1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
prohojiy [21]
3 years ago
6

Use the scenario to answer the question. Tyrone was asked to write the recursive and explicit formula for the sequence {−4,−7,−1

0,−13,…}. He wrote the recursive formula as a1=−4 and an=−4−3. He wrote the explicit formula as an=−4−3(n−1). Where did he make a mistake?
In his recursive formula, he should have added 3 instead of subtracting 3 to get an=−4+3.

In his explicit formula, the −4 is not necessary. He should have written an=−3(n−1).

In his explicit formula, he should have used n−3 instead of n−1 to get an=−4−3(n−3).

In his recursive formula, he should have the term an−1 instead of −4 to get an=an−1−3.
Mathematics
1 answer:
Anton [14]3 years ago
7 0

The last option: In his recursive formula, he should have the term an−1 instead of −4 to get an=an−1−3.

You might be interested in
Akemi plants a tree that is 4 feet tall. The tree is expected to grow 1/2 of a foot each month. Akemi wants to know after how ma
Anton [14]

Answer: 12 Months

Step-by-step explanation:

5 0
3 years ago
Read 2 more answers
Let A be a given matrix below. First, find the eigenvalues and their corresponding eigenspaces for the following matrices. Then,
Rama09 [41]

It looks like given matrices are supposed to be

\begin{array}{ccccccc}\begin{bmatrix}3&2\\2&3\end{bmatrix} & & \begin{bmatrix}1&-1\\2&-1\end{bmatrix} & & \begin{bmatrix}1&2&3\\0&2&3\\0&0&3\end{bmatrix} & & \begin{bmatrix}3&1&1\\1&3&1\\1&1&3\end{bmatrix}\end{array}

You can find the eigenvalues of matrix A by solving for λ in the equation det(A - λI) = 0, where I is the identity matrix. We also have the following facts about eigenvalues:

• tr(A) = trace of A = sum of diagonal entries = sum of eigenvalues

• det(A) = determinant of A = product of eigenvalues

(a) The eigenvalues are λ₁ = 1 and λ₂ = 5, since

\mathrm{tr}\begin{bmatrix}3&2\\2&3\end{bmatrix} = 3 + 3 = 6

\det\begin{bmatrix}3&2\\2&3\end{bmatrix} = 3^2-2^2 = 5

and

λ₁ + λ₂ = 6   ⇒   λ₁ λ₂ = λ₁ (6 - λ₁) = 5

⇒   6 λ₁ - λ₁² = 5

⇒   λ₁² - 6 λ₁ + 5 = 0

⇒   (λ₁ - 5) (λ₁ - 1) = 0

⇒   λ₁ = 5 or λ₁ = 1

To find the corresponding eigenvectors, we solve for the vector v in Av = λv, or equivalently (A - λI) v = 0.

• For λ = 1, we have

\begin{bmatrix}3-1&2\\2&3-1\end{bmatrix}v = \begin{bmatrix}2&2\\2&2\end{bmatrix}v = 0

With v = (v₁, v₂)ᵀ, this equation tells us that

2 v₁ + 2 v₂ = 0

so that if we choose v₁ = -1, then v₂ = 1. So Av = v for the eigenvector v = (-1, 1)ᵀ.

• For λ = 5, we would end up with

\begin{bmatrix}-2&2\\2&-2\end{bmatrix}v = 0

and this tells us

-2 v₁ + 2 v₂ = 0

and it follows that v = (1, 1)ᵀ.

Then the decomposition of A into PDP⁻¹ is obtained with

P = \begin{bmatrix}-1 & 1 \\ 1 & 1\end{bmatrix}

D = \begin{bmatrix}1 & 0 \\ 0 & 5\end{bmatrix}

where the n-th column of P is the eigenvector associated with the eigenvalue in the n-th row/column of D.

(b) Consult part (a) for specific details. You would find that the eigenvalues are i and -i, as in i = √(-1). The corresponding eigenvectors are (1 + i, 2)ᵀ and (1 - i, 2)ᵀ, so that A = PDP⁻¹ if

P = \begin{bmatrix}1+i & 1-i\\2&2\end{bmatrix}

D = \begin{bmatrix}i&0\\0&i\end{bmatrix}

(c) For a 3×3 matrix, I'm not aware of any shortcuts like above, so we proceed as usual:

\det(A-\lambda I) = \det\begin{bmatrix}1-\lambda & 2 & 3 \\ 0 & 2-\lambda & 3 \\ 0 & 0 & 3-\lambda\end{bmatrix} = 0

Since A - λI is upper-triangular, the determinant is exactly the product the entries on the diagonal:

det(A - λI) = (1 - λ) (2 - λ) (3 - λ) = 0

and it follows that the eigenvalues are λ₁ = 1, λ₂ = 2, and λ₃ = 3. Now solve for v = (v₁, v₂, v₃)ᵀ such that (A - λI) v = 0.

