Eureka Math Precalculus Module 2 Mid Module Assessment Answer Key

Engage NY Eureka Math Precalculus Module 2 Mid Module Assessment Answer Key

Eureka Math Precalculus Module 2 Mid Module Assessment Task Answer Key

Question 1.
Kyle wishes to expand his business and is entertaining four possible options. If he builds a new store, he expects to make a profit of 9 million dollars if the market remains strong; however, if market growth declines, he could incur a loss of 5 million dollars. If Kyle invests in a franchise, he could profit 4 million dollars in a strong market but lose 3 million dollars in a declining market. If he modernizes his current facilities, he could profit 4 million dollars in a strong market but lose 2 million dollars in a declining one. If he sells his business, he will make a profit of 2 million dollars irrespective of the state of the market.

a. Write down a 4×2 payoff matrix P summarizing the profits and losses Kyle could expect to see with all possible scenarios. (Record a loss as a profit in a negative amount.) Explain how to interpret your matrix.
Answer:
We have P = \(\left[\begin{array}{cc}
9 & -5 \\
4 & -3 \\
4 & -2 \\
2 & 2
\end{array}\right]\)
Here the four rows correspond to, in turn, the options of building a new store, investing in a franchise, modernizing, and selling. The first column gives the payoffs in a strong market, and the second column gives the payoffs in a declining market.
All entries are in units of millions of dollars.
Note: Other presentations for the matrix P.are possible.

b. Kyle realizes that all his figures need to be adjusted by 10% in magnitude due to inflation costs. What is the appropriate value of a real number λ so that the matrix λP represents a correctly adjusted payoff matrix? Explain your reasoning. Write down the new payoff matrix λP.
Answer:
Each entry in the matrix needs to increase 10% in magnitude. This can be accomplished by multiplying each entry by 1.10. If we set λ = 1.1 , then
λP = \(\left[\begin{array}{cc}
9.9 & -5.5 \\
4.4 & -3.3 \\
4.4 & -2.2 \\
2.2 & 2.2
\end{array}\right]\) is the appropriate new payoff matrix.

c. Kyle hopes to receive a cash donation of 1 million dollars. If he does, all the figures in his payoff matrix will increase by 1 million dollars.
Write down a matrix Q so that if Kyle does receive this donation, his new payoff matrix is given by Q+λP. Explain your thinking.
Answer:
Set Q = \(\left[\begin{array}{ll}
1 & 1 \\
1 & 1 \\
1 & 1 \\
1 & 1
\end{array}\right]\). Then Q + λP is the matrix λP with 1 added to each entry. This is the effect we seek, increasing each expected payoff by 1 million dollars.
Question 2.
The following diagram shows a map of three land masses, numbered region 1, region 2, and region 3, connected via bridges over water. Each bridge can be traversed in either direction.

a. Write down a 3×3 matrix A with a_ij, for i = 1, 2, or 3 and j = 1, 2, or 3, equal to the number of ways to walk from region i to region j by crossing exactly one bridge. Notice that there are no paths that start and end in the same region crossing exactly one bridge.
Answer:
A = \(\left[\begin{array}{lll}
0 & 1 & 1 \\
1 & 0 & 3 \\
1 & 3 & 0
\end{array}\right]\)

b. Compute the matrix product A².
Answer:
A² = \(\left[\begin{array}{ccc}
2 & 3 & 3 \\
3 & 10 & 1 \\
3 & 1 & 10
\end{array}\right]\)

c. Show that there are 10 walking routes that start and end in region 2, crossing over water exactly twice. Assume each bridge, when crossed, is fully traversed to the next land mass.
Answer:
The entries of A² give the number of paths via two bridges between land regions. As the row 2, column 2 entry of A² is 10 , this is the count of two-bridge journeys that start and end in region 2.

