1st Iteration;
D x 0 = Dx_0= D x 0 = ( 1 1 2 2 1 1 1 1 3 ) ( 1 0 0 ) = ( 1 2 1 ) \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} ⎝ ⎛ 1 2 1 1 1 1 2 1 3 ⎠ ⎞ ⎝ ⎛ 1 0 0 ⎠ ⎞ = ⎝ ⎛ 1 2 1 ⎠ ⎞
x 1 = 1 2 ( 1 2 1 ) = ( 0.5 1 0.5 ) x_1=\frac{1}{2}\begin{pmatrix}
1 \\
2\\
1
\end{pmatrix}=\begin{pmatrix}
0.5\\
1\\
0.5
\end{pmatrix} x 1 = 2 1 ⎝ ⎛ 1 2 1 ⎠ ⎞ = ⎝ ⎛ 0.5 1 0.5 ⎠ ⎞
2nd Iteration
D x 1 = ( 1 1 2 2 1 1 1 1 3 ) ( 0.5 1 0.5 ) = ( 2.5 2.5 3 ) 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} D x 1 = ⎝ ⎛ 1 2 1 1 1 1 2 1 3 ⎠ ⎞ ⎝ ⎛ 0.5 1 0.5 ⎠ ⎞ = ⎝ ⎛ 2.5 2.5 3 ⎠ ⎞
x 2 = 1 3 ( 2.5 2.5 3 ) = ( 0.83 0.83 1 ) x_2=\frac{1}{3}\begin{pmatrix}
2.5 \\
2.5 \\
3
\end{pmatrix}=\begin{pmatrix}
0.83 \\
0.83\\
1
\end{pmatrix} x 2 = 3 1 ⎝ ⎛ 2.5 2.5 3 ⎠ ⎞ = ⎝ ⎛ 0.83 0.83 1 ⎠ ⎞
3rd Iteration
D x 2 = ( 1 1 2 2 1 1 1 1 3 ) ( 0.83 0.83 1 ) = ( 3.67 3.5 4.67 ) 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} D x 2 = ⎝ ⎛ 1 2 1 1 1 1 2 1 3 ⎠ ⎞ ⎝ ⎛ 0.83 0.83 1 ⎠ ⎞ = ⎝ ⎛ 3.67 3.5 4.67 ⎠ ⎞
x 3 = 1 4.67 ( 3.67 3.5 4.67 ) = ( 0.79 0.75 1 ) 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} x 3 = 4.67 1 ⎝ ⎛ 3.67 3.5 4.67 ⎠ ⎞ = ⎝ ⎛ 0.79 0.75 1 ⎠ ⎞
4th Iteration
D x 3 = ( 1 1 2 2 1 1 1 1 3 ) ( 0.79 0.75 1 ) = ( 3.54 3.32 4.54 ) 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} D x 3 = ⎝ ⎛ 1 2 1 1 1 1 2 1 3 ⎠ ⎞ ⎝ ⎛ 0.79 0.75 1 ⎠ ⎞ = ⎝ ⎛ 3.54 3.32 4.54 ⎠ ⎞
x 4 = 1 4.54 ( 3.54 3.32 4.54 ) = ( 0.78 0.73 1 ) 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} x 4 = 4.54 1 ⎝ ⎛ 3.54 3.32 4.54 ⎠ ⎞ = ⎝ ⎛ 0.78 0.73 1 ⎠ ⎞
5th Iteration
D x 4 = ( 1 1 2 2 1 1 1 1 3 ) ( 0.78 0.73 1 ) = ( 3.51 3.29 4.51 ) 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} D x 4 = ⎝ ⎛ 1 2 1 1 1 1 2 1 3 ⎠ ⎞ ⎝ ⎛ 0.78 0.73 1 ⎠ ⎞ = ⎝ ⎛ 3.51 3.29 4.51 ⎠ ⎞
x 5 = 1 4.51 ( 3.51 3.29 4.51 ) = ( 0.78 0.73 1 ) 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} x 5 = 4.51 1 ⎝ ⎛ 3.51 3.29 4.51 ⎠ ⎞ = ⎝ ⎛ 0.78 0.73 1 ⎠ ⎞
6th Iteration
D x 5 = ( 1 1 2 2 1 1 1 1 3 ) ( 0.78 0.73 1 ) = ( 3.51 3.29 4.51 ) 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} D x 5 = ⎝ ⎛ 1 2 1 1 1 1 2 1 3 ⎠ ⎞ ⎝ ⎛ 0.78 0.73 1 ⎠ ⎞ = ⎝ ⎛ 3.51 3.29 4.51 ⎠ ⎞
