## Engage NY Eureka Math Precalculus Module 2 Lesson 13 Answer Key

### Eureka Math Precalculus Module 2 Lesson 13 Opening Exercise Answer Key

Opening:

A common way to send coded messages is to assign each letter of the alphabet to a number 1 – 26 and send the message as a string of integers. For example, If we encode the message “THE CROW FLIES AT MIDNIGHT” according to the chart below, we get the string of numbers.

20, 8, 5, 0, 3, 18, 15, 23, 0, 6, 12, 9, 5, 19, 0, 1, 20, 0, 13, 9, 4, 14, 9, 7, 8, 20.

However, codes such as these are easily broken using an analysis of the frequency of numbers that appear in the coded messages.

We can instead encode a message using matrix multiplication. If a matrix E has an inverse, then we can encode a message as follows.

→ First, convert the characters of the message to integers between 1 and 26 using the chart above.

→ If the encoding matrix E is an n × n matrix, then break up the numerical message into n rows of the same length. If needed, add extra zeros to make the rows the same length.

→ Place the rows into a matrix M.

→ Compute the product EM to encode the message.

→ The message is sent as the numbers in the rows of the matrix EM.

### Eureka Math Precalculus Module 2 Lesson 13 Exercise Answer Key

Exercises:

Exercise 1.

You have received an encoded message: 34, 101, 13, 16, 23, 45, 10, 8, 15, 50, 8, 12. You know that the message was encoded using matrix E = \(\left[\begin{array}{lll}

1 & 2 & 3 \\

0 & 2 & 1 \\

1 & 1 & 1

\end{array}\right]\).

a. Store your message in a matrix C. What are the dimensions of C?

Answer:

There are 12 numbers in the coded message, and it was encoded using a 3 × 3 matrix. Thus, the matrix C needs to have three rows. That means C has four columns, so C is a 3 × 4 matrix.

b. You have forgotten whether the proper decoding matrix is matrix X, Y, or Z as shown below. Determine which of these is the correct matrix to use to decode this message.

Answer:

Matrices used to encode and decode messages must be inverses of each other. Thus, the correct decoding matrix is the matrix D so that D . E = I. We can find the correct decoding matrix by multiplying X . E, Y . E, and Z . E.

Since Z . E = I, we know that Z is the decoding matrix we need.

c. Decode the message.

Answer:

Using matrix Z to decode, we have

The decoded message is “ARCHIMEDES”.

Exercise 2.

You have been assigned a group number. The message your group receives is listed below. This message is TOP

SECRET! It Is of such importance that it has been encoded four times.

Your group’s portion of the coded message is listed below.

Group 1:

1500, 3840, 0, 3444, 3420, 4350, 0, 4824, 3672, 3474, -2592, -6660, 0, -5976, -5940, -7560, 0, -8388, -6372, -6048

Group 2:

2424, 3024, -138, 396, -558, -1890, -1752, 1512, -2946, 1458, 438, 540, -24, 72, -90, -324, -300, 270, -510, 270

Group 3:

489, 1420, 606, 355, 1151, 33, 1002, 829, 99, 1121, 180, 520, 222, 130, 422, 12, 366, 304, 36,410

Group 4:

-18, 10, -18, 44, -54, 42, -6, -74, -98, -124, 0, 10, -12, 46, -26, 42, -4, -36, -60, -82

Group 5:

-120, 0, -78, -54, -84, -30, 0, -6, -108, -30, -120, 114, 42, 0, -12, 42, 0, 36, 0, 0

Group 6:

126, 120, 60, 162, 84, 120, 192,42, 84, 192, -18, -360, -90, -324, 0, -18, -216, -36, -90, -324

a. Store your message in a matrix C with two rows. How many columns does matrix C have?

Answer:

(Sample responses are provided for Group 1.) Our message is stored in a matrix with ten columns:

C =

b. Begin at the station of your group number, and apply the decoding matrix at this first station.

Answer:

c. Proceed to the next station in numerical order; if you are at Station 6, proceed to Station 1. Apply the decoding matrix at this second station.

Answer:

d. Proceed to the next station in numerical order; if you are at Station 6, proceed to Station 1. Apply the decoding matrix at this third station.