d. How many walking routes are there from region 3 to region 2 that cross over water exactly three times? Again, assume each bridge is fully traversed to the next land mass.
Answer:
The entries of A³ give the counts of three-bridge journeys between land masses. We seek the row 3, column 2 entry of the product.
\(\left[\begin{array}{ccc}
2 & 3 & 3 \\
3 & 10 & 1 \\
3 & 1 & 10
\end{array}\right]\left[\begin{array}{lll}
0 & 1 & 1 \\
1 & 0 & 3 \\
1 & 3 & 0
\end{array}\right]\)
This entry is (3 ⋅ 1) + (1 ⋅ 0) + (10 ⋅ 3) = 33 . Therefore, there are 33 such routes.

e. If the number of bridges between each pair of land masses is doubled, how does the answer to part (d) change? That is, what would be the new count of routes from region 3 to region 2 that cross over water exactly three times?
Answer:
We are now working with the matrix 2A . The number of routes from region 3 to region 2 via three bridges is the row 3, column 2 entry of (2A)³ = 8 A³ . As all the entries are multiplied by eight, there are 8 × 33 = 264 routes of the particular type we seek.

Question 3.
Let P = \(\left[\begin{array}{cc}
3 & -5 \\
5 & 3
\end{array}\right]\) and Q = \(\left[\begin{array}{cc}
-2 & 1 \\
-1 & -2
\end{array}\right]\).

a. Show the work needed and compute 2P-3Q.
Answer:
2P-3Q = \(\left[\begin{array}{cc}
6 & -10 \\
10 & 6
\end{array}\right]\)–\(\left[\begin{array}{cc}
-6 & 3 \\
-3 & -6
\end{array}\right]\) = \(\left[\begin{array}{cc}
12 & -13 \\
13 & 12
\end{array}\right]\)

b. Show the work needed and compute PQ.
Answer:
PQ = \(\left[\begin{array}{cc}
-6+5 & 3+10 \\
-10-3 & 5-6
\end{array}\right]\) = \(\left[\begin{array}{cc}
-1 & 13 \\
-13 & -1
\end{array}\right]\)

c. Show that P²Q = PQP.
Answer:
PQ = \(\left[\begin{array}{cc}
-1 & 13 \\
-13 & -1
\end{array}\right]\)
P² Q = P (PQ ) = \(\left[\begin{array}{cc}
62 & 44 \\
-44 & 62
\end{array}\right]\)
PQP = (PQ )P = \(\left[\begin{array}{cc}
-1 & 13 \\
-13 & -1
\end{array}\right]\left[\begin{array}{cc}
3 & -5 \\
5 & 3
\end{array}\right]\) = \(\left[\begin{array}{cc}
62 & 44 \\
-44 & 62
\end{array}\right]\)
These are identical matrices.

Question 4.
a. Show that if the matrix equation (A + B)² = A² + 2AB + B² holds for two square matrices A and B of the same dimension, then these two matrices commute under multiplication.
Answer:
We have (A + B )² = (A + B)(A + B) .
By the distributive rule, which does hold for matrices, this equals A (A + B ) + B(A + B) , which, again by the distributive rule, equals A² + AB + BA + B² .
On the other hand, A ² + 2AB + B ² equals A² + AB + AB + B² .
So, if (A + B )² = A² + 2AB + B² , then we have A² + AB + BA + B² = A² + AB + AB + B² .
Adding -A² and -AB and -B² to each side of this equation gives BA = AB .
This shows that A and B commute under multiplication in this special case when (A+B)² = A² + 2AB + B² , but in general, matrix multiplication is not commutative.

