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Home > Matrix & Vector calculators > Moore-Penrose Pseudoinverse of a Matrix example
18. Moore-Penrose Pseudoinverse of a Matrix example ( Enter your problem )
( Enter your problem )
  1. Example `[[4,0],[3,-5]]` `("Formula " A^(+)=V Sigma^(+) U^T)`
  2. Example `[[1,0,1,0],[0,1,0,1]]` `("Formula " A^(+)=V Sigma^(+) U^T)`
  3. Example `[[4,0],[3,-5]]` `("Formula " A^(+)=A^T * (A*A^T)^(-1))`
  4. Example `[[1,0,1,0],[0,1,0,1]]` `("Formula " A^(+)=A^T * (A*A^T)^(-1))`
  5. Example `[[1,-2,3],[5,8,-1],[2,1,1],[-1,4,-3]]` `("Formula " A^(+)=(A^T*A)^(-1) * A^T)`
 		Other related methods
  1. Transforming matrix to Row Echelon Form
  2. Transforming matrix to Reduced Row Echelon Form
  3. Rank of matrix
  4. Characteristic polynomial of matrix
  5. Eigenvalues
  6. Eigenvectors (Eigenspace)
  7. Triangular Matrix
  8. LU decomposition using Gauss Elimination method of matrix
  9. LU decomposition using Doolittle's method of matrix
  10. LU decomposition using Crout's method of matrix
  11. Diagonal Matrix
  12. Cholesky Decomposition
  13. QR Decomposition (Gram Schmidt Method)
  14. QR Decomposition (Householder Method)
  15. LQ Decomposition
  16. Pivots
  17. Singular Value Decomposition (SVD)
  18. Moore-Penrose Pseudoinverse
  19. Power Method for dominant eigenvalue
  20. Determinant by gaussian elimination
  21. Expanding determinant along row / column
  22. Determinants using montante (bareiss algorithm)
  23. Leibniz formula for determinant
  24. determinants using Sarrus Rule
  25. determinants using properties of determinants
  26. Row Space
  27. Column Space
  28. Null Space
4. Example `[[1,0,1,0],[0,1,0,1]]` `("Formula " A^(+)=A^T * (A*A^T)^(-1))`
(Previous example)		19. Power Method for dominant eigenvalue
(Next method)

5. Example `[[1,-2,3],[5,8,-1],[2,1,1],[-1,4,-3]]` `("Formula " A^(+)=(A^T*A)^(-1) * A^T)`


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`


This material is intended as a summary. Use your textbook for detail explanation.
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4. Example `[[1,0,1,0],[0,1,0,1]]` `("Formula " A^(+)=A^T * (A*A^T)^(-1))`
(Previous example)		19. Power Method for dominant eigenvalue
(Next method)


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