Determine whether A is similar B .If A sim B, give an invertible matrix P such that P^{-1}AP = B A=begin{bmatrix}1 & 1&0 0&1&10&0&1 end{bmatrix} , B=begin{bmatrix}1&1&0 0&1&00&0&1 end{bmatrix}

Burhan Hopper

Burhan Hopper

Answered question

2020-10-26

Determine whether A is similar B .If AB, give an invertible matrix P such that P1AP=B
A=[110011001],B=[110010001]

Answer & Explanation

cyhuddwyr9

cyhuddwyr9

Skilled2020-10-27Added 90 answers

Step 1
The given matrices are A=[110011001] and B=[110010001]
Here, it is observed that the matrices A and B are diagonal matrix.
Therefore, the eigenvalues are the entries on the main diagonal values.
Hence, the eigenvalues of both A, B are λ1=1,λ2=1and λ3=1
Step 2
Obtain the eigenvector corresponding to the eigenvalue λ=1 by solving the system (A-I)X=0. Let X=[x1x2x3]
Then, ([110011001][110010001])[x1x2x3]=0
[010001000][x1x2x3]=0
The system of equations corresponding to the above system is x1=x1(free variable)
x2=0
x3=0
If x1=1 then v1=[100]
Step 3
Obtain the eigenvector corresponding to the eigenvalue λ=1 by solving the system (B-I)X=0.
Then ([110010001][100010001])[x1x2x3]=0
[010000000][x1x2x3]=0
The system of equations corresponding to the above system is x1=x1(free variable)
x2=0
x3=x3(free variable)
If x1=1,x3=0 then v1=[100]
Choose the arbitrary value x1=0,x3=1
Then the eigenvector is v2=[001]
It is observed that the eigenspace corresponding to λ=1 of a matrix B has dimension 2 and the eigenspace corresponding to λ=1 of a matrix A has dimension 1.
Hence, the matrices are not similar.

Jeffrey Jordon

Jeffrey Jordon

Expert2022-01-27Added 2605 answers

Answer is given below (on video)

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