Question

Directions: Answer the following questions on separate paper. Clearly indicate your solutions for each problem. Scan and upload your problems as a single pdf to Canvas. 1. For each of the matrices use multiplication to check if the given vectors or eigenvectors. Determine the eigenvalue for each eigenvector. $\begin{pmatrix} 2 & 3 \\ 3 & -6 \end{pmatrix}$, and vectors $u = \begin{pmatrix} 3 \\ 1 \end{pmatrix}$, $v = \begin{pmatrix} 1 \\ 1 \end{pmatrix}$, and $w = \begin{pmatrix} 1 \\ 3 \end{pmatrix}$ a. M = $\begin{pmatrix} 7 & -4 & 4 \\ 8 & -5 & 8 \\ 0 & 0 & 3 \end{pmatrix}$, and vectors $u = \begin{pmatrix} 1 \\ 2 \\ 0 \end{pmatrix}$, $v = \begin{pmatrix} -1 \\ 0 \\ 1 \end{pmatrix}$, and $w = \begin{pmatrix} 0 \\ 1 \\ 1 \end{pmatrix}$ b. M = 2. For each of the matrices check that the given $\lambda$ is and eigenvalue by determining if $M - \lambda I$ is dependent. Then find a basis for the calculating the Null Space of $M - \lambda I$. $\begin{pmatrix} 1 & 2 \\ 2 & 1 \end{pmatrix}$, and and $\lambda = -1$. a. M = $\begin{pmatrix} 1 & 3 & 3 \\ -3 & -5 & -3 \\ 3 & 3 & 1 \end{pmatrix}$, and $\lambda = -2$. b. M =

          Directions: Answer the following questions on separate paper. Clearly indicate your solutions for each
problem. Scan and upload your problems as a single pdf to Canvas.
1. For each of the matrices use multiplication to check if the given vectors or eigenvectors. Determine
the eigenvalue for each eigenvector.
$\begin{pmatrix} 2 & 3 \\ 3 & -6 \end{pmatrix}$, and vectors $u = \begin{pmatrix} 3 \\ 1 \end{pmatrix}$, $v = \begin{pmatrix} 1 \\ 1 \end{pmatrix}$, and $w = \begin{pmatrix} 1 \\ 3 \end{pmatrix}$
a. M = 
$\begin{pmatrix} 7 & -4 & 4 \\ 8 & -5 & 8 \\ 0 & 0 & 3 \end{pmatrix}$, and vectors $u = \begin{pmatrix} 1 \\ 2 \\ 0 \end{pmatrix}$, $v = \begin{pmatrix} -1 \\ 0 \\ 1 \end{pmatrix}$, and $w = \begin{pmatrix} 0 \\ 1 \\ 1 \end{pmatrix}$
b. M =
2. For each of the matrices check that the given $\lambda$ is and eigenvalue by determining if $M - \lambda I$ is dependent.
Then find a basis for the calculating the Null Space of $M - \lambda I$.
$\begin{pmatrix} 1 & 2 \\ 2 & 1 \end{pmatrix}$, and and $\lambda = -1$.
a. M =
$\begin{pmatrix} 1 & 3 & 3 \\ -3 & -5 & -3 \\ 3 & 3 & 1 \end{pmatrix}$, and $\lambda = -2$.
b. M =

Show more…

Directions: Answer the following questions on separate paper. Clearly indicate your solutions for each
problem. Scan and upload your problems as a single pdf to Canvas.
1. For each of the matrices use multiplication to check if the given vectors or eigenvectors. Determine
the eigenvalue for each eigenvector.
< p m a t r i x >, and vectors u =
< p m a t r i x >, v =
< p m a t r i x >, and w =
< p m a t r i x >
a. M =
< p m a t r i x >, and vectors u =
< p m a t r i x >, v =
< p m a t r i x >, and w =
< p m a t r i x >
b. M =
2. For each of the matrices check that the given λ is and eigenvalue by determining if M - λ I is dependent.
Then find a basis for the calculating the Null Space of M - λ I.
< p m a t r i x >, and and λ = -1.
a. M =
< p m a t r i x >, and λ = -2.
b. M =

Added by Sara A.