Answer:

e. Proceed to the next station in numerical order; If you are at Station 6, proceed to Station 1. Apply the decoding matrix at this fourth station.

Answer:

f. If we know the original message had 21 characters and the last letter was an H, decode your message.

Answer:

The numerical message is 4, 15, 0, 14, 15, 20, 0, 23, 15, 18, 18, 25, 0, 20, 15, 15, 0, 13, 21, 3, which represents the characters “DO NOT WORRY TOO MUCH.”

Exercise 3.

Sydnie was in Group 1 and tried to decode her message by calculating the matrix (D_{1} . D_{2} . D_{3} . D_{4}) and then multiplying (D_{1} D_{2} D_{3} D_{4}) C. This produced the matrix.

M =

a. How did she know that she made a mistake?

Answer:

If Sydnie had properly decoded her message, all entries in the matrix M would be integers between 0 and 26.

b. Matrix C was encoded using matrices E_{1}, E_{2}, E_{3}, and E_{4}, where D_{1} decodes a message encoded by E_{1}, D_{2} decodes a message encoded by E_{2}, and so on. What is the relationship between matrices E_{1} and D_{1,} between E_{2} and D_{2}, etc.?

Answer:

Matrices E_{1} and D_{1} are inverse matrices, as are E_{2} and D_{2}, E_{3}, and D_{3}, and so on.

c. The matrix that Sydnie received was encoded by C = E_{1} . E_{2} . E_{3} . E_{4} M. Explain to Sydnie how the decoding process works to recover the original matrix M, and devise a correct method for decoding using multiplication by a single decoding matrix.

Answer:

Since C = E_{1} . E_{2} . E_{3} . E_{4} . M, we can recover the original matrix M by multiplying both sides of this equation by the proper decoding matrix at each step, remembering that D_{1} . E_{1} = I, D_{1} . E_{1} = I, etc.

C = E_{1} . E_{2} . E_{3} . E_{4} . M

D_{1} . C = D_{1} . (E_{1} . E_{2} . E_{3} . E_{4} .

M)

= (D_{1} . E_{1}) . (E_{2} . E_{3} . E_{4} . M)

= I . (E_{2} . E_{3} . E_{4} . M)

= (E_{2} . E_{3} . E_{4} . M)

D_{2} . D_{1} . C = D_{2} . (E_{2} . E_{3} . E_{4} . M)

= (D_{2} . E_{2}) . (E_{3} . E_{4} . M)

= I . (E_{3} . E_{4} . M)

= (E_{3} . E_{4} . M)

D_{3} . D_{2} . D_{1} . C = D_{3} . (E_{3} . E_{4} . M)

= (D_{3} . E_{3}) . (E_{4} . M)

= I . (E_{4} . M)

= E_{4} . M

D_{4} . D_{3} . D_{2} . D_{1}. C = D_{4} . (E_{4} . M)

= (D_{4} . E_{4}) . M

= I . M

= M

Since Matrix Multiplication is associative, this means that M = (D_{4} . D_{3} . D_{2} . D_{1}). C

d. Apply the method you devised in part (c) to your group’s message to verify that it works.

Answer:

Exercise 4.

You received a coded message in the matrix C = \(\left[\begin{array}{ccc}

30 & 30 & 69 \\

2 & 1 & 15 \\

9 & 14 & 20

\end{array}\right]\). However, the matrix D that will decode this message has been corrupted, and you do not know the value of entry d_{12}. You know that all entries in matrix D are integers. Using x to represent this unknown entry, the decoding matrix D is given by D = \(\left[\begin{array}{ccc}

2 & x & -4 \\

-1 & 2 & 3 \\

1 & -1 & -2

\end{array}\right]\). Decode the message in matrix C.

Answer:

Decoding the message requires that we multiply D . C:

Since we know all entries are integers and that the entries represent letters, we know that

0 ≤ 24 + 2x ≤ 26

0 ≤ 4 + x ≤ 26

0 ≤ 58 + 15x ≤ 26.

