Determine whether the given matrix is diagonalizable; if so,...

Name: Determine whether the given matrix is diagonalizable; if so, find a matrix P and a diagonal matrix D such that A = PDP?1. (If the matrix is not diagonalizable, enter DNE in any cell.) A = [ 4 -2; 1 1 ] [P D] = [ ]
Uploaded: 2023-03-24T06:08:10-08:00
Duration: 6 min 34 s
Channel: Supreeta N
Description: Determine whether the given matrix is diagonalizable; if so, find a matrix P and a diagonal matrix D such that A = PDP?1. (If the matrix is not diagonalizable, enter DNE in any cell.) A = [ 4 -2; 1 1 ] [P D] = [ ]

Determine whether the given matrix is diagonalizable; if so, find a matrix P and a diagonal matrix D such that A = PDP?1. (If the matrix is not diagonalizable, enter DNE in any cell.) A = [ 4 -2; 1 1 ] [P D] = [ ]

Transcript

00:01 Hi, let's start the solution.

00:02 In this question we have given matrix a is 4 1 minus 2 1.

00:07 Then question says that check matrix a is diagonalizable or not.

00:11 So to check matrix is diagonalizable or not, first we find the eigenvalues of matrix a.

00:17 So, determinant a minus lambda i is 0.

00:22 So we get determinant a is 4 1 minus 2 1 minus 2 .1 minus 1 minus lambda identity of 2 cross 2 matrix is 1 .0.

00:30 0 1 equal to 0 so here we get determinant 4 minus lambda 1 minus 2 1 minus lambda equal to 0 we know that determinant of 2 cross 2 matrix is a cross multiplication so we get 4 minus lambda into 1 minus lambda plus 2 equal to 0 by simplifying this here we get this 1 is 4 into 1 is 4 minus 4 lambda plus lambda plus lambda equal to 0 so here we get lambda square minus 5 lambda plus 6 equal to 0 now factorize that characteristic equation then we get lambda minus 2 into lambda minus 3 equal to 0 so that means here we have lambda minus 2 equal to 0 a lambda minus 3 equal to 0 so we get lambda is equal to 2 and lambda equal to 3 so here we see that eigen values of matrix a is 2 and 3 which is distinct.

01:42 So by result when eigenvalues of matrix a is distinct then matrix a is diagonalizable.

01:58 Next question says that next we find eigen factors.

02:07 So for eigenvctors put first lambda is 3 then we get 4 minus lambda 1.

02:14 Minus 2 1 minus lambda which is equal to 1 1 minus 2 minus minus minus so now we reduce this matrix in a row reduced accolent form.

02:35 So here we get 1 minus 2 0 0 into x1 x2 is 0 0.

02:45 So because for eigenvactor a minus a minus lambda are into x is 0 so from here we get x1 is 2 t and x2 is t so we get eigenvactor x is 2 t so that means 2 .1 t 2 .1 is a null space of this matrix next for lambda equal to 2 we get 4 minus lambda 1 minus 2 1 minus lambda is equal to 2 1 minus 2 1 minus 2 1.

03:37 Now reduce this matrix in a row echolent form.

03:42 Then we get 1 minus 1 0 0 into x 1 x2 is 0.

03:53 So from here we get factor x is t...

Question

Determine whether the given matrix is diagonalizable; if so, find a matrix P and a diagonal matrix D such that A = PDP?1. (If the matrix is not diagonalizable, enter DNE in any cell.) A = [ 4 -2; 1 1 ] [P D] = [ ]

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