Calculus: Early Transcendentals

James Stewart 8th Edition

Instant Answer

Solved by Expert Zhumagali Shomanov

10/31/2022

Step 1

Step 1: For matrix M=[[2,3],[3,-6]] and vector u=[[3],[1]]: Calculate Mu: Mu = [[2,3],[3,-6]] * [[3],[1]] = [[2*3 + 3*1], [3*3 + (-6)*1]] = [[9],[3]] Now, calculate λu: λu = λ * u = λ * [[3],[1]] = -1 * [[3],[1]] = [[-3],[-1]] Check if Mu = λu: [[9],[3]] ≠ Show more…

Show all steps

lock

Thanks for your feedback!

Directions: Answer the following questions on separate paper. Clearly indicate your solutions for each problem. Scan and upload your problems as a single PDF to Canvas. 1. For each of the matrices, use multiplication to check if the given vectors are eigenvectors. Determine the eigenvalue for each eigenvector. a. M=[[2,3],[3,-6]], and vectors u=[[3],[1]], v=[[1],[1]], and w=[[1],[3]]. b. M=[[7,-4,4],[8,-5,8],[0,0,3]], and vectors u=[[1],[2],[0]], v=[[-1],[0],[1]], and w=[[0],[1],[1]]. 2. For each of the matrices, check that the given λ is an eigenvalue by determining if M-λI is dependent. Then find a basis for calculating the Null Space of M-λI. a. M=[[1,2],[2,1]], and λ=-1. b. M=[[1,3,3],[-3,-5,-3],[3,3,1]], and λ=-2.

Play audio

Feedback

Powered by NumerAI

Zhumagali Shomanov and 83 other subject Calculus 3 educators are ready to help you.

Ask a new question

Labs

Want to see this concept in action?

NEW

Explore this concept interactively to see how it behaves as you change inputs.

View Labs

Key Concepts

Recommended Videos

Recommended Textbooks

Transcript

00:01 In this question, we are asked to find the eigenvalues of the matrix a, and then also find the eigenvectors.

00:08 To find the eigenvalues, we need to solve the equation, determinant a minus lambda i equals 0.

00:18 And to get the matrix a minus lambda i, we simply need to subtract lambda from the entries on the main diagonal for the matrix a.

00:27 We are going to get 3 minus lambda, negative 2, 6, 0, or 0.

00:33 1 minus lambda 0 and 0 01 minus lambda.

00:43 Now we need to calculate the determinant of this matrix.

00:48 So determinant a minus lambda i equals to the determinant of this matrix and should be lambda 1 minus lambda and we need to put vertical bars since we are calculating the determinant.

01:31 And this matrix is a triangular matrix which means that its determinant determinant is simply the product of the entries on the main diagonal.

01:43 And this equals to 3 minus lambda, multiplied by 1 minus lambda squared.

01:53 And we want this to be equal to 0 to find the eigenvalues.

02:02 And it's easy to see that the roots are lambda equals to 3 and lambda equals to 1.

02:10 Now we need, so these are the eigenvalues.

02:15 There are two eigenvalues.

02:17 Now let's calculate the corresponding eigenvectors.

02:24 To find the eigenvectors, we need to solve, so let's call the eigenvalues lambda 1 and lambda 2.

02:31 To find the eigenvalues corresponding to lambda 1, we need to solve the equation a x, a minus lambda 1i times x equals 0, or it's equivalent to a minus 3i equals x equals 0.

02:54 To calculate a minus 3i, we need to subtract 3 from the entries on the main diagonal of the matrix a.

03:11 So we simply need to replace lambda by 3 here.

03:19 We are going to get 3 minus 3, 1 minus 3 and 1 minus 3.

03:26 So this is a minus 3i, and it simplifies to 0, negative 2, 6, 0 negative 2, 0, 0, 0 .000, 0.

03:38 And 0, 0, negative 2.

03:44 Now, the augmented matrix of the equation, a minus 3 i .x equals 0 ,000, is going to be 0 negative 2, 6, 0 ,0 negative 2, 0 ,000, and 0 ,0 negative 2.

04:10 And what we're going to do is we will subtract the second row from the first row, and multiply the third row by 3 and add it to the first row...

Ace is your personal tutor. It breaks down any question with clear steps so you can learn.

Start Using Ace

Continue solving this problem

Create a free account to:

View full step-by-step solution
Ask follow-up questions with Ace AI
Save progress and study later