

1

\end{array}\right)\) It is reflected about the vertical plane through the line y=z on the yz-plane.

i. \(\left[\begin{array}{lll}

0 & 0 & 1 \\

0 & 1 & 0 \\

1 & 0 & 0

\end{array}\right]\)

Answer:

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

2 \\

1 \\

3

\end{array}\right)\) It is reflected about the vertical plane through the line x=z on the xz-plane.

Question 6.

Find the matrix that will transform the point x=\(\left(\begin{array}{l}

1 \\

3 \\

2

\end{array}\right)\) to the following point:

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

-4 \\

-12 \\

-8

\end{array}\right)\)

Answer:

\(\left[\begin{array}{ccc}

-4 & 0 & 0 \\

0 & -4 & 0 \\

0 & 0 & -4

\end{array}\right]\)

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

3 \\

1 \\

2

\end{array}\right)\)

Answer:

\(\left[\begin{array}{lll}

0 & 1 & 0 \\

1 & 0 & 0 \\

0 & 0 & 1

\end{array}\right]\)

Question 7.

Find the matrix/matrices that will transform the point x=\(\left(\begin{array}{l}

2 \\

3 \\

1

\end{array}\right)\) to the following point:

a. x’=\(\left(\begin{array}{l}

6 \\

4 \\

2

\end{array}\right)\)

Answer:

\(\left[\begin{array}{lll}

0 & 2 & 0 \\

2 & 0 & 0 \\

0 & 0 & 2

\end{array}\right]\)

b. x’=\(\left(\begin{array}{c}

-1 \\

3 \\

2

\end{array}\right)\)

Answer:

\(\left[\begin{array}{ccc}

0 & 0 & -1 \\

0 & 1 & 0 \\

1 & 0 & 0

\end{array}\right]\)

### Eureka Math Precalculus Module 2 Lesson 6 Exit Ticket Answer Key

Question 1.

Given x=\(\left(\begin{array}{l}

1 \\

2

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

4 \\

2

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

3 \\

1

\end{array}\right)\), and k=-2

a. Verify the associative property holds: x+(y+z)=(x+y)+z.

Answer:

b. Verify the distributive property holds: k(x+y)=kx+ky.

Answer:

Question 2.

Describe the geometric effect of the transformation on the 3×3 identity matrix given by the following matrices.

a. \(\left[\begin{array}{lll}

2 & 0 & 0 \\

0 & 2 & 0 \\

0 & 0 & 2

\end{array}\right]\)

Answer:

It dilates the point by a factor of 2.

b. \(\left[\begin{array}{ccc}

-1 & 0 & 0 \\

0 & 1 & 0 \\

0 & 0 & 1

\end{array}\right]\)

Answer:

It reflects the point about the yz-plane.

c. \(\left[\begin{array}{ccc}

1 & 0 & 0 \\

0 & -1 & 0 \\

0 & 0 & 1

\end{array}\right]\)

Answer:

It reflects the point about the xz-plane.

d. \(\left[\begin{array}{ccc}

1 & 0 & 0 \\

0 & 1 & 0 \\

0 & 0 & -1

\end{array}\right]\)

Answer:

It reflects the point about the xy-plane.