Question #264229

Consider the linear eigenproblem Ax=λx for the matrix



D=[1 1 2



2 1 1



1 1 3 ]



1. Solve for the largest (in magnitude) eigenvalue of the matrix and the corresponding eigenvector by the power method with 𝑥(0)T=[1 0 0]



2. Solve for the smallest eigenvalue of the matrix and the corresponding eigenvector by the inverse power method using the matrix inverse. Use Gauss-Jordan elimination to find the matrix inverse.

1
Expert's answer
2021-11-12T06:48:32-0500

1st Iteration;


Dx0=Dx_0= (112211113)(100)=(121)\begin{pmatrix} 1&1&2 \\ 2&1& 1\\ 1&1&3 \end{pmatrix}\begin{pmatrix} 1 \\ 0\\ 0 \end{pmatrix}=\begin{pmatrix} 1 \\ 2 \\ 1 \end{pmatrix}



x1=12(121)=(0.510.5)x_1=\frac{1}{2}\begin{pmatrix} 1 \\ 2\\ 1 \end{pmatrix}=\begin{pmatrix} 0.5\\ 1\\ 0.5 \end{pmatrix}



2nd Iteration



Dx1=(112211113)(0.510.5)=(2.52.53)Dx_1=\begin{pmatrix} 1&1&2 \\ 2&1& 1\\ 1&1&3 \end{pmatrix}\begin{pmatrix} 0.5 \\ 1 \\ 0.5 \end{pmatrix}=\begin{pmatrix} 2.5 \\ 2.5 \\ 3 \end{pmatrix}


x2=13(2.52.53)=(0.830.831)x_2=\frac{1}{3}\begin{pmatrix} 2.5 \\ 2.5 \\ 3 \end{pmatrix}=\begin{pmatrix} 0.83 \\ 0.83\\ 1 \end{pmatrix}



3rd Iteration



Dx2=(112211113)(0.830.831)=(3.673.54.67)Dx_2=\begin{pmatrix} 1&1&2 \\ 2&1 & 1\\ 1&1&3 \end{pmatrix}\begin{pmatrix} 0.83 \\ 0.83 \\ 1 \end{pmatrix}=\begin{pmatrix} 3.67 \\ 3.5 \\ 4.67 \end{pmatrix}



x3=14.67(3.673.54.67)=(0.790.751)x_3=\frac{1}{4.67}\begin{pmatrix} 3.67 \\ 3.5\\ 4.67 \end{pmatrix}=\begin{pmatrix} 0.79 \\ 0.75 \\ 1 \end{pmatrix}




4th Iteration



Dx3=(112211113)(0.790.751)=(3.543.324.54)Dx_3=\begin{pmatrix} 1&1 & 2 \\ 2&1 & 1\\ 1&1&3 \end{pmatrix}\begin{pmatrix} 0.79 \\ 0.75\\ 1 \end{pmatrix}=\begin{pmatrix} 3.54 \\ 3.32 \\ 4.54 \end{pmatrix}




x4=14.54(3.543.324.54)=(0.780.731)x_4=\frac{1}{4.54}\begin{pmatrix} 3.54 \\ 3.32 \\ 4.54 \end{pmatrix}=\begin{pmatrix} 0.78 \\ 0.73\\ 1 \end{pmatrix}




5th Iteration



Dx4=(112211113)(0.780.731)=(3.513.294.51)Dx_4=\begin{pmatrix} 1&1 & 2 \\ 2&1 & 1\\ 1&1&3 \end{pmatrix}\begin{pmatrix} 0.78\\ 0.73 \\ 1 \end{pmatrix}=\begin{pmatrix} 3.51 \\ 3.29\\ 4.51 \end{pmatrix}




x5=14.51(3.513.294.51)=(0.780.731)x_5=\frac{1}{4.51}\begin{pmatrix} 3.51 \\ 3.29 \\ 4.51 \end{pmatrix}=\begin{pmatrix} 0.78 \\ 0.73 \\ 1 \end{pmatrix}




6th Iteration



Dx5=(112211113)(0.780.731)=(3.513.294.51)Dx_5=\begin{pmatrix} 1&1 & 2\\ 2&1& 1\\ 1&1&3 \end{pmatrix}\begin{pmatrix} 0.78 \\ 0.73 \\ 1 \end{pmatrix}=\begin{pmatrix} 3.51 \\ 3.29\\ 4.51 \end{pmatrix}




x6=14.51(3.513.294.51)=(0.780.731)x_6=\frac{1}{4.51}\begin{pmatrix} 3.51 \\ 3.29 \\ 4.51 \end{pmatrix}=\begin{pmatrix} 0.78 \\ 0.73\\ 1 \end{pmatrix}




\therefore The dominant eigenvalue λ=4.51\lambda=4.51

and the dominant eigenvector is


(0.780.731)\begin{pmatrix} 0.78 \\ 0.73 \\ 1 \end{pmatrix}



D-1 by Gaus - Jordan Elimination


Augmenting DD with a 3×33×3 identity matrix


[112100211010113001]\begin{bmatrix} 1&1&2| 1&0&0 \\ 2&1&1|0&1&0 \\ 1&1&3|0&0&1 \end{bmatrix}


12R2R1R2\frac{1}{2}R_2-R_1\to\>R_2


[112100012321120113001]\begin{bmatrix} 1&1 & 2| 1&0&0\\ 0&\frac{-1}{2}& \frac{-3}{2}\>\>\>|-1&\frac{1}{2}&0\\ -1&1&3|0&0&1 \end{bmatrix}



R3R1R3R_3-R_1\to\>R_3


[112200012321120001101]\begin{bmatrix} 1&1&2\>\>\>| 2&0&0\\ 0&\frac{-1}{2} &\>\> \frac{-3}{2}\>\>\>\>\>|-1&\frac{1}{2}&0\\ 0&0&\>\>1\>\>\>\>\>\>\>\>|-1&0&1 \end{bmatrix}



2R23R3R2-2R_2-3R_3\to\>R_2


[112100010513001101]\begin{bmatrix} 1&1&2 |1&0&0 \\ 0&1&0 |5&-1&-3\\ 0&0&\>1\>\>\>\>\>|-1&0&1 \end{bmatrix}



R12R3R1R_1-2R_3\to\>R_1


[110302010513001101]\begin{bmatrix} 1&1 & 0|3&0&-2\\ 0&1&0| 5&-1&-3\\ 0&0&\>\>-1|-1&0&1 \end{bmatrix}



R1R2R1R_1-R_2\to\>R_1


[100211010513001101]\begin{bmatrix} 1&0&\>\>\>\>0 | \quad-2&1&1\\ 0&1&0|\quad 5&-1&-3\\ 0&0&1|-1&0&1 \end{bmatrix}



D-1 is the right side of Augmented matrix


D1=(211513101)D^{-1}=\begin{pmatrix} -2&1&1 \\ 5&-1&-3 \\ -1&0&1 \end{pmatrix}




Solving for the largest eigenvalue of D-1


1st Iteration


D1x0=(211513101)(100)=(251)D^{-1}x_0=\begin{pmatrix} -2&1 & 1\\ 5&-1 & -3\\ -1&0&1 \end{pmatrix}\begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix}=\begin{pmatrix} -2 \\ 5 \\ -1 \end{pmatrix}




x1=15(251)=(0.410.2)x_1=\frac{1}{5}\begin{pmatrix} -2 \\ 5 \\ -1 \end{pmatrix}=\begin{pmatrix} -0.4 \\ 1\\ -0.2 \end{pmatrix}



2nd Iteration


D1x1=(211513101)(0.410.2)=(1.62.40.2)D^{-1}x_1=\begin{pmatrix} -2&1 & 1\\ 5&-1 & -3\\ -1&0&1 \end{pmatrix}\begin{pmatrix} -0.4 \\ 1\\ -0.2 \end{pmatrix}=\begin{pmatrix} 1.6 \\ -2.4 \\ 0.2 \end{pmatrix}



x2=12.4(1.62.40.2)=(0.6710.08)x_2=\frac{1}{2.4}\begin{pmatrix} 1.6 \\ -2.4 \\ 0.2 \end{pmatrix}=\begin{pmatrix} 0.67 \\ -1 \\ 0.08 \end{pmatrix}



3rd Iteration


Dx2=(211513101)(0.6710.08)=(2.254.080.58)D^-x_2=\begin{pmatrix} -2&1&1 \\ 5&-1 & -3\\ -1&0&1 \end{pmatrix}\begin{pmatrix} 0.67 \\ -1 \\ 0.08 \end{pmatrix}=\begin{pmatrix} -2.25 \\ 4.08\\ -0.58 \end{pmatrix}



x3=14.08(2.254.080.58)=(0.5510.14)x_3=\frac{1}{4.08}\begin{pmatrix} -2.25 \\ 4.08 \\ -0.58 \end{pmatrix}=\begin{pmatrix} -0.55 \\ 1 \\ -0.14 \end{pmatrix}




4th Iteration


Dx3=(211513101)(0.5510.14)=(1.963.330.41)D^-x_3=\begin{pmatrix} -2&1 & 1 \\ 5&-1& -3\\ -1&0&1 \end{pmatrix}\begin{pmatrix} -0.55 \\ 1 \\ -0.14 \end{pmatrix}=\begin{pmatrix} 1.96 \\ -3.33 \\ 0.41 \end{pmatrix}




x4=13.33(1.963.330.41)=(0.5910.12)x_4=\frac{1}{3.33}\begin{pmatrix} 1.96 \\ -3.33\\ 0.41 \end{pmatrix}=\begin{pmatrix} 0.59 \\ -1 \\ 0.12 \end{pmatrix}



5th Iteration


Dx4=(211513101)(0.5910.12)=(2.063.580.47)D^-x_4=\begin{pmatrix} -2&1 & 1 \\ 5&-1& -3\\ -1&0&1 \end{pmatrix}\begin{pmatrix} 0.59 \\ -1\\ 0.12 \end{pmatrix}=\begin{pmatrix} -2.06 \\ 3.58 \\ -0.47 \end{pmatrix}




x5=13.58(2.063.580.47)=(0.5710.13)x_5=\frac{1}{3.58}\begin{pmatrix} -2.06 \\ 3.58 \\ -0.47 \end{pmatrix}=\begin{pmatrix} -0.57 \\ 1 \\ -0.13 \end{pmatrix}



6th Iteration


D1x5=(211513101)(0.5710.13)=(2.023.480.44)D^{-1}x_5=\begin{pmatrix} -2&1 & 1 \\ 5&-1& -3\\ -1&0&1 \end{pmatrix}\begin{pmatrix} -0.57 \\ 1 \\ -0.13 \end{pmatrix}=\begin{pmatrix} 2.02 \\ -3.48 \\ 0.44 \end{pmatrix}



x6=13.48(2.023.480.44)=(0.5810.13)x_6=\frac{1}{3.48}\begin{pmatrix} 2.02\\ -3.48 \\ 0.44 \end{pmatrix}=\begin{pmatrix} 0.58 \\ -1 \\ 0.13 \end{pmatrix}



8th Iteration


D1x7=(211513101)(0.5810.13)=(2.033.50.45)D^{-1}x_7=\begin{pmatrix} -2&1 & 1 \\ 5&-1 & -3\\ -1&0&1 \end{pmatrix}\begin{pmatrix} -0.58 \\ 1\\ -0.13 \end{pmatrix}=\begin{pmatrix} 2.03 \\ -3.5 \\ 0.45 \end{pmatrix}




x8=13.5(2.033.50.45)=(0.5810.13)x_8=\frac{1}{3.5}\begin{pmatrix} 2.03 \\ -3.5 \\ 0.45 \end{pmatrix}=\begin{pmatrix} 0.58 \\ -1 \\ 0.13 \end{pmatrix}



9th Iteration


D1x8=(211513101)(0.5810.13)=(2.033.510.45)D^{-1}x_8=\begin{pmatrix} -2&1& 1\\ 5&-1 & -3\\ -1&0&1 \end{pmatrix}\begin{pmatrix} 0.58 \\ -1 \\ 0.13 \end{pmatrix}=\begin{pmatrix} -2.03 \\ 3.51 \\ -0.45 \end{pmatrix}



x9=13.51(2.033.510.45)=(0.5810.13)x_9=\frac{1}{3.51}\begin{pmatrix} -2.03 \\ 3.51 \\ -0.45 \end{pmatrix}=\begin{pmatrix} 0.58 \\ 1 \\ -0.13 \end{pmatrix}



10th Iteration


Dx9=(211513101)(0.5810.13)=(2.033.510.45)^-x_9=\begin{pmatrix} -2&1 & 1\\ 5&-1 & -3\\ -1&0&1 \end{pmatrix}\begin{pmatrix} 0.58 \\ -1 \\ 0.13 \end{pmatrix}=\begin{pmatrix} 2.03 \\ -3.51\\ 0.45 \end{pmatrix}



x10=13.51(2.033.510.45)=(0.5810.13)x_{10}=\frac{1}{3.51}\begin{pmatrix} 2.03 \\ -3.51 \\ 0.45 \end{pmatrix}=\begin{pmatrix} 0.58 \\ -1\\ 0.13 \end{pmatrix}




The dominant eigenvalue of D- 3.513.51


The magnitude of the smallest eigenvalues of D=13.51=0.2849D=\frac{1}{3.51}=0.2849

and the corresponding eigenvector of


D=(0.5810.13)D=\begin{pmatrix} 0.58\\ -1 \\ 0.13 \end{pmatrix}





Need a fast expert's response?

Submit order

and get a quick answer at the best price

for any assignment or question with DETAILED EXPLANATIONS!

Comments

No comments. Be the first!
LATEST TUTORIALS
APPROVED BY CLIENTS