B consider the matrix a beginbmatrix 7 1 1 11 3 2 18 2 4...

Name: b consider the matrix a beginbmatrix 7 1 1 11 3 2 18 2 4 endbmatrix i find the eigenvalue and the fundamental eigenvectors of a ii is a diagonalizable justify your answer c find the reduced 94087
Uploaded: 2021-04-24T19:40:13-08:00
Duration: 15 min 48 s
Channel: Chris Trentman
Description: b consider the matrix a beginbmatrix 7 1 1 11 3 2 18 2 4 endbmatrix i find the eigenvalue and the fundamental eigenvectors of a ii is a diagonalizable justify your answer c find the reduced 94087

(b) Consider the matrix: $$A = \begin{bmatrix} 7 & 1 & -1 \\ -11 & -3 & 2 \\ 18 & 2 & -4 \end{bmatrix}$$ (i) Find the eigenvalue and the fundamental eigenvectors of A. (ii) Is A diagonalizable? Justify your answer. (c) Find the reduced row echelon form matrix B of the following matrix : $$A = \begin{bmatrix} 1 & 2 & -2 & -11 \\ 2 & 4 & -1 & -10 \\ 3 & 6 & -4 & -25 \end{bmatrix}$$ and then give a sequence of row operations that convert B back to A.

Transcript

00:01 We're given a matrix a.

00:06 A is the 2x2 matrix 2, negative 1, negative 2, 3.

00:18 In part a, we're asked to find the eigenvalues in corresponding eigenvectors of this matrix.

00:25 To do this, let's find the characteristic polynomial.

00:28 For a 2x2 polynomial, this is 2x2 matrix.

00:32 This is t squared minus the trace of a, which is 5 times t, plus the determinant of a, which is 8.

00:45 And we can factor this as so it should be instead of plus 8 plus 4.

01:12 And so this can actually be factored as t minus 1 times t minus 4.

01:22 And therefore the zeros of the characteristic polynomial are our eigenvalues.

01:33 Lambda 1 equals 1 and lambda 2 equals 4.

01:44 We'll begin with the first eigenvalue, lambda 1 equals 1.

01:48 I'll subtract 1 down the diagonal of a.

01:51 We get the matrix m, which is a minus i.

01:56 This is the 2x2 matrix 1, negative 1, negative 2.

02:03 2.

02:07 And this corresponds to the homogenous system x minus y equals 0 and negative 2x plus 2y equals 0.

02:16 And this corresponds to the single equation x minus y equals 0.

02:21 So take y equals 1, then x that equal 1.

02:25 There's only one independent eigenvector.

02:31 I'll call it u.

02:32 So that u equals 1 -1 is an eigenvector belonging to the eigenvalue lambda 1, which is 1.

02:55 Now, with their second eigenvalue, 4, i'm going to subtract 4 down the diagonal of a.

03:03 So we get our matrix m, which is a minus 4i.

03:09 And this is the 2x2 matrix, negative 2, negative 1, negative 1, negative 2, negative 1.

03:18 This gives us the homogenous system, negative 2x minus y equals 0, and negative 2x minus y equals 0, which is of course just the single equation, negative 2x minus y equals 0.

03:35 And if you take x to be positive 1, then y is negative 2, and so we get the vector v, which is 1, negative 2, and this is an eigenvector belonging to the eigenvalue lambda 2 which was 4.

04:18 Then in part b we're asked to find a non -singular matrix p such that matrix d which is p inverse ap is diagonal.

04:32 Well because we have a maximal set of a nearly independent eigenvectors which is u and v for 1 -1 and 1 -2.

05:00 So it follows that the matrix p, whose columns are these hyginvectors is non -singular and diagonalizes our matrix a.

05:15 So we'll take p to be the matrix with columns 1 -1, and 1 negative 2.

05:27 And as a result we have that a is diagonalizable and d is equal to p inverse ap where d is a diagonal matrix of corresponding eigenvalues for entries so 1 -004 then in part c we're asked to find a to the 8th power and find f of a, where f of t is a certain polynomial.

06:10 First i'll find a to the eighth power...

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