Eureka Math Precalculus Module 1 Lesson 29 Answer Key

Engage NY Eureka Math Precalculus Module 1 Lesson 29 Answer Key

Eureka Math Precalculus Module 1 Lesson 29 Example Answer Key

Example 1.
Find the determinant of $$\left[\begin{array}{ll} 1 & 2 \\ 2 & 4 \end{array}\right]$$.
The determinant is 0.

Eureka Math Precalculus Module 1 Lesson 29 Exercise Answer Key

Opening Exercise

Find the inverse of $$\left[\begin{array}{cc} -7 & -2 \\ 4 & 1 \end{array}\right]$$. Show your work. Confirm that the matrices are inverses.
$$\left[\begin{array}{ll} a & c \\ b & d \end{array}\right]\left[\begin{array}{cc} -7 & -2 \\ 4 & 1 \end{array}\right]$$=$$\left[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right]$$
-7a+4c=1,-2a+c=0,-7b+4d=0,-2b+d=1
a=1,b=-4,c=2, and d=-7
$$\left[\begin{array}{cc} 1 & 2 \\ -4 & -7 \end{array}\right]\left[\begin{array}{cc} -7 & -2 \\ 4 & 1 \end{array}\right]$$ = $$\left[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right]$$

Exercises

Exercise 1.
Find the inverse of $$\left[\begin{array}{ll} 5 & 3 \\ 2 & 4 \end{array}\right]$$. Confirm your answer.
$$\left[\begin{array}{cc} 4 & -3 \\ -2 & 5 \end{array}\right]\left[\begin{array}{ll} 5 & 3 \\ 2 & 4 \end{array}\right]$$=$$\left[\begin{array}{cc} 14 & 0 \\ 0 & 14 \end{array}\right]$$

→ Was the matrix that you found using the pattern the inverse? What was missing?
→ No. Where we needed 1’s, we had 14’s.
→ Let’s look at this a little further. Look at the matrices in the table. Find the determinant of the matrices. (Assign different groups/pairs different matrices from above.)
→ All of the determinants were 1.
→ Do you think that makes a difference? What was the determinant of the matrix in Exercise 1?
→ The determinant was 14.
→ How does that compare to the matrix that resulted from multiplying the matrices in Exercise 1?
→ That was the number that was in the position that should have been a 1.
→ How do you think this ties into the way we find an inverse matrix?
→ We can still use our pattern, but we need to divide each term by the determinant. (Answers may vary, but let students try out their hypotheses to come up with the right answer.)
→ Try it on the inverse matrix in Exercise 1. Write the inverse matrix.
→ $$\left[\begin{array}{cc} \frac{4}{14} & -\frac{3}{14} \\ -\frac{2}{14} & \frac{5}{14} \end{array}\right]$$
→ Verify that is the inverse. Were you correct?
→ $$\left[\begin{array}{cc} \frac{4}{14} & -\frac{3}{14} \\ -\frac{2}{14} & \frac{5}{14} \end{array}\right]\left[\begin{array}{cc} 5 & 3 \\ 2 & 4 \end{array}\right]$$ = $$\left[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right]$$
→ Yes. We get the identify matrix again.
→ Explain to your neighbor how to find the inverse of a matrix.
→ Switch the numbers in the top left and bottom right. Change the signs of the numbers in the top right and bottom left. Divide all of the terms by the determinant of the original matrix.

Find the inverse matrix and verify.

Exercise 2.
$$\left[\begin{array}{cc} 3 & -3 \\ 1 & 4 \end{array}\right]$$
Determinant =(3)(4)-(-3)(1)=12+3=15
$$\left[\begin{array}{cc} \frac{4}{15} & \frac{3}{15} \\ \frac{-1}{15} & \frac{3}{15} \end{array}\right]\left[\begin{array}{cc} 3 & -3 \\ 1 & 4 \end{array}\right]$$ = $$\left[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right]$$

Exercise 3.
$$\left[\begin{array}{ll} 5 & -2 \\ 4 & -3 \end{array}\right]$$
Determinant =(5)(-3)-(-2)(4)=-15+8=-7
$$\left[\begin{array}{ll} \frac{3}{7} & -\frac{2}{7} \\ \frac{4}{7} & -\frac{5}{7} \end{array}\right]\left[\begin{array}{ll} 5 & -2 \\ 4 & -3 \end{array}\right]$$ = $$\left[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right]$$

Exercise 4.
$$\left[\begin{array}{ll} a & c \\ b & d \end{array}\right]$$ Eureka Math Precalculus Module 1 Lesson 29 Problem Set Answer Key

Find the inverse matrix of the following.

