Question #292571

Question. Given the matrix A = 3 1 1

2 4 2

1 1 3


a. Solve A for its eigen values and given vectors

b. Constant Similar matrix for A if possible


1
Expert's answer
2022-02-01T14:52:35-0500

a.


A=[311242113]A=\begin{bmatrix} 3 & 1 & 1 \\ 2 & 4 & 2 \\ 1 & 1 & 3 \end{bmatrix}

AλI=[3λ1124λ2113λ]A-\lambda I=\begin{bmatrix} 3-\lambda & 1 & 1 \\ 2 & 4-\lambda & 2 \\ 1 & 1 & 3-\lambda \end{bmatrix}

detAλI=3λ1124λ2113λ\det A-\lambda I=\begin{vmatrix} 3-\lambda & 1 & 1 \\ 2 & 4-\lambda & 2 \\ 1 & 1 & 3-\lambda \end{vmatrix}

=(3λ)2(4λ)+2+2(4λ)2(3λ)=(3-\lambda)^2(4-\lambda)+2+2-(4-\lambda)-2(3-\lambda)

2(3λ)=3624λ+4λ29λ+6λ2λ3-2(3-\lambda)=36-24\lambda+4\lambda^2-9\lambda+6\lambda^2-\lambda^3

+44+λ12+4λ=λ3+10λ228λ+24=0+4-4+\lambda-12+4\lambda=-\lambda^3+10\lambda^2-28\lambda+24=0

λ2(λ2)+8λ(λ2)12(λ2)=0-\lambda^2(\lambda-2)+8\lambda(\lambda-2)-12(\lambda-2)=0

(λ2)2(λ6)=0-(\lambda-2)^2(\lambda-6)=0

λ1=6,λ2=2,λ3=2\lambda_1=6, \lambda_2=2, \lambda_3=2

These are the eigenvalues.


b. Find the eigenvectors.

λ=6\lambda=6


[3λ1124λ2113λ]=[311222113]\begin{bmatrix} 3-\lambda & 1 & 1 \\ 2 & 4-\lambda & 2 \\ 1 & 1 & 3-\lambda \end{bmatrix}=\begin{bmatrix} -3 & 1 & 1 \\ 2 & -2 & 2 \\ 1 & 1 & -3 \end{bmatrix}

R1=R1/(3)R_1=R_1/(-3)


[11/31/3222113]\begin{bmatrix} 1 & -1/3 & -1/3 \\ 2 & -2 & 2 \\ 1 & 1 & -3 \end{bmatrix}

R2=R22R1R_2=R_2-2R_1


[11/31/304/38/3113]\begin{bmatrix} 1 & -1/3 & -1/3 \\ 0 & -4/3 & 8/3 \\ 1 & 1 & -3 \end{bmatrix}

R3=R3R1R_3=R_3-R_1


[11/31/304/38/304/38/3]\begin{bmatrix} 1 & -1/3 & -1/3 \\ 0 & -4/3 & 8/3 \\ 0 & 4/3 & -8/3 \end{bmatrix}

R2=3R2/4R_2=-3R_2/4


[11/31/301204/38/3]\begin{bmatrix} 1 & -1/3 & -1/3 \\ 0 & 1 & -2 \\ 0 & 4/3 & -8/3 \end{bmatrix}

R1=R1+R2/3R_1=R_1+R_2/3


[10101204/38/3]\begin{bmatrix} 1 & 0 & -1 \\ 0 & 1 & -2 \\ 0 & 4/3 & -8/3 \end{bmatrix}

R3=R34R2/3R_3=R_3-4R_2/3


[101012000]\begin{bmatrix} 1 & 0 & -1 \\ 0 & 1 & -2 \\ 0 & 0 & 0 \end{bmatrix}

[101012000][x1x2x3]=[000]\begin{bmatrix} 1 & 0 & -1 \\ 0 & 1 & -2 \\ 0 & 0 & 0 \end{bmatrix}\begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix}=\begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix}

If we take x3=t,x_3=t, then x1=t,x2=2t.x_1=t, x_2=2t.

The eigenvector is [121]\begin{bmatrix} 1 \\ 2 \\ 1 \end{bmatrix}



λ=2\lambda=2


[3λ1124λ2113λ]=[111222111]\begin{bmatrix} 3-\lambda & 1 & 1 \\ 2 & 4-\lambda & 2 \\ 1 & 1 & 3-\lambda \end{bmatrix}=\begin{bmatrix} 1 & 1 & 1 \\ 2 & 2 & 2 \\ 1 & 1 & 1 \end{bmatrix}

R2=R22R1R_2=R_2-2R_1


[111000111]\begin{bmatrix} 1 &1 & 1\\ 0 & 0 & 0 \\ 1 & 1 & 1\end{bmatrix}

R3=R3R1R_3=R_3-R_1


[111000000]\begin{bmatrix} 1 & 1 & 1 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix}




[111000000][x1x2x3]=[000]\begin{bmatrix} 1 & 1 & 1 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix}\begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix}=\begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix}

If we take x2=t,x3=s,x_2=t,x_3=s, then x1=ts.x_1=-t-s.


The eigenvectors are [110],[101].\begin{bmatrix} -1 \\ 1 \\ 0 \end{bmatrix},\begin{bmatrix} -1 \\ 0 \\ 1 \end{bmatrix}.



P=[111210101],D=[600020002]P=\begin{bmatrix} 1 & -1 & -1 \\ 2 & 1 & 0 \\ 1 & 0 & 1 \end{bmatrix}, D=\begin{bmatrix} 6 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 2 \end{bmatrix}


PDP1=APDP^{-1}=A

The matrices A=[311242113]A=\begin{bmatrix} 3 & 1 & 1 \\ 2 & 4 & 2 \\ 1 & 1 & 3 \end{bmatrix} and D=[600020002]D=\begin{bmatrix} 6 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 2 \end{bmatrix} are similar.



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