1. Find Inverse Power Method for finding dominant eigenvalue ...
`[[2,3],[5,4]]`
`x_0` =

Solution:

`A`

=

`2` `3` `5` `4`

1. Find `A^-1`

`|A|`

=

`2`   `3`   `5`   `4`

`=2 xx 4 - 3 xx 5`

`=8 -15`

`=-7`

`Adj(A)`

=

Adj `2` `3` `5` `4`

=	`+(4)` `-(5)` `-(3)` `+(2)` T

=	`4` `-5` `-3` `2` T

=	`4` `-3` `-5` `2`

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

=

`1/(-7) xx`

`4` `-3` `-5` `2`

=	`-0.5714` `0.4286` `0.7143` `-0.2857`

`A^-1=`

-0.5714 0.4286 0.7143 -0.2857

`x_0=`

1 1

`1^(st)` iteration :

Multiply the matrix by the vector

`A^-1 x_0 =`

-0.5714 0.4286 0.7143 -0.2857

1 1

`=`

-0.1429 0.4286

Normalize the resulting vector
To normalize, divide each element of vector by its largest absolute value, which is `0.4286`

`x_1=`

`1/0.4286`

-0.1429 0.4286

`=`

-0.3333 1

`2^(nd)` iteration :

Repeat the multiplication

`A^-1 x_1 =`

-0.5714 0.4286 0.7143 -0.2857

-0.3333 1

`=`

0.619 -0.5238

Normalize again
The largest absolute value is `0.619`

`x_2=`

`1/0.619`

0.619 -0.5238

`=`

1 -0.8462

`3^(rd)` iteration :

Repeat the multiplication

`A^-1 x_2 =`

-0.5714 0.4286 0.7143 -0.2857

1 -0.8462

`=`

-0.9341 0.956

Normalize again
The largest absolute value is `0.956`

`x_3=`

`1/0.956`

-0.9341 0.956

`=`

-0.977 1

`4^(th)` iteration :

Repeat the multiplication

`A^-1 x_3 =`

-0.5714 0.4286 0.7143 -0.2857

-0.977 1

`=`

0.9869 -0.9836

Normalize again
The largest absolute value is `0.9869`

`x_4=`

`1/0.9869`

0.9869 -0.9836

`=`

1 -0.9967

`5^(th)` iteration :

Repeat the multiplication

`A^-1 x_4 =`

-0.5714 0.4286 0.7143 -0.2857

1 -0.9967

`=`

-0.9986 0.999

Normalize again
The largest absolute value is `0.999`

`x_5=`

`1/0.999`

-0.9986 0.999

`=`

-0.9995 1

`6^(th)` iteration :

Repeat the multiplication

`A^-1 x_5 =`

-0.5714 0.4286 0.7143 -0.2857

-0.9995 1

`=`

0.9997 -0.9997

Normalize again
The largest absolute value is `0.9997`

`x_6=`

`1/0.9997`

0.9997 -0.9997

`=`

1 -0.9999

`7^(th)` iteration :

Repeat the multiplication

`A^-1 x_6 =`

-0.5714 0.4286 0.7143 -0.2857

1 -0.9999

`=`

-1 1

Normalize again
The largest absolute value is `1`

`x_7=`

`1/1`

-1 1

`=`

-1 1

`:.` The dominant eigenvalue `lamda=1~=1`

and the dominant eigenvector is :

`=`

-1 1

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