b. Give an example of a pair of 2×2 matrices A and B for which (A+B)² ≠ A²+2AB+B².
Answer:
A pair of matrices that do not commute under multiplication, such as A = \(\left[\begin{array}{ll}
1 & 0 \\
1 & 1
\end{array}\right]\) and B = \(\left[\begin{array}{ll}
1 & 1 \\
0 & 1
\end{array}\right]\), should do the trick.
To check: (A + B )² = \(\left[\begin{array}{ll}
2 & 1 \\
1 & 2
\end{array}\right]\)² = \(\left[\begin{array}{ll}
5 & 4 \\
4 & 5
\end{array}\right]\)
A ² + 2AB + B ² = \(\left[\begin{array}{ll}
1 & 0 \\
1 & 1
\end{array}\right]\)² + 2 \(\left[\begin{array}{ll}
1 & 0 \\
1 & 1
\end{array}\right]\left[\begin{array}{ll}
1 & 1 \\
0 & 1
\end{array}\right]\) + \(\left[\begin{array}{ll}
1 & 1 \\
0 & 1
\end{array}\right]\)²
= \(\left[\begin{array}{ll}
1 & 0 \\
2 & 1
\end{array}\right]+2\left[\begin{array}{ll}
1 & 1 \\
1 & 2
\end{array}\right]+\left[\begin{array}{ll}
1 & 2 \\
0 & 1
\end{array}\right]\)
= \(\left[\begin{array}{ll}
4 & 4 \\
4 & 6
\end{array}\right]\)
These are indeed different.

c. In general, does AB = BA? Explain.
Answer:
No. Since matrix multiplication can represent linear transformations, we know that they will not always commute since linear transformations do not always commute.

d. In general, does A(B + C) = AB + AC? Explain.
Answer:
Yes. Consider the effect on the point x made by both sides of the equation. On the left-hand side, the transformation B + C is applied to the point x , but we know that this is the same as Bx + Cx from our work with linear transformations. Applying the transformation represented by A to either Bx + Cx or (B + C )x now is ABx + ACx because they work like linear maps.

e. In general, does A(BC) = (AB)C? Explain.
Answer:
Yes. If we consider the effect the matrices on both sides make on a point x , the matrices are applied in the exact same order, C , then B , then A , regardless of whether AB is computed first or BC is computed first.

Question 5.
Let I be the 3×3 identity matrix and A the 3×3 zero matrix. Let the 3×1 column x = \(\left[\begin{array}{l}
x \\
y \\
z
\end{array}\right]\) represent a point in three-dimensional space. Also, set P = \(\left[\begin{array}{lll}
2 & 0 & 5 \\
0 & 0 & 0 \\
0 & 0 & 4
\end{array}\right]\).

a. Use examples to illustrate how matrix A plays the same role in matrix addition that the number 0 plays in real number addition. Include an explanation of this role in your response.
Answer:
The sum of two 3 × 3 matrices is determined by adding entries in corresponding positions of the two matrices to produce a new 3 × 3 matrix. Each and every entry of matrix A is zero, so a sum of the form A + P , where P is another 3 × 3 matrix, is given by adding zero to each entry of P . Thus, A + P = P. For example:
\(\left[\begin{array}{lll}
0 & 0 & 0 \\
0 & 0 & 0 \\
0 & 0 & 0
\end{array}\right]+\left[\begin{array}{lll}
1 & 4 & 7 \\
2 & 5 & 8 \\
3 & 6 & 9
\end{array}\right] = \left[\begin{array}{lll}
0+1 & 0+4 & 0+7 \\
0+2 & 0+5 & 0+8 \\
0+3 & 0+6 & 0+9
\end{array}\right] = \left[\begin{array}{lll}
1 & 4 & 7 \\
2 & 5 & 8 \\
3 & 6 & 9
\end{array}\right]\)
This is analogous to the role of zero in the real number system: 0 + p = p for every real number p.
In the same way, P + A = P for all 3 × 3 matrices P, analogous to p + 0 = p for all real numbers p.

b. Use examples to illustrate how matrix I plays the same role in matrix multiplication that the number 1 plays in real number multiplication. Include an explanation of this role in your response.
Answer:
We have I = \(\left[\begin{array}{lll}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1
\end{array}\right]\). By the definition of matrix multiplication, we see, for example:

We have, in general, that I × P = P for every 3 × 3 matrix P. We also have P × I = P .
Thus, the matrix I plays the role of the number I in real number arithmetic where 1 × p = p and p × 1 = p for each real number p.

