## Engage NY Eureka Math Precalculus Module 2 Lesson 12 Answer Key

### Eureka Math Precalculus Module 2 Lesson 12 Example Answer Key

Example

In three-dimensional space, let A represent a rotation of 90° about the x-axis, B represent a reflection about the

yz-plane, and C represent a rotation of 180° about the z-axis. Let X=\(\left[\begin{array}{l}

1 \\

1 \\

1

\end{array}\right]\).

a. As best you can, sketch a three-dimensional set of axes and the location of the point X.

Answer:

b. Using only your geometric intuition, what are the coordinates of BX? CX? Explain your thinking.

Answer:

BX=\(\left[\begin{array}{l}

-1 \\

1 \\

1

\end{array}\right]\); CX=\(\left[\begin{array}{l}

-1 \\

-1 \\

1

\end{array}\right]\); answers will vary but could include that when rotating about the x-axis 90°, only the x-coordinate would change signs; however, when rotating about the z-axis 180°, the x- and

y-coordinates would change signs.

c. Write down matrices B and C, and verify or disprove your answers to part (b).

Answer:

d. What is the sum of BX+CX?

Answer:

BX+CX=\(\left[\begin{array}{c}

-2 \\

0 \\

2

\end{array}\right]\)

e. Write down matrix A, and compute A(BX+CX).

Answer:

f. Compute AB and AC.

Answer:

g. Compute (AB)X, (AC)X, and their sum. Compare your result to your answer to part (e). What do you notice?

Answer:

A(BX+CX)=(AB)X+(AC)X

h. In general, must A(B+C) and AB+AC have the same geometric effect on a point, no matter what matrices A, B, and C are? Explain.

Answer:

Yes. See full explanation in questions above.

### Eureka Math Precalculus Module 2 Lesson 12 Exercise Answer Key

Opening Exercise

Write the 3×3 matrix that would represent the transformation listed.

a. No change when multiplying (the multiplicative identity matrix)

Answer:

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

1 & 0 & 0 \\

0 & 1 & 0 \\

0 & 0 & 1

\end{array}\right]\)

b. No change when adding (the additive identity matrix)

Answer:

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

0 & 0 & 0 \\

0 & 0 & 0 \\

0 & 0 & 0

\end{array}\right]\)

c. A rotation about the x-axis of θ degrees

Answer:

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

1 & 0 & 0 \\

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

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

\end{array}\right]\)

d. A rotation about the y-axis of θ degrees

Answer:

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

\cos (\theta) & 0 & \sin (\theta) \\

0 & 1 & 0 \\

-\sin (\theta) & 0 & \cos (\theta)

\end{array}\right]\)

e. A rotation about the z-axis of θ degrees

Answer:

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

\cos (\theta) & -\sin (\theta) & 0 \\

\sin (\theta) & \cos (\theta & 0 \\

0 & 0 & 1

\end{array}\right]\)

f. A reflection over the xy-plane

Answer:

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

1 & 0 & 0 \\

0 & 1 & 0 \\

0 & 0 & -1

\end{array}\right]\)

g. A reflection over the yz-plane

Answer:

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

-1 & 0 & 0 \\

0 & 1 & 0 \\

0 & 0 & 1

\end{array}\right]\)

h. A reflection over the xz-plane

Answer:

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

1 & 0 & 0 \\

0 & -1 & 0 \\

0 & 0 & 1

\end{array}\right]\)

i. A reflection over y=x in the xy-plane

Answer:

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

0 & 1 & 0 \\

1 & 0 & 0 \\

0 & 0 & 1

\end{array}\right]\)

Exercises

Exercise 1.

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

x & z \\

y & w

\end{array}\right]\), B=\(\left[\begin{array}{ll}

a & c \\

b & d

\end{array}\right]\), and C=\(\left[\begin{array}{ll}

e & g \\

f & h

\end{array}\right]\).

a. Write down the products AB, AC, and A(B+C).

Answer:

b. Verify that A(B+C)=AB+AC.

Answer:

Therefore, A(B+C)=AB+AC.

Exercise 2.

Suppose A, B, and C are 3×3 matrices and X is a point in three-dimensional space.

a. Explain why the point (A(BC))X must be the same point as ((AB)C)X.

Answer:

(A(BC))X=(A)(B)(C)X. Applying BC and then A is the same as applying C, then B, and then A.

