Problem 115 find all possible values of the unknown...

Problem 11.5. Find all possible values of the unknown constants given the condition. 1. $A = \begin{bmatrix} 0 & a \\ 1 & b \end{bmatrix}$ and the eigenvalues are 4, 7. 2. $A = \begin{bmatrix} 0 & 0 & a \\ 1 & 0 & b \\ 0 & 1 & c \end{bmatrix}$ and the eigenvalues are 2, 3, 4. 3. $A = \begin{bmatrix} 2 & -2 & 0 \\ -2 & x & -2 \\ -2 & -2 & 0 \end{bmatrix}$, $B = \begin{bmatrix} 2 \\ 2 \\ y \end{bmatrix}$, and A, B have the same characteristic polynomial. 4. $A = \begin{bmatrix} 2 & 1 & 1 \\ 1 & 2 & 1 \\ 1 & 1 & a \end{bmatrix}$ is invertible, and $x = \begin{bmatrix} 1 \\ b \\ 1 \end{bmatrix}$ is an eigenvector of A. 5. $A = \begin{bmatrix} 3 & 2 & -2 \\ -k & -1 & k \\ 4 & 2 & -3 \end{bmatrix}$ is diagonalizable.

Transcript

00:02 We're given a matrix a.

00:04 A has elements 2213.

00:13 Yeah.

00:14 In part a, we're asked to find all the eigenvalues in corresponding eigenvectors of this matrix.

00:20 Well, first we'll find the characteristic polynomial of a.

00:26 So delta of t, this is, because this is a 2x2 matrix, t, squared minus the trace of a times t plus the determinant of a which is t squared minus five t plus four which you can factor as t minus one times t minus four which you can factor as t minus one times t minus four she had a stroke from being like bulimic or something i thought she now the roots, lambda equals 1 and the lambda equals 4 of delta t.

01:13 These are in fact the eigenvalues of our matrix a.

01:18 Do you know the darwin? you put your limp dick in a girl's mouth and then it evolves into a boner.

01:24 Now let's find the corresponding eigenvectors.

01:30 So first we'll find the one's corresponding to lambda equals 1.

01:34 We'll subtract 1 down the diagonal of a.

01:37 So we get the matrix m, which is a minus i.

01:44 And the corresponding homogenous system, m times x equals 0, will yield the eigenvectors belonging to lambda equals 1.

01:53 So we have the matrix m is 1, 212.

02:00 A lot of black soldiers fall.

02:02 And so the corresponding homogenous system is x plus 2y equals 0.

02:10 X plus 2y equals 0, which of course is just the same as x plus 2y equals 0.

02:19 So we have one degree of freedom.

02:20 And one of the non -zero solutions is we take x to be some kid came up to me.

02:32 I was watching your heavy weights.

02:34 Two and y to be negative one.

02:39 So for example, v1, the matrix 2 negative 1, it is a non -zero solution.

02:47 And therefore, it's also an eigenvector belonging to.

02:59 Eigenvalue lambda equals 1 now consider a bunch of now consider the eigenvalue lambda equals 4 we'll subtract this down the diagonal of a to get the matrix m this is m is a minus 4i which gives us the matrix negative 2 -2 -1 -negative 1.

03:28 This corresponds to the homogenous system negative 2x plus 2y equals 0, and x minus y equals 0, which corresponds to the system x minus y equals 0, which has only one independent solution.

03:46 So, for example, you can take x and y to both be 1, and therefore b2, which is the vector 1 -1, is an eigenvector.

03:58 And then they cross that with retar juice.

04:02 Belonging to the eigen value, lambda equals 4.

04:08 And then they crossed beetle juice with just retard.

04:12 All right.

04:16 Then in part b, we were to find a non -singular matrix p, such that the matrix d, which is p inverse times a -p, is diagonal.

04:30 And to also find the matrix p inverse, which we know exists because p will be non -singular.

04:39 Well, let p be the matrix whose columns are our eigenvectors, v1 and v2, then p is the matrix.

04:47 To negative 1, 1, 1.

04:52 And the matrix d, which is p inverse ap.

04:57 Well, this is simply the matrix whose diagonal entries will be the eigenvalues.

05:03 Corresponding eigenvalues, 1 and 4.

05:06 I still have anxiety and panic, but it's like, yeah, but it's because i'm high.

05:10 And then it just gives you an excuse to feel as shitty as normally dead.

05:14 I feel like it's what i use it for is to when i'm anxious and i'm freaking out...

Question

Please give Ace some feedback