c. What is the row 3, column 3 entry of (AP+I)²? Explain how you obtained your answer.
Answer:
Since A is the zero matrix, AP equals the zero matrix. That is, AP = A.
Thus, (AP + I )² = (A + I )² = I ² = I .
So, (AP + I )² is just the 3 × 3 identity matrix. The row 3, column 3 entry is thus 1 .

d. Show that (P – 1)(P + 1)equals P² – I.
Answer:
We have (P – I )(P + I ) = P² – IP + PI – I² (using the distributive property, which holds for matrices). Since PI = P , IP = P , and I² = I , this equals P² – P + P – I = P² -I .

e. Show that Px is sure to be a point in the xz-plane in three-dimensional space.
Answer:
We have
Px = \(\left[\begin{array}{lll}
2 & 0 & 5 \\
0 & 0 & 0 \\
0 & 0 & 4
\end{array}\right]\left[\begin{array}{l}
x \\
y \\
z
\end{array}\right]\) = \(\left[\begin{array}{c}
2 x+5 z \\
0 \\
4 z
\end{array}\right]\)
The image is a point with y -coordinate zero, and so it is a point in the xz -plane in three-dimensional space.

f. Is there a 3×3 matrix Q, not necessarily the matrix inverse for P, for which QPx = x for every 3×1 column x representing a point? Explain your answer.
Answer:
If there were such a matrix Q , then QPx = x for x = \(\left[\begin{array}{l}
0 \\
1 \\
0
\end{array}\right]\). But Px = \(\left[\begin{array}{lll}
2 & 0 & 5 \\
0 & 0 & 0 \\
0 & 0 & 4
\end{array}\right]\left[\begin{array}{l}
0 \\
1 \\
0
\end{array}\right]\) = \(\left[\begin{array}{l}
0 \\
0 \\
0
\end{array}\right]\), and so QPx = Q \(\left[\begin{array}{l}
0 \\
0 \\
0
\end{array}\right]\) = \(\left[\begin{array}{l}
0 \\
0 \\
0
\end{array}\right]\), which is not x = \(\left[\begin{array}{l}
0 \\
1 \\
0
\end{array}\right]\) after all. There can be no such matrix Q .

g. Does the matrix P have a matrix inverse? Explain your answer.
Answer:
If P had a matrix inverse P^-1, then we would have P^-1 P = I and so P^-1 Px = x for all 3 × 1 columns x representing a point. By part (d), there is no such matrix.
OR
By part (c), P takes all points in the three-dimensional space and collapses them to a plane. So there are points that are taken to the same image point by P. Thus, no inverse transformation, P^-1, can exist.

h. What is the determinant of the matrix P?
Answer:
The unit cube is mapped onto a plane, and so the image of the unit cube under P has zero volume. The determinant of P is thus 0.
OR
By part (e), P has no multiplicative inverse, and so its determinant must be 0.

Question 6.
What is the image of the point given by the 3×1 column matrix \(\left[\begin{array}{c}
1 \\
2 \\
-1
\end{array}\right]\) when it is rotated 45° about the z-axis in the counterclockwise direction (as according to the orientation of the xy-plane) and then 180° about the y-axis?
Answer:
A rotation about the z -axis of 45° is effected by multiplication by the matrix:

So, the image of the point under the first rotation is
\(R_{1}\left[\begin{array}{c}
1 \\
2 \\
-1
\end{array}\right]=\left[\begin{array}{c}
-\frac{1}{\sqrt{2}} \\
\frac{3}{\sqrt{2}} \\
-1
\end{array}\right]\)
A rotation of 180° about the y -axis is effected by multiplication by

The final image point we seek is thus
\(R_{2}\left[\begin{array}{c}
-\frac{1}{\sqrt{2}} \\
\frac{3}{\sqrt{2}} \\
-1
\end{array}\right] = \left[\begin{array}{c}
\frac{1}{\sqrt{2}} \\
\frac{3}{\sqrt{2}} \\
1
\end{array}\right]\)