a. $$\left[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right]$$
Determinant =1-0=1 Inverse matrix: $$\left[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right]$$
Verify:
$$\left[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right]\left[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right]$$ = $$\left[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right]$$

b. $$\left[\begin{array}{ll} 0 & 1 \\ 1 & 0 \end{array}\right]$$
Determinant =0-1=-1 Inverse matrix: $$\left[\begin{array}{ll} 0 & 1 \\ 1 & 0 \end{array}\right]$$
Verify:
$$\left[\begin{array}{ll} 0 & 1 \\ 1 & 0 \end{array}\right]\left[\begin{array}{ll} 0 & 1 \\ 1 & 0 \end{array}\right]$$ = $$\left[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right]$$

c. $$\left[\begin{array}{ll} 1 & 1 \\ 1 & 1 \end{array}\right]$$
Determinant =1-1=0 No inverse matrix

d. $$\left[\begin{array}{ll} 1 & 0 \\ 1 & 0 \end{array}\right]$$
Determinant =0 No inverse matrix

e. $$\left[\begin{array}{ll} 0 & 1 \\ 0 & 1 \end{array}\right]$$
Determinant =0 No inverse matrix

f. $$\left[\begin{array}{ll} -2 & 2 \\ -5 & 4 \end{array}\right]$$
Determinant =-8+10=2 Inverse matrix: $$\left[\begin{array}{ll} 2 & -1 \\ \frac{5}{2} & -1 \end{array}\right]$$
Verify:
$$\left[\begin{array}{ll} -2 & 2 \\ -5 & 4 \end{array}\right]\left[\begin{array}{ll} 2 & -1 \\ \frac{5}{2} & -1 \end{array}\right]$$ = $$\left[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right]$$

g. $$\left[\begin{array}{ll} 4 & 6 \\ 5 & 8 \end{array}\right]$$
Determinant =32-30=2 Inverse matrix: $$\left[\begin{array}{cc} 4 & -3 \\ \frac{5}{2} & 2 \end{array}\right]$$
Verify:
$$\left[\begin{array}{ll} 4 & 6 \\ 5 & 8 \end{array}\right]\left[\begin{array}{cc} 4 & -3 \\ 5 & 2 \end{array}\right]$$ = $$\left[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right]$$

h. $$\left[\begin{array}{ll} 6 & -9 \\ 5 & -7 \end{array}\right]$$
Determinant =-42+45=3 Inverse matrix: $$\left[\begin{array}{ll} -\frac{7}{3} & 3 \\ -\frac{5}{3} & 2 \end{array}\right]$$
Verify:
$$\left[\begin{array}{cc} 6 & -9 \\ 5 & -7 \end{array}\right]\left[\begin{array}{cc} -\frac{7}{3} & 3 \\ -\frac{5}{3} & 2 \end{array}\right]$$=$$\left[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right]$$

i. $$\left[\begin{array}{cc} \frac{1}{2} & -\frac{2}{3} \\ -6 & 4 \end{array}\right]$$
Determinant =2-4=-2 Inverse matrix: $$\left[\begin{array}{ll} -2 & -\frac{1}{3} \\ -3 & \frac{1}{4} \end{array}\right]$$
Verify:
$$\left[\begin{array}{cc} \frac{1}{2} & -\frac{2}{3} \\ -6 & 4 \end{array}\right]\left[\begin{array}{cc} -2 & -\frac{1}{3} \\ -3 & \frac{1}{4} \end{array}\right]$$=$$\left[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right]$$

j.$$\left[\begin{array}{cc} 0.8 & 0.4 \\ -0.75 & -0.5 \end{array}\right]$$
Determinant =-0.4+0.3=-0.1 Inverse matrix: $$\left[\begin{array}{cc} 5 & -4 \\ -7.5 & -8 \end{array}\right]$$
Verify:
$$\left[\begin{array}{cc} 0.8 & 0.4 \\ -0.75 & -0.5 \end{array}\right]\left[\begin{array}{cc} 5 & -4 \\ -7.5 & -8 \end{array}\right]$$ = $$\left[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right]$$

Eureka Math Precalculus Module 1 Lesson 29 Exit Ticket Answer Key

A=$$\left[\begin{array}{cc} 4 & -2 \\ -1 & 3 \end{array}\right]$$

Question 1.
$$\left[\begin{array}{cc} \frac{3}{10} & \frac{2}{10} \\ \frac{1}{10} & \frac{4}{10} \end{array}\right]\left[\begin{array}{cc} 4 & -2 \\ -1 & 3 \end{array}\right]$$ = $$\left[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right]$$
Explain why the matrix $$\left[\begin{array}{ll} 6 & 3 \\ 4 & 2 \end{array}\right]$$ has no inverse.