Find Moore-Penrose Pseudoinverse ...
`[[1,-2,3],[5,8,-1],[2,1,1],[-1,4,-3]]`

Solution:
Pseudoinverse of a matrix A is `A^(+) = (A^T*A)^(-1) * A^T`

1. Find `A'`

`A^T`

=

`1` `-2` `3` `5` `8` `-1` `2` `1` `1` `-1` `4` `-3` T

=

`1` `5` `2` `-1` `-2` `8` `1` `4` `3` `-1` `1` `-3`

2. Find `A'*A`

`A'×A`

=

`1` `5` `2` `-1` `-2` `8` `1` `4` `3` `-1` `1` `-3`

×

`1` `-2` `3` `5` `8` `-1` `2` `1` `1` `-1` `4` `-3`

=	`1×1+5×5+2×2-1×-1` `1×-2+5×8+2×1-1×4` `1×3+5×-1+2×1-1×-3` `-2×1+8×5+1×2+4×-1` `-2×-2+8×8+1×1+4×4` `-2×3+8×-1+1×1+4×-3` `3×1-1×5+1×2-3×-1` `3×-2-1×8+1×1-3×4` `3×3-1×-1+1×1-3×-3`

=	`1+25+4+1` `-2+40+2-4` `3-5+2+3` `-2+40+2-4` `4+64+1+16` `-6-8+1-12` `3-5+2+3` `-6-8+1-12` `9+1+1+9`

=	`31` `36` `3` `36` `85` `-25` `3` `-25` `20`

3. Find the inverse matrix `(A'*A)^(-1)`

`|A'*A|`

=

`31`   `36`   `3`   `36`   `85`   `-25`   `3`   `-25`   `20`

=

`31`   ×    `85`   `-25`   `-25`   `20`

`-36`   ×    `36`   `-25`   `3`   `20`

`+3`   ×    `36`   `85`   `3`   `-25`

`=31 xx (85 × 20 - (-25) × (-25)) -36 xx (36 × 20 - (-25) × 3) +3 xx (36 × (-25) - 85 × 3)`

`=31 xx (1700 -625) -36 xx (720 +75) +3 xx (-900 -255)`

`=31 xx (1075) -36 xx (795) +3 xx (-1155)`

`= 33325 -28620 -3465`

`=1240`

`Adj(A'*A)`

=

Adj `31` `36` `3` `36` `85` `-25` `3` `-25` `20`

=	+   `85`   `-25`   `-25`   `20`   -   `36`   `-25`   `3`   `20`   +   `36`   `85`   `3`   `-25`   -   `36`   `3`   `-25`   `20`   +   `31`   `3`   `3`   `20`   -   `31`   `36`   `3`   `-25`   +   `36`   `3`   `85`   `-25`   -   `31`   `3`   `36`   `-25`   +   `31`   `36`   `36`   `85`  T

=	`+(85 × 20 - (-25) × (-25))` `-(36 × 20 - (-25) × 3)` `+(36 × (-25) - 85 × 3)` `-(36 × 20 - 3 × (-25))` `+(31 × 20 - 3 × 3)` `-(31 × (-25) - 36 × 3)` `+(36 × (-25) - 3 × 85)` `-(31 × (-25) - 3 × 36)` `+(31 × 85 - 36 × 36)` T

=	`+(1700 -625)` `-(720 +75)` `+(-900 -255)` `-(720 +75)` `+(620 -9)` `-(-775 -108)` `+(-900 -255)` `-(-775 -108)` `+(2635 -1296)` T

=	`1075` `-795` `-1155` `-795` `611` `883` `-1155` `883` `1339` T

=	`1075` `-795` `-1155` `-795` `611` `883` `-1155` `883` `1339`

`"Now, "A'*A^(-1)=1/|A'*A| × Adj(A'*A)`

=

`1/(1240)` ×

`1075` `-795` `-1155` `-795` `611` `883` `-1155` `883` `1339`

=	`0.8669` `-0.6411` `-0.9315` `-0.6411` `0.4927` `0.7121` `-0.9315` `0.7121` `1.0798`

4. Find the inverse matrix `(A'*A)^(-1) * A'`

`((A*A')^-1)×A'`

=

`0.8669` `-0.6411` `-0.9315` `-0.6411` `0.4927` `0.7121` `-0.9315` `0.7121` `1.0798`

×

`1` `5` `2` `-1` `-2` `8` `1` `4` `3` `-1` `1` `-3`

=

`0.8669×1-0.6411×-2-0.9315×3` `0.8669×5-0.6411×8-0.9315×-1` `0.8669×2-0.6411×1-0.9315×1` `0.8669×-1-0.6411×4-0.9315×-3` `-0.6411×1+0.4927×-2+0.7121×3` `-0.6411×5+0.4927×8+0.7121×-1` `-0.6411×2+0.4927×1+0.7121×1` `-0.6411×-1+0.4927×4+0.7121×-3` `-0.9315×1+0.7121×-2+1.0798×3` `-0.9315×5+0.7121×8+1.0798×-1` `-0.9315×2+0.7121×1+1.0798×1` `-0.9315×-1+0.7121×4+1.0798×-3`

=	`0.8669+1.2823-2.7944` `4.3347-5.129+0.9315` `1.7339-0.6411-0.9315` `-0.8669-2.5645+2.7944` `-0.6411-0.9855+2.1363` `-3.2056+3.9419-0.7121` `-1.2823+0.4927+0.7121` `0.6411+1.971-2.1363` `-0.9315-1.4242+3.2395` `-4.6573+5.6968-1.0798` `-1.8629+0.7121+1.0798` `0.9315+2.8484-3.2395`

=	`-0.6452` `0.1371` `0.1613` `-0.6371` `0.5097` `0.0242` `-0.0774` `0.4758` `0.8839` `-0.0403` `-0.071` `0.5403`

`:.` Moore-Penrose pseudoinverse `A^(+)=`

`-0.6452` `0.1371` `0.1613` `-0.6371` `0.5097` `0.0242` `-0.0774` `0.4758` `0.8839` `-0.0403` `-0.071` `0.5403`

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