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anonymous
 4 years ago
Is it possible to find an inverse of a nonsquare matrix? I got a problem from a professor with Matrix A = 4x5 matrix. The Matrix A multiplied by B gives a matrix 4x5. I'm given the Matrix A
[1 1 2 5 4
3 2 1 4 2
0 1 2 1 3
2 1 5 0 1]
AB is
[2 1 6 5 8
6 2 4 4 4
0 1 6 1 6
4 1 15 0 2]
Solve for B? If I can't find an inverse, can I find a solution for B?
anonymous
 4 years ago
Is it possible to find an inverse of a nonsquare matrix? I got a problem from a professor with Matrix A = 4x5 matrix. The Matrix A multiplied by B gives a matrix 4x5. I'm given the Matrix A [1 1 2 5 4 3 2 1 4 2 0 1 2 1 3 2 1 5 0 1] AB is [2 1 6 5 8 6 2 4 4 4 0 1 6 1 6 4 1 15 0 2] Solve for B? If I can't find an inverse, can I find a solution for B?

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anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Here is a nice trick: If you multiply \[A^{T}\] on the left of each side of your equality, you will have: \[A^{T}\times A \times B = A^{T}\times AB\] Well, if A is a 4x5 matrix, then \[A^{T}\] will be a 5x4 matrix, furthermore, the product \[A^{T}\times A \] will be a 5x5 matrix, is that ok? We shall call \[A^{T}\times A \ = \Omega\] Now, we can rewrite our initial problem as