Solving these inequalities gives

-12 < x < 1

-4 ≤ x ≤ 22

58 32

\(-\frac{58}{15}\) ≤ x ≤ \(-\frac{32}{15}\).

Because we know that x is an integer, the third Inequality becomes -3 ≤ x ≤ -3, so we know that x = -3. Then, the decoded message is

D . C = \(\left[\begin{array}{ccc}

24+2(-3) & 4+(-3) & 58+15(-3) \\

1 & 14 & 21 \\

10 & 1 & 14

\end{array}\right]\);

thus, D . C = \(\left[\begin{array}{ccc}

18 & 1 & 13 \\

1 & 14 & 21 \\

10 & 1 & 14

\end{array}\right]\),

and the decodedd message is “RAMANUJAN”.

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

Question 1.

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

1 & 3 \\

2 & 5

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

-2 & 7 \\

3 & -4

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

-5 & 3 \\

2 & -1

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

0 & 0 \\

0 & 0

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

1 & 0 \\

0 & 1

\end{array}\right]\). Evaluate the following.

a. A + B

Answer:

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

-1 & 10 \\

5 & 1

\end{array}\right]\)

b. B + A

Answer:

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

-1 & 10 \\

5 & 1

\end{array}\right]\)

c. A + (B + C)

Answer:

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

-6 & 13 \\

7 & 0

\end{array}\right]\)

d. (A + B) + C

Answer:

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

-6 & 13 \\

7 & 0

\end{array}\right]\)

e. A + I

Answer:

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

2 & 3 \\

2 & 6

\end{array}\right]\)

f. A + Z

Answer:

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

1 & 3 \\

2 & 5

\end{array}\right]\)

g. A . Z

Answer:

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

0 & 0 \\

0 & 0

\end{array}\right]\)

h. Z . A

Answer:

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

0 & 0 \\

0 & 0

\end{array}\right]\)

i. I . A

Answer:

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

1 & 3 \\

2 & 5

\end{array}\right]\)

j. A . B

Answer:

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

7 & -5 \\

11 & -6

\end{array}\right]\)

k. B . A

Answer:

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

12 & 29 \\

-5 & -11

\end{array}\right]\)

l. A . C

Answer:

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

1 & 0 \\

0 & 1

\end{array}\right]\)

m. C . A

Answer:

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

1 & 0 \\

0 & 1

\end{array}\right]\)

n. A . B + A . C

Answer:

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

8 & -5 \\

11 & -5

\end{array}\right]\)

o. A . (B + C)

Answer:

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

8 & -5 \\

11 & -5

\end{array}\right]\)

p. A . B . C

Answer:

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

-45 & 26 \\

-67 & 39

\end{array}\right]\)

q. C . B . A

Answer:

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

-75 & -178 \\

29 & 69

\end{array}\right]\)

r. A . C . B

Answer:

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

-2 & 7 \\

3 & -4

\end{array}\right]\)

s. det(A)

Answer:

-1

t. det(B)

Answer:

-13

u. det(C)

Answer:

-1

v. det(Z)

Answer:

0

w. det(I)

Answer:

1

x. det(A . B . C)

Answer:

-13

y. det(C . B . A)

Answer:

-13

Question 2.

For any 2 × 2 matrix A and any real number k, show that if kA = \(\left[\begin{array}{ll}

0 & 0 \\

0 & 0

\end{array}\right]\). then k = 0 or A = \(\left[\begin{array}{ll}

0 & 0 \\

0 & 0

\end{array}\right]\).

Answer:

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

\boldsymbol{a} & \boldsymbol{b} \\

\boldsymbol{c} & \boldsymbol{d}

\end{array}\right]\); then kA = \left[\begin{array}{ll}

\boldsymbol{a} & \boldsymbol{b} \\

\boldsymbol{c} & \boldsymbol{d}

\end{array}\right]. Suppose that kA = \(\left[\begin{array}{ll}

0 & 0 \\

0 & 0

\end{array}\right]\).