• For λ = 1,

\begin{bmatrix}0&2&3\\0&1&3\\0&0&2\end{bmatrix}v = 0

tells us we can freely choose v₁ = 1, while the other components must be v₂ = v₃ = 0. Then v = (1, 0, 0)ᵀ.

• For λ = 2,

\begin{bmatrix}-1&2&3\\0&0&3\\0&0&1\end{bmatrix}v = 0

tells us we need to fix v₃ = 0. Then -v₁ + 2 v₂ = 0, so we can choose, say, v₂ = 1 and v₁ = 2. Then v = (2, 1, 0)ᵀ.

• For λ = 3,

\begin{bmatrix}-2&2&3\\0&-1&3\\0&0&0\end{bmatrix}v = 0

tells us if we choose v₃ = 1, then it follows that v₂ = 3 and v₁ = 9/2. To make things neater, let's scale these components by a factor of 2, so that v = (9, 6, 2)ᵀ.

Then we have A = PDP⁻¹ for

P = \begin{bmatrix}1&2&9\\0&1&6\\0&0&2\end{bmatrix}

D = \begin{bmatrix}1&0&0\\0&2&0\\0&0&3\end{bmatrix}

(d) Consult part (c) for all the details. Or, we can observe that λ₁ = 2 is an eigenvalue, since subtracting 2I from A gives a matrix of only 1s and det(A - 2I) = 0. Then using the eigen-facts,

• tr(A) = 3 + 3 + 3 = 9 = 2 + λ₂ + λ₃   ⇒   λ₂ + λ₃ = 7

• det(A) = 20 = 2 λ₂ λ₃   ⇒   λ₂ λ₃ = 10

and we find λ₂ = 2 and λ₃ = 5.

I'll omit the details for finding the eigenvector associated with λ = 5; I ended up with v = (1, 1, 1)ᵀ.

• For λ = 2,

\begin{bmatrix}1&1&1\\1&1&1\\1&1&1\end{bmatrix}v = 0

tells us that if we fix v₃ = 0, then v₁ + v₂ = 0, so that we can pick v₁ = 1 and v₂ = -1. So v = (1, -1, 0)ᵀ.

• For the repeated eigenvalue λ = 2, we find the generalized eigenvector such that (A - 2I)² v = 0.

\begin{bmatrix}1&1&1\\1&1&1\\1&1&1\end{bmatrix}^2 v = \begin{bmatrix}3&3&3\\3&3&3\\3&3&3\end{bmatrix}v = 0

This time we fix v₂ = 0, so that 3 v₁ + 3 v₃ = 0, and we can pick v₁ = 1 and v₃ = -1. So v = (1, 0, -1)ᵀ.

Then A = PDP⁻¹ if

P = \begin{bmatrix}1 & 1 & 1 \\ 1 & -1 & 0 \\ 1 & 0 & -1\end{bmatrix}

D = \begin{bmatrix}5&0&0\\0&2&0\\0&2&2\end{bmatrix}

3 0
3 years ago
Point S is located at (-3,2) on the coordinate plane. Point S is reflected over the y-
grigory [225]

Answer:

(3,2)

Step-by-step explanation:

3 0
3 years ago
Write an equation parallel to the y-axis through the point (3,−2)
MatroZZZ [7]

do you have a picture of the graph?

8 0
3 years ago
NEED HELP ASAP!!!!!
Gennadij [26K]

Answer:

Step-by-step explanation:

5x-1=49

5x=50

X=50/5

X=10

2x+5=1

2x=-4

X= -4/2=-2

1-x=10

-x=10+1

X= -11

9x - 3 = 15

9x =12

X =12 /9

X=4/3=1.33

9-6x= 17

-6x= 8

X=8/-6

X=-1.33

3 0
3 years ago
Other questions:
  • Ms. Mincey wants to spend less than $80 on presents for her 2 brothers. What is the most amount of money she can spend on each o
    13·2 answers
  • Sylvia needs $110 for a concert ticket. She already has $16 and she can earn the rest by working 10 hours at her job. If h repre
    7·1 answer
  • April claims that 1 + cos2(theta) / sin2(theta)= 1 / sin2(theta) is an identity for all real numbers thetatheta that follows fro
    7·1 answer
  • Which number sentences show ways to solve the problem?
    12·1 answer
  • Consider the following function: f(x) = x² - 4
    13·1 answer
  • I need to find the resultant matrix
    11·1 answer
  • The school sold 300 tickets to the theatre performance. They sold adult tickets for $9 each and student tickets for $6 each. The
    12·1 answer
  • Beth works as a sales rep and receives $43 for every $210 she sells. What is her rate of commission? Round to the nearest tenths
    13·1 answer
  • Help please !! thank youu
    8·1 answer
  • A survey was given to people who own a certain type of car. What percent of the people surveyed were completely satisfied with t
    11·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!