x 6 = 1 4.51 ( 3.51 3.29 4.51 ) = ( 0.78 0.73 1 ) 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} x 6 = 4.51 1 ⎝ ⎛ 3.51 3.29 4.51 ⎠ ⎞ = ⎝ ⎛ 0.78 0.73 1 ⎠ ⎞
∴ \therefore ∴ The dominant eigenvalue λ = 4.51 \lambda=4.51 λ = 4.51
and the dominant eigenvector is
( 0.78 0.73 1 ) \begin{pmatrix}
0.78 \\
0.73 \\
1
\end{pmatrix} ⎝ ⎛ 0.78 0.73 1 ⎠ ⎞
D -1 by Gaus - Jordan Elimination
Augmenting D D D with a 3 × 3 3×3 3 × 3 identity matrix
[ 1 1 2 ∣ 1 0 0 2 1 1 ∣ 0 1 0 1 1 3 ∣ 0 0 1 ] \begin{bmatrix}
1&1&2| 1&0&0 \\
2&1&1|0&1&0 \\
1&1&3|0&0&1
\end{bmatrix} ⎣ ⎡ 1 2 1 1 1 1 2∣1 1∣0 3∣0 0 1 0 0 0 1 ⎦ ⎤
1 2 R 2 − R 1 → R 2 \frac{1}{2}R_2-R_1\to\>R_2 2 1 R 2 − R 1 → R 2
[ 1 1 2 ∣ 1 0 0 0 − 1 2 − 3 2 ∣ − 1 1 2 0 − 1 1 3 ∣ 0 0 1 ] \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} ⎣ ⎡ 1 0 − 1 1 2 − 1 1 2∣1 2 − 3 ∣ − 1 3∣0 0 2 1 0 0 0 1 ⎦ ⎤
R 3 − R 1 → R 3 R_3-R_1\to\>R_3 R 3 − R 1 → R 3
[ 1 1 2 ∣ 2 0 0 0 − 1 2 − 3 2 ∣ − 1 1 2 0 0 0 1 ∣ − 1 0 1 ] \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} ⎣ ⎡ 1 0 0 1 2 − 1 0 2 ∣2 2 − 3 ∣ − 1 1 ∣ − 1 0 2 1 0 0 0 1 ⎦ ⎤
− 2 R 2 − 3 R 3 → R 2 -2R_2-3R_3\to\>R_2 − 2 R 2 − 3 R 3 → R 2
[ 1 1 2 ∣ 1 0 0 0 1 0 ∣ 5 − 1 − 3 0 0 1 ∣ − 1 0 1 ] \begin{bmatrix}
1&1&2 |1&0&0 \\
0&1&0 |5&-1&-3\\
0&0&\>1\>\>\>\>\>|-1&0&1
\end{bmatrix} ⎣ ⎡ 1 0 0 1 1 0 2∣1 0∣5 1 ∣ − 1 0 − 1 0 0 − 3 1 ⎦ ⎤
R 1 − 2 R 3 → R 1 R_1-2R_3\to\>R_1 R 1 − 2 R 3 → R 1
[ 1 1 0 ∣ 3 0 − 2 0 1 0 ∣ 5 − 1 − 3 0 0 − 1 ∣ − 1 0 1 ] \begin{bmatrix}
1&1 & 0|3&0&-2\\
0&1&0| 5&-1&-3\\
0&0&\>\>-1|-1&0&1
\end{bmatrix} ⎣ ⎡ 1 0 0 1 1 0 0∣3 0∣5 − 1∣ − 1 0 − 1 0 − 2 − 3 1 ⎦ ⎤
R 1 − R 2 → R 1 R_1-R_2\to\>R_1 R 1 − R 2 → R 1
[ 1 0 0 ∣ − 2 1 1 0 1 0 ∣ 5 − 1 − 3 0 0 1 ∣ − 1 0 1 ] \begin{bmatrix}
1&0&\>\>\>\>0 | \quad-2&1&1\\
0&1&0|\quad 5&-1&-3\\
0&0&1|-1&0&1
\end{bmatrix} ⎣ ⎡ 1 0 0 0 1 0 0∣ − 2 0∣ 5 1∣ − 1 1 − 1 0 1 − 3 1 ⎦ ⎤
D-1 is the right side of Augmented matrix
D − 1 = ( − 2 1 1 5 − 1 − 3 − 1 0 1 ) D^{-1}=\begin{pmatrix}
-2&1&1 \\
5&-1&-3 \\
-1&0&1
\end{pmatrix} D − 1 = ⎝ ⎛ − 2 5 − 1 1 − 1 0 1 − 3 1 ⎠ ⎞