Case 1:

Suppose k ≠ 0. Then, ka = 0, kb = 0, kc = 0, and kd = 0; all imply that a = b = c = d = 0. Thus, if k ≠ 0,then A = \(\left[\begin{array}{ll}

0 & 0 \\

0 & 0

\end{array}\right]\)

Case 2:

Suppose that A ≠ \(\left[\begin{array}{ll}

0 & 0 \\

0 & 0

\end{array}\right]\). Then, at least one of a, b, c, and d is not zero, so ka = 0, kb = 0, kc = 0, and kd = 0 imply that k = 0.

Thus, if kA = \(\left[\begin{array}{ll}

0 & 0 \\

0 & 0

\end{array}\right]\). then either k = 0 or A = \(\left[\begin{array}{ll}

0 & 0 \\

0 & 0

\end{array}\right]\).

Question 3.

Claire claims that she multiplied A = \(\left[\begin{array}{cc}

-3 & 2 \\

0 & 4

\end{array}\right]\) by another matrix X and obtained \(\left[\begin{array}{cc}

-3 & 2 \\

0 & 4

\end{array}\right]\) as her result. What matrix did she multiply by? How do you know?

Answer:

She multiplied A by the multiplicative identit y matrix I = \(\left[\begin{array}{ll}

1 & 0 \\

0 & 1

\end{array}\right]\). Since the product is a 2 × 2 matrix, we know that X is a 2 × 2 matrix of the form = \(\left[\begin{array}{ll}

\boldsymbol{a} & \boldsymbol{b} \\

\boldsymbol{c} & \boldsymbol{d}

\end{array}\right]\). Multiplying A . X gives A . X = \(\left[\begin{array}{cc}

-3 a+2 c & -3 b+2 d \\

0 a+4 c & 0 b+4 d

\end{array}\right]\) Since A . X = A, we have the following system of equations:

-3a + 2c = -3

-3b + 2d = 2

0a +4c = 0

0b + 4d = 4.

The third and fourth equations give c = 0 and d = 1, respectively, and substituting into the first two equations gives

-3a = -3 and -3b + 2 = 2. Thus,a = 1 and b = 0, and the matrix X must be X = I.

Question 4.

Show that the only matrix B such that A + B = A Is the zero matrix.

Answer:

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

a & b \\

c & d

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

x & y \\

z & w

\end{array}\right]\) then, we have A + B = \(\left[\begin{array}{ll}

a+x & b+y \\

c+z & d+w

\end{array}\right]\), a + x = a, b + y = b, c + z = c, and d + w = d. In each case, solving for the elements of B, we find that B = \(\left[\begin{array}{ll}

0 & 0 \\

0 & 0

\end{array}\right]\)

Question 5.

A 2 × 2 matrix of the form \(\left[\begin{array}{ll}

\boldsymbol{a} & \mathbf{0} \\

\mathbf{0} & \boldsymbol{b}

\end{array}\right]\) is a diagonal matrix. Daniel calculated

and concluded that if X is a diagonal matrix and A is any other matrix, then X . A = A . X.

a. Is there anything wrong with Daniel’s reasoning? Prove or disprove that If X is a diagonal 2 × 2 matrix, then X . A = A . X for any other matrix A.

Answer:

Yes, there is something wrong with Daniel’s reasoning. A single example does not establish that a statement is true, and the example he calculated used a special case of a diagonal matrix in which the entries on the main diagonal are equal.

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

1 & 2 \\

3 & 4

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

2 & 0 \\

0 & 3

\end{array}\right]\), then X . A = \(\left[\begin{array}{cc}

2 & 4 \\

9 & 12

\end{array}\right]\) andA X = \(\left[\begin{array}{cc}

2 & 6 \\

6 & 12

\end{array}\right]\). Thus, it is nor true that X . A = A . X for all diagonal matrices X and all other matrices A.

b. For 3 × 3 matrices, Elda claims that only diagonal matrices of the form X = \(\left[\begin{array}{lll}

\boldsymbol{C} & 0 & 0 \\

0 & \boldsymbol{C} & 0 \\

0 & 0 & \boldsymbol{C}

\end{array}\right]\) satisfy X . A = A . X for any other 3 × 3 matrix A. Is her claim correct?

