Find Inverse Power Method for finding dominant eigenvalue ...
`[[3,2],[1,4]]`
`x_0` = -1,1

Solution:

`A`

=

`3` `2` `1` `4`

1. Find `A^-1`

`|A|`

=

`3`   `2`   `1`   `4`

`=3 xx 4 - 2 xx 1`

`=12 -2`

`=10`

`Adj(A)`

=

Adj `3` `2` `1` `4`

=	`+(4)` `-(1)` `-(2)` `+(3)` T

=	`4` `-1` `-2` `3` T

=	`4` `-2` `-1` `3`

`"Now, "A^(-1)=1/|A| xx Adj(A)`

=

`1/(10) xx`

`4` `-2` `-1` `3`

=	`0.4` `-0.2` `-0.1` `0.3`

`A^-1=`

0.4 -0.2 -0.1 0.3

`x_0=`

-1 1

`1^(st)` iteration :

Multiply the matrix by the vector

`A^-1 x_0 =`

0.4 -0.2 -0.1 0.3

-1 1

`=`

-0.6 0.4

Normalize the resulting vector
To normalize, divide each element of vector by its largest absolute value, which is `0.6`

`x_1=`

`1/0.6`

-0.6 0.4

`=`

-1 0.6667

`2^(nd)` iteration :

Repeat the multiplication

`A^-1 x_1 =`

0.4 -0.2 -0.1 0.3

-1 0.6667

`=`

-0.5333 0.3

Normalize again
The largest absolute value is `0.5333`

`x_2=`

`1/0.5333`

-0.5333 0.3

`=`

-1 0.5625

`3^(rd)` iteration :

Repeat the multiplication

`A^-1 x_2 =`

0.4 -0.2 -0.1 0.3

-1 0.5625

`=`

-0.5125 0.2688

Normalize again
The largest absolute value is `0.5125`

`x_3=`

`1/0.5125`

-0.5125 0.2688

`=`

-1 0.5244

`4^(th)` iteration :

Repeat the multiplication

`A^-1 x_3 =`

0.4 -0.2 -0.1 0.3

-1 0.5244

`=`

-0.5049 0.2573

Normalize again
The largest absolute value is `0.5049`

`x_4=`

`1/0.5049`

-0.5049 0.2573

`=`

-1 0.5097

`5^(th)` iteration :

Repeat the multiplication

`A^-1 x_4 =`

0.4 -0.2 -0.1 0.3

-1 0.5097

`=`

-0.5019 0.2529

Normalize again
The largest absolute value is `0.5019`

`x_5=`

`1/0.5019`

-0.5019 0.2529

`=`

-1 0.5038

`6^(th)` iteration :

Repeat the multiplication

`A^-1 x_5 =`

0.4 -0.2 -0.1 0.3

-1 0.5038

`=`

-0.5008 0.2512

Normalize again
The largest absolute value is `0.5008`

`x_6=`

`1/0.5008`

-0.5008 0.2512

`=`

-1 0.5015

`7^(th)` iteration :

Repeat the multiplication

`A^-1 x_6 =`

0.4 -0.2 -0.1 0.3

-1 0.5015

`=`

-0.5003 0.2505

Normalize again
The largest absolute value is `0.5003`

`x_7=`

`1/0.5003`

-0.5003 0.2505

`=`

-1 0.5006

`8^(th)` iteration :

Repeat the multiplication

`A^-1 x_7 =`

0.4 -0.2 -0.1 0.3

-1 0.5006

`=`

-0.5001 0.2502

Normalize again
The largest absolute value is `0.5001`

`x_8=`

`1/0.5001`

-0.5001 0.2502

`=`

-1 0.5002

`9^(th)` iteration :

Repeat the multiplication

`A^-1 x_8 =`

0.4 -0.2 -0.1 0.3

-1 0.5002

`=`

-0.5 0.2501

Normalize again
The largest absolute value is `0.5`

`x_9=`

`1/0.5`

-0.5 0.2501

`=`

-1 0.5001

`10^(th)` iteration :

Repeat the multiplication

`A^-1 x_9 =`

0.4 -0.2 -0.1 0.3

-1 0.5001

`=`

-0.5 0.25

Normalize again
The largest absolute value is `0.5`

`x_10=`

`1/0.5`

-0.5 0.25

`=`

-1 0.5

`:.` The dominant eigenvalue `lamda=0.5~=0.5`

and the dominant eigenvector is :

`=`

-1 0.5

`~=`

-1 0.5

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