Find LU decomposition using Crout's method of Matrix ...
`[[1,1,1],[-1,-3,-3],[2,4,4]]`

Solution:
Crout's method for LU decomposition
Let `A=LU`

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

=

`l_(11)` `0` `0` `l_(21)` `l_(22)` `0` `l_(31)` `l_(32)` `l_(33)`

`xx`

`1` `u_(12)` `u_(13)` `0` `1` `u_(23)` `0` `0` `1`

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

=

`l_(11)` `l_(11)u_(12)` `l_(11)u_(13)` `l_(21)` `l_(21)u_(12) + l_(22)` `l_(21)u_(13) + l_(22)u_(23)` `l_(31)` `l_(31)u_(12) + l_(32)` `l_(31)u_(13) + l_(32)u_(23) + l_(33)`

This implies
`l_(11)=1`

`l_(11)u_(12)=1=>1xxu_(12)=1=>u_(12)=1`

`l_(11)u_(13)=1=>1xxu_(13)=1=>u_(13)=1`



`l_(21)=-1`

`l_(21)u_(12) + l_(22)=-3=>(-1)xx1 + l_(22)=-3=>l_(22)=-2`

`l_(21)u_(13) + l_(22)u_(23)=-3=>(-1)xx1 + (-2)xxu_(23)=-3=>u_(23)=1`



`l_(31)=2`

`l_(31)u_(12) + l_(32)=4=>2xx1 + l_(32)=4=>l_(32)=2`

`l_(31)u_(13) + l_(32)u_(23) + l_(33)=4=>2xx1 + 2xx1 + l_(33)=4=>l_(33)=0`



Now checking `A=LU` ?

`LU`

=

`1` `0` `0` `-1` `-2` `0` `2` `2` `0`

`xx`

`1` `1` `1` `0` `1` `1` `0` `0` `1`

=

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

And `A`

=

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

Solution is possible.

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