Answer:

Elda is correct since X = \(\left[\begin{array}{lll}

\boldsymbol{C} & 0 & 0 \\

0 & \boldsymbol{C} & 0 \\

0 & 0 & \boldsymbol{C}

\end{array}\right]\). Then, X . A = cI. A = c(I . A) = cA = Ac = (A . I)c = A . (cI) = A . X for all matrices A.

Question 6.

Calvin encoded a message using E = \(\left[\begin{array}{cc}

2 & 2 \\

-1 & 3

\end{array}\right]\), giving the coded message 4, 28, 42, 56, 2, -6, -1, 52. Decode the message, or explain why the original message cannot be recovered.

Answer:

Putting the message 2 × 4 matrix, we have C = \(\left[\begin{array}{cccc}

4 & 28 & 42 & 56 \\

2 & -6 & -1 & 52

\end{array}\right]\). We can decode the message with

D = E^{-1} = \(\frac{1}{6-(-2)}\)\(\left[\begin{array}{cc}

3 & -2 \\

1 & 2

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

\frac{3}{8} & -\frac{2}{8} \\

\frac{1}{8} & \frac{2}{8}

\end{array}\right]\). Then, the original message is found in message M:

M = D C = \(\left[\begin{array}{cc}

\frac{3}{8} & -\frac{2}{8} \\

\frac{1}{8} & \frac{2}{8}

\end{array}\right]\) . \(\left[\begin{array}{cccc}

4 & 28 & 42 & 56 \\

2 & -6 & -1 & 52

\end{array}\right]\) = \(\left[\begin{array}{cccc}

1 & 12 & 16 & 8 \\

1 & 2 & 5 & 20

\end{array}\right]\)

The original message is “ALPHABET.”

Question 7.

Decode the message below using the matrix D = \(\left[\begin{array}{ccc}

1 & 1 & -1 \\

-1 & 0 & 2 \\

1 & 2 & 1

\end{array}\right]\):

22, 17, 24, 9, -1, 14, -9, 34, 44, 64, 47, 77.

Answer:

The decoded message is found by multiplying . Then, the message is “CRYPTOGRAPHY.”

Question 8.

Brandon encoded his name with the matrix E = \(\left[\begin{array}{ll}

1 & 1 \\

2 & 2

\end{array}\right]\) producing the matrix C = \(\left[\begin{array}{cccc}

6 & 33 & 15 & 14 \\

12 & 66 & 30 & 28

\end{array}\right]\) Decode the message, or explain why the original message cannot be recovered.

Answer:

Brandon used a matrix that is not invertible. The oriqinal matrix cannot be recovered.

Question 9.

Janelle used the encoding matrix E = \(\left[\begin{array}{cc}

1 & 2 \\

1 & -1

\end{array}\right]\) to encode the message “FROG” by multiplying

C = . When taylor decoded it, she computed

M = . What went wrong.

Answer:

Janelle multiplied her matrices in the wrong order. When Janelle triedto decode the matrix C = \(\left[\begin{array}{ll}

24 & 30 \\

22 & 37

\end{array}\right]\) using the decoding matrix D = \(\left[\begin{array}{cc}

1 & 2 \\

1 & -1

\end{array}\right]^{-1}\), she ended up calculating

D . C = D . M . E

= E^{-1} . M . E

Because matrix multiplication is not commutative, E^{-1} . M . E ≠ M, Taylor was unable to recover the original message.

### Eureka Math Precalculus Module 2 Lesson 13 Exit Ticket Answer Key

Question 1.

Morgan used matrix E = \(\left[\begin{array}{cc}

1 & -2 \\

-1 & 3

\end{array}\right]\) to encide the name of her favourite mathematician in the message -32, 7, 14, 1, 52, 2, -13, -1.

a. How can you tell whether or not her message can be decoded?

Answer:

Since the matrix E has determinant det(E) = 3 – 2 = 1, we know that det(E) ≠ 0, so then a decoding matrix D = E^{-1} exists.

b. Decode the message, or explain why the original message cannot be recovered.

Answer:

First, we place the coded message into a 2 × 4 matrix C. Using D = E^{-1} = \(\left[\begin{array}{ll}

3 & 2 \\

1 & 1

\end{array}\right]\), we have

M = D . C

The decoded message is “HYPATIA”