Solving for the largest eigenvalue of D-1
1st Iteration
D − 1 x 0 = ( − 2 1 1 5 − 1 − 3 − 1 0 1 ) ( 1 0 0 ) = ( − 2 5 − 1 ) 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} D − 1 x 0 = ⎝ ⎛ − 2 5 − 1 1 − 1 0 1 − 3 1 ⎠ ⎞ ⎝ ⎛ 1 0 0 ⎠ ⎞ = ⎝ ⎛ − 2 5 − 1 ⎠ ⎞
x 1 = 1 5 ( − 2 5 − 1 ) = ( − 0.4 1 − 0.2 ) x_1=\frac{1}{5}\begin{pmatrix}
-2 \\
5 \\
-1
\end{pmatrix}=\begin{pmatrix}
-0.4 \\
1\\
-0.2
\end{pmatrix} x 1 = 5 1 ⎝ ⎛ − 2 5 − 1 ⎠ ⎞ = ⎝ ⎛ − 0.4 1 − 0.2 ⎠ ⎞
2nd Iteration
D − 1 x 1 = ( − 2 1 1 5 − 1 − 3 − 1 0 1 ) ( − 0.4 1 − 0.2 ) = ( 1.6 − 2.4 0.2 ) 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} D − 1 x 1 = ⎝ ⎛ − 2 5 − 1 1 − 1 0 1 − 3 1 ⎠ ⎞ ⎝ ⎛ − 0.4 1 − 0.2 ⎠ ⎞ = ⎝ ⎛ 1.6 − 2.4 0.2 ⎠ ⎞
x 2 = 1 2.4 ( 1.6 − 2.4 0.2 ) = ( 0.67 − 1 0.08 ) 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} x 2 = 2.4 1 ⎝ ⎛ 1.6 − 2.4 0.2 ⎠ ⎞ = ⎝ ⎛ 0.67 − 1 0.08 ⎠ ⎞
3rd Iteration
D − x 2 = ( − 2 1 1 5 − 1 − 3 − 1 0 1 ) ( 0.67 − 1 0.08 ) = ( − 2.25 4.08 − 0.58 ) 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} D − x 2 = ⎝ ⎛ − 2 5 − 1 1 − 1 0 1 − 3 1 ⎠ ⎞ ⎝ ⎛ 0.67 − 1 0.08 ⎠ ⎞ = ⎝ ⎛ − 2.25 4.08 − 0.58 ⎠ ⎞
x 3 = 1 4.08 ( − 2.25 4.08 − 0.58 ) = ( − 0.55 1 − 0.14 ) 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} x 3 = 4.08 1 ⎝ ⎛ − 2.25 4.08 − 0.58 ⎠ ⎞ = ⎝ ⎛ − 0.55 1 − 0.14 ⎠ ⎞
4th Iteration
D − x 3 = ( − 2 1 1 5 − 1 − 3 − 1 0 1 ) ( − 0.55 1 − 0.14 ) = ( 1.96 − 3.33 0.41 ) 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} D − x 3 = ⎝ ⎛ − 2 5 − 1 1 − 1 0 1 − 3 1 ⎠ ⎞ ⎝ ⎛ − 0.55 1 − 0.14 ⎠ ⎞ = ⎝ ⎛ 1.96 − 3.33 0.41 ⎠ ⎞
x 4 = 1 3.33 ( 1.96 − 3.33 0.41 ) = ( 0.59 − 1 0.12 ) 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} x 4 = 3.33 1 ⎝ ⎛ 1.96 − 3.33 0.41 ⎠ ⎞ = ⎝ ⎛ 0.59 − 1 0.12 ⎠ ⎞
5th Iteration
D − x 4 = ( − 2 1 1 5 − 1 − 3 − 1 0 1 ) ( 0.59 − 1 0.12 ) = ( − 2.06 3.58 − 0.47 ) 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} D − x 4 = ⎝ ⎛ − 2 5 − 1 1 − 1 0 1 − 3 1 ⎠ ⎞ ⎝ ⎛ 0.59 − 1 0.12 ⎠ ⎞ = ⎝ ⎛ − 2.06 3.58 − 0.47 ⎠ ⎞
x 5 = 1 3.58 ( − 2.06 3.58 − 0.47 ) = ( − 0.57 1 − 0.13 ) 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} x 5 = 3.58 1 ⎝ ⎛ − 2.06 3.58 − 0.47 ⎠ ⎞ = ⎝ ⎛ − 0.57 1 − 0.13 ⎠ ⎞