(A)(B)(C)X=((AB)C)X. Applying C, then B, and then A is the same as applying C and then AB.

b. Explain why matrix multiplication must be associative.

Answer:

Matrix multiplication is associative because performing the transformation B and then A on a point X is the same as applying the product of AB to point X.

c. Verify using the matrices from Exercise 1 that A(BC)=(AB)C.

Answer:

### Eureka Math Precalculus Module 2 Lesson 12 Problem Set Answer Key

Question 1.

Let matrix A=\(\left(\begin{array}{cc}

3 & -2 \\

-1 & 0

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

4 & 4 \\

3 & 9

\end{array}\right)\), and matrix C=\(\left(\begin{array}{cc}

8 & 2 \\

7 & -5

\end{array}\right)\). Calculate the following:

a. AB

Answer:

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

6 & -6 \\

-4 & -4

\end{array}\right)\)

b. AC

Answer:

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

10 & 16 \\

-8 & -2

\end{array}\right)\)

c. A(B+C)

Answer:

We have that A(B+C)=AB+AC.

d. AB+AC

Answer:

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

16 & 10 \\

-12 & -6

\end{array}\right)\)

e. (A+B)C

Answer:

(A+B)C=AC+BC, so BC has not been calculated yet. We get

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

60 & -12 \\

87 & -39

\end{array}\right)\).

So,

(A+B)C=\(\left(\begin{array}{cc}

70 & 4 \\

79 & -41

\end{array}\right)\).

f. A(BC)

Answer:

A(BC)=A\(\left(\begin{array}{ll}

60 & -12 \\

87 & -39

\end{array}\right)\)

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

6 & 42 \\

-60 & 12

\end{array}\right)\)

Question 2.

Apply each of the transformations you found in Problem 1 to the points x=\(\left(\begin{array}{l}

1 \\

1

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

-3 \\

2

\end{array}\right)\), and x+y.

Answer:

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

\mathbf{0} \\

-8

\end{array}\right)\)

(AB)y=\(\left(\begin{array}{c}

-30 \\

4

\end{array}\right)\)

(AB)(x+y)=\(\left(\begin{array}{c}

-30 \\

-4

\end{array}\right)\)

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

26 \\

-10

\end{array}\right)\)

(AC)y=\(\left(\begin{array}{c}

2 \\

20

\end{array}\right)\)

(AC)(x+y)=\(\left(\begin{array}{c}

28 \\

10

\end{array}\right)\)

c. (A(B+C))x=\(\left(\begin{array}{c}

26 \\

-18

\end{array}\right)\)

(A(B+C))y=\(\left(\begin{array}{c}

-28 \\

24

\end{array}\right)\)

(A(B+C))(x+y)=\(\left(\begin{array}{c}

-2 \\

6

\end{array}\right)\)

d. Same as part (c)

(AB+AC)x=\(\left(\begin{array}{c}

26 \\

-18

\end{array}\right)\)

(AB+AC)y=\(\left(\begin{array}{c}

-28 \\

24

\end{array}\right)\)

(AB+AC)(x+y)=\(\left(\begin{array}{c}

-2 \\

6

\end{array}\right)\)

e. ((A+B)C)x=\(\left(\begin{array}{c}

74 \\

38

\end{array}\right)\)

((A+B)C)y=\(\left(\begin{array}{c}

-202 \\

-319

\end{array}\right)\)

((A+B)C)(x+y)=\(\left(\begin{array}{c}

-128 \\

-281

\end{array}\right)\)

f. (A(BC))x=\(\left(\begin{array}{c}

48 \\

-48

\end{array}\right)\)

(A(BC))y=\(\left(\begin{array}{c}

66 \\

204

\end{array}\right)\)

(A(BC))(x+y)=\(\left(\begin{array}{c}

114 \\

156

\end{array}\right)\)

Question 3.

Let A, B, C, and D be any four square matrices of the same dimensions. Use the distributive property to evaluate the following:

a. (A+B)(C+D)

Answer:

(A+B)C+(A+B)D=AC+BC+AD+BD

b. (A+B)(A+B)

Answer:

AA+AB+BA+BB

c. What conditions need to be true for part (b) to equal AA+2AB+BB?

Answer:

AB=BA needs to be true.

Question 4.

Let A be a 2×2 matrix and B, C be the scalar matrices B=\(\left(\begin{array}{ll}

2 & 0 \\

0 & 2

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

3 & 0 \\

0 & 3

\end{array}\right)\). Answer the following questions.

a. Evaluate the following:

i. BC

Answer:

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

6 & 0 \\

0 & 6

\end{array}\right)\)

ii. CB

Answer:

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

6 & 0 \\

0 & 6

\end{array}\right)\)

iii. B+C

Answer:

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

5 & 0 \\

0 & 5

\end{array}\right)\)

iv. B-C

Answer:

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

-1 & 0 \\

0 & -1

\end{array}\right)\)

b. Are your answers to part (a) what you expected? Why or why not?

Answer:

Answers may vary. Students should expect that the matrices behave like real numbers since they represent scalars.

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

x & y \\

z & w

\end{array}\right)\); does AB=BA? Does AC=CA?

Answer:

Yes, AB=\(\left(\begin{array}{cc}

2 x & 2 y \\

2 z & 2 w

\end{array}\right)\)=BA, and AC=\(\left(\begin{array}{cc}

3 x & 3 y \\

3 z & 3 w

\end{array}\right)\)=CA.

d. What is (A+B)(A+C)? Write the matrix A with the letter and not in matrix form. How does this compare to (x+2)(x+3)?

Answer:

(A+B)(A+C)=AA+BA+AC+BC

=AA+2A+3A+\(\left(\begin{array}{ll}

6 & 0 \\

0 & 6

\end{array}\right)\)

=AA+5A+\(\left(\begin{array}{ll}

6 & 0 \\

0 & 6

\end{array}\right)\)

This is identical to (x+2)(x+3), with x=A.

e. With B and C given as above, is it possible to factor AA-A-BC?

Answer:

Yes. We need factors of -BC that add to -1. It looks like B and -C work, so we get (A+B)(A-C).

Question 5.

Define the sum of any two functions with the same domain to be the function f+g such that for each x in the domain of f and g, (f+g)(x)=f(x)+g(x). Define the product of any two functions to be the function fg, such that for each x in the domain of f and g, (fg)(x)=(f(x))(g(x)).

Let f, g, and h be real-valued functions defined by the equations f(x)=3x+1, g(x)=-\(\frac{1}{2}\) x+2, and

h(x)=x^{2}-4.

a. Does f(g+h)=fg+fh?

Answer:

Yes. If we can show that (f(g+h))(x)=(fg)(x)+(fh)(x), then we will have shown that the functions are equal to each other.

(f(g+h))(x)=(f(x))((g+h)(x))

=(3x+1)(-\(\frac{1}{2}\) x+2+x^{2}-4)

=(3x+1)((-\(\frac{1}{2}\) x+2)+(x^{2}-4))

=(3x+1)(-\(\frac{1}{2}\) x+2)+(3x+1)(x^{2}-4)

=(f(x))(g(x))+(f(x))(h(x))

=(fg)(x)+(fh)(x)

b. Show that this is true for any three functions with the same domains.

Answer:

(f(g+h))(x)=(f(x))((g+h)(x))

=(f(x))(g(x)+h(x))

=(f(x))(g(x))+(f(x))(h(x))

=(fg)(x)+(fh)(x)

c. Does f∘(g+h)=f∘g+f∘h for the functions described above?

Answer:

No

(f∘(g+h))(x)=f(-\(\frac{1}{2}\) x+2+x^{2}-4)

=3⋅(-\(\frac{1}{2}\) x+2+x^{2}-4)+1

=3⋅(-\(\frac{1}{2}\) x+2)+3⋅(x^{2}-4)+1

The addition by 1 prevented it from working. If f(x) would have been a proportion, then the composition would have worked.

### Eureka Math Precalculus Module 2 Lesson 12 Exit Ticket Answer Key

In three-dimensional space, matrix A represents a 180° rotation about the y-axis, matrix B represents a reflection about the xz-plane, and matrix C represents a reflection about the xy-plane. Answer the following:

a. Write matrices A, B, and C.

Answer:

b. If X=\(\left[\begin{array}{l}

2 \\

2 \\

2

\end{array}\right]\), compute A(BX+CX).

Answer:

c. What matrix operations are equivalent to A(BX+CX)? What property is shown?

Answer:

A(BX+CX)=(AB)X+(AC)X; matrix multiplication is distributive.

d. Would (A(BC))X=((AB)C)X? Why?

Answer:

Yes; matrix multiplication is associative.