1. Find LU decomposition using Crout's method of Matrix ...
`[[8,-6,2],[-6,7,-4],[2,-4,3]]`

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

`8` `-6` `2` `-6` `7` `-4` `2` `-4` `3`

=

`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`

`8` `-6` `2` `-6` `7` `-4` `2` `-4` `3`

=

`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)=8`

`l_(11)u_(12)=-6=>8xxu_(12)=-6=>u_(12)=-0.75`

`l_(11)u_(13)=2=>8xxu_(13)=2=>u_(13)=0.25`



`l_(21)=-6`

`l_(21)u_(12) + l_(22)=7=>(-6)xx(-0.75) + l_(22)=7=>l_(22)=2.5`

`l_(21)u_(13) + l_(22)u_(23)=-4=>(-6)xx0.25 + 2.5xxu_(23)=-4=>u_(23)=-1`



`l_(31)=2`

`l_(31)u_(12) + l_(32)=-4=>2xx(-0.75) + l_(32)=-4=>l_(32)=-2.5`

`l_(31)u_(13) + l_(32)u_(23) + l_(33)=3=>2xx0.25 + (-2.5)xx(-1) + l_(33)=3=>l_(33)=0`



Now checking `A=LU` ?

`LU`

=

`8` `0` `0` `-6` `2.5` `0` `2` `-2.5` `0`

`xx`

`1` `-0.75` `0.25` `0` `1` `-1` `0` `0` `1`

=

`8` `-6` `2` `-6` `7` `-4` `2` `-4` `3`

And `A`

=

`8` `-6` `2` `-6` `7` `-4` `2` `-4` `3`

Solution is possible.

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