Solution to Determine all eigenvalues and the corresponding eigenspaces for the matrix 𝐴 = [ βˆ’9 4 … - Sikademy
Author Image

Archangel Macsika

Determine all eigenvalues and the corresponding eigenspaces for the matrix 𝐴 = [ βˆ’9 4 4 βˆ’8 3 4 βˆ’16 8 7 ]

The Answer to the Question
is below this banner.

Can't find a solution anywhere?

NEED A FAST ANSWER TO ANY QUESTION OR ASSIGNMENT?

Get the Answers Now!

You will get a detailed answer to your question or assignment in the shortest time possible.

Here's the Solution to this Question

\det(A-\lambda I)=\begin{vmatrix} -9-\lambda & 4 & 4 \\ -8& 3-\lambda & 4 \\ -16 & 8 & 7-\lambda \\ \end{vmatrix}


=(-9-\lambda)\begin{vmatrix} 3-\lambda & 4 \\ 8 & 7-\lambda \end{vmatrix}-4\begin{vmatrix} -8 & 4 \\ -16 & 7-\lambda \end{vmatrix}

+4\begin{vmatrix} -8 & 3-\lambda \\ -16 & 8 \end{vmatrix}=(-9-\lambda)(21-10\lambda+\lambda^2-32)

-4(-56+8\lambda+64)+4(-64+48-16\lambda)

=99+11\lambda+90\lambda+10\lambda^2-9\lambda^2-\lambda^3

-32-32\lambda-64-64\lambda

=-\lambda^3+\lambda^2+5\lambda+3=0

-\lambda^2(\lambda+1)+2\lambda(\lambda+1)+3(\lambda+1)=0

-(\lambda+1)(\lambda^2-2\lambda-3)=0

-(\lambda+1)^2(\lambda-3)=0



\lambda_1=-1, \lambda_2=-1, \lambda_3=3


These are the eigenvalues:Β -1, -1, 3.


\lambda=-1


A-\lambda I=\begin{pmatrix} -9+1 & 4 & 4 \\ -8& 3+1 & 4 \\ -16 & 8 & 7+1 \\ \end{pmatrix}

=\begin{pmatrix} -8 & 4 & 4 \\ -8 & 4 & 4 \\ -16 & 8 & 8 \\ \end{pmatrix}

R_2=R_2-R_1


\begin{pmatrix} -8 & 4 & 4 \\ 0 & 0 & 0 \\ -16 & 8 & 8 \\ \end{pmatrix}

R_3=R_3-2R_1


\begin{pmatrix} -8 & 4 & 4 \\ 0 & 0 & 0 \\ 0 & 0 &0 \\ \end{pmatrix}

R_1=R_1/(-8)


\begin{pmatrix} 1 & -1/2 & -1/2 \\ 0 & 0 & 0 \\ 0 & 0 &0 \\ \end{pmatrix}

\begin{pmatrix} 1 & -1/2 & -1/2 \\ 0 & 0 & 0 \\ 0 & 0 &0 \\ \end{pmatrix}\begin{pmatrix} x_1 \\ x_2 \\ x_3 \\ \end{pmatrix}=\begin{pmatrix} 0 \\ 0 \\ 0 \\ \end{pmatrix}

If we takeΒ x_2=t, x_3=s,Β thenΒ x_1=\dfrac{1}{2}t+\dfrac{1}{2}s

Thus


\vec x=\begin{pmatrix} s/2+t/2 \\ 1 \\ 1 \\ \end{pmatrix}s=\begin{pmatrix} 1/2 \\ 1 \\ 0 \\ \end{pmatrix}t+\begin{pmatrix} 1/2 \\ 0 \\ 1 \\ \end{pmatrix}s

The eigenvectors are


\begin{pmatrix} 1/2 \\ 1 \\ 0 \\ \end{pmatrix},\begin{pmatrix} 1/2 \\ 0 \\ 1 \\ \end{pmatrix}

\lambda=3

A-\lambda I=\begin{pmatrix} -9-3 & 4 & 4 \\ -8& 3-3 & 4 \\ -16 & 8 & 7-3 \\ \end{pmatrix}

=\begin{pmatrix} -12 & 4 & 4 \\ -8 & 0 & 4 \\ -16 & 8 & 4 \\ \end{pmatrix}

R_2=R_2-2R_1/3

\begin{pmatrix} -12 & 4 & 4 \\ 0 & -8/3 & 4/3 \\ -16 & 8 & 4 \\ \end{pmatrix}

R_3=R_3-4R_1/3

\begin{pmatrix} -12 & 4 & 4 \\ 0 & -8/3 & 4/3 \\ 0 & 8/3 & -4/3 \\ \end{pmatrix}

R_3=R_3+R_2

\begin{pmatrix} -12 & 4 & 4 \\ 0 & -8/3 & 4/3 \\ 0 & 0 & 0 \\ \end{pmatrix}

R_2=-3R_2/8


\begin{pmatrix} -12 & 4 & 4 \\ 0 & 1 & -1/2 \\ 0 & 0 & 0 \\ \end{pmatrix}

R_1=-R_1/12


\begin{pmatrix} 1 &-1/3 & -1/3 \\ 0 & 1 & -1/2 \\ 0 & 0 & 0 \\ \end{pmatrix}

R_1=R_1+R_2/3


\begin{pmatrix} 1 &0 & -1/2 \\ 0 & 1 & -1/2 \\ 0 & 0 & 0 \\ \end{pmatrix}


\begin{pmatrix} 1 &0 & -1/2 \\ 0 & 1 & -1/2 \\ 0 & 0 & 0 \\ \end{pmatrix}\begin{pmatrix} x_1 \\ x_2 \\ x_3 \\ \end{pmatrix}=\begin{pmatrix} 0 \\ 0 \\ 0 \\ \end{pmatrix}

If we takeΒ x_3=t,Β thenΒ x_1=\dfrac{1}{2}t, x_2=\dfrac{1}{2}t

Thus


\vec x=\begin{pmatrix} t/2 \\ t/2 \\ t \\ \end{pmatrix}=\begin{pmatrix} 1/2 \\ 1/2 \\ 1 \\ \end{pmatrix}t


The eigenvector is


\begin{pmatrix} 1/2 \\ 1/2 \\ 1 \\ \end{pmatrix}



Eigenvalue:Β βˆ’1, multiplicity:Β 2, eigenvectors:Β \begin{pmatrix} 1/2 \\ 1 \\ 0 \\ \end{pmatrix},\begin{pmatrix} 1/2 \\ 0 \\ 1 \\ \end{pmatrix}


Eigenvalue:Β 3, multiplicity:Β 1, eigenvector:Β \begin{pmatrix} 1/2 \\ 1/2 \\ 1 \\ \end{pmatrix}



Related Answers

Was this answer helpful?

Join our Community to stay in the know

Get updates for similar and other helpful Answers

Question ID: mtid-5-stid-8-sqid-2619-qpid-1089