6th Iteration
D − 1 x 5 = ( − 2 1 1 5 − 1 − 3 − 1 0 1 ) ( − 0.57 1 − 0.13 ) = ( 2.02 − 3.48 0.44 ) 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} D − 1 x 5 = ⎝ ⎛ − 2 5 − 1 1 − 1 0 1 − 3 1 ⎠ ⎞ ⎝ ⎛ − 0.57 1 − 0.13 ⎠ ⎞ = ⎝ ⎛ 2.02 − 3.48 0.44 ⎠ ⎞
x 6 = 1 3.48 ( 2.02 − 3.48 0.44 ) = ( 0.58 − 1 0.13 ) 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} x 6 = 3.48 1 ⎝ ⎛ 2.02 − 3.48 0.44 ⎠ ⎞ = ⎝ ⎛ 0.58 − 1 0.13 ⎠ ⎞
8th Iteration
D − 1 x 7 = ( − 2 1 1 5 − 1 − 3 − 1 0 1 ) ( − 0.58 1 − 0.13 ) = ( 2.03 − 3.5 0.45 ) 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} D − 1 x 7 = ⎝ ⎛ − 2 5 − 1 1 − 1 0 1 − 3 1 ⎠ ⎞ ⎝ ⎛ − 0.58 1 − 0.13 ⎠ ⎞ = ⎝ ⎛ 2.03 − 3.5 0.45 ⎠ ⎞
x 8 = 1 3.5 ( 2.03 − 3.5 0.45 ) = ( 0.58 − 1 0.13 ) 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} x 8 = 3.5 1 ⎝ ⎛ 2.03 − 3.5 0.45 ⎠ ⎞ = ⎝ ⎛ 0.58 − 1 0.13 ⎠ ⎞
9th Iteration
D − 1 x 8 = ( − 2 1 1 5 − 1 − 3 − 1 0 1 ) ( 0.58 − 1 0.13 ) = ( − 2.03 3.51 − 0.45 ) 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} D − 1 x 8 = ⎝ ⎛ − 2 5 − 1 1 − 1 0 1 − 3 1 ⎠ ⎞ ⎝ ⎛ 0.58 − 1 0.13 ⎠ ⎞ = ⎝ ⎛ − 2.03 3.51 − 0.45 ⎠ ⎞
x 9 = 1 3.51 ( − 2.03 3.51 − 0.45 ) = ( 0.58 1 − 0.13 ) 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} x 9 = 3.51 1 ⎝ ⎛ − 2.03 3.51 − 0.45 ⎠ ⎞ = ⎝ ⎛ 0.58 1 − 0.13 ⎠ ⎞
10th Iteration
D− x 9 = ( − 2 1 1 5 − 1 − 3 − 1 0 1 ) ( 0.58 − 1 0.13 ) = ( 2.03 − 3.51 0.45 ) ^-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} − x 9 = ⎝ ⎛ − 2 5 − 1 1 − 1 0 1 − 3 1 ⎠ ⎞ ⎝ ⎛ 0.58 − 1 0.13 ⎠ ⎞ = ⎝ ⎛ 2.03 − 3.51 0.45 ⎠ ⎞
x 10 = 1 3.51 ( 2.03 − 3.51 0.45 ) = ( 0.58 − 1 0.13 ) 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} x 10 = 3.51 1 ⎝ ⎛ 2.03 − 3.51 0.45 ⎠ ⎞ = ⎝ ⎛ 0.58 − 1 0.13 ⎠ ⎞
The dominant eigenvalue of D- 3.51 3.51 3.51
The magnitude of the smallest eigenvalues of D = 1 3.51 = 0.2849 D=\frac{1}{3.51}=0.2849 D = 3.51 1 = 0.2849
and the corresponding eigenvector of
D = ( 0.58 − 1 0.13 ) D=\begin{pmatrix}
0.58\\
-1 \\
0.13
\end{pmatrix} D = ⎝ ⎛ 0.58 − 1 0.13 ⎠ ⎞
Comments