# Eureka Math Precalculus Module 2 Lesson 12 Answer Key

## 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. b. Using only your geometric intuition, what are the coordinates of BX? CX? Explain your thinking.
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). d. What is the sum of BX+CX?
BX+CX=$$\left[\begin{array}{c} -2 \\ 0 \\ 2 \end{array}\right]$$

e. Write down matrix A, and compute A(BX+CX). f. Compute AB and AC. g. Compute (AB)X, (AC)X, and their sum. Compare your result to your answer to part (e). What do you notice? 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.
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)
$$\left[\begin{array}{lll} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right]$$

$$\left[\begin{array}{lll} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{array}\right]$$

c. A rotation about the x-axis of θ degrees
$$\left[\begin{array}{ccc} 1 & 0 & 0 \\ 0 & \cos (\theta) & -\sin (\theta) \\ 0 & \sin (\theta) & \cos (\theta) \end{array}\right]$$

d. A rotation about the y-axis of θ degrees
$$\left[\begin{array}{ccc} \cos (\theta) & 0 & \sin (\theta) \\ 0 & 1 & 0 \\ -\sin (\theta) & 0 & \cos (\theta) \end{array}\right]$$

e. A rotation about the z-axis of θ degrees
$$\left[\begin{array}{ccc} \cos (\theta) & -\sin (\theta) & 0 \\ \sin (\theta) & \cos (\theta & 0 \\ 0 & 0 & 1 \end{array}\right]$$

f. A reflection over the xy-plane
$$\left[\begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & -1 \end{array}\right]$$

g. A reflection over the yz-plane
$$\left[\begin{array}{ccc} -1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right]$$

h. A reflection over the xz-plane
$$\left[\begin{array}{ccc} 1 & 0 & 0 \\ 0 & -1 & 0 \\ 0 & 0 & 1 \end{array}\right]$$

i. A reflection over y=x in the xy-plane
$$\left[\begin{array}{lll} 0 & 1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 1 \end{array}\right]$$

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). b. Verify that A(B+C)=AB+AC. 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.
(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.
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. ### 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
$$\left(\begin{array}{cc} 6 & -6 \\ -4 & -4 \end{array}\right)$$

b. AC
$$\left(\begin{array}{cc} 10 & 16 \\ -8 & -2 \end{array}\right)$$

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

d. AB+AC
$$\left(\begin{array}{cc} 16 & 10 \\ -12 & -6 \end{array}\right)$$

e. (A+B)C
(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)
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.
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)

b. (A+B)(A+B)
AA+AB+BA+BB

c. What conditions need to be true for part (b) to equal AA+2AB+BB?
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
$$\left(\begin{array}{ll} 6 & 0 \\ 0 & 6 \end{array}\right)$$

ii. CB
$$\left(\begin{array}{ll} 6 & 0 \\ 0 & 6 \end{array}\right)$$

iii. B+C
$$\left(\begin{array}{ll} 5 & 0 \\ 0 & 5 \end{array}\right)$$

iv. B-C
$$\left(\begin{array}{cc} -1 & 0 \\ 0 & -1 \end{array}\right)$$

b. Are your answers to part (a) what you expected? Why or why not?
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?
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)?
(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?
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)=x2-4.

a. Does f(g+h)=fg+fh?
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+x2-4)
=(3x+1)((-$$\frac{1}{2}$$ x+2)+(x2-4))
=(3x+1)(-$$\frac{1}{2}$$ x+2)+(3x+1)(x2-4)
=(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.
(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?
No
(f∘(g+h))(x)=f(-$$\frac{1}{2}$$ x+2+x2-4)
=3⋅(-$$\frac{1}{2}$$ x+2+x2-4)+1
=3⋅(-$$\frac{1}{2}$$ x+2)+3⋅(x2-4)+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. b. If X=$$\left[\begin{array}{l} 2 \\ 2 \\ 2 \end{array}\right]$$, compute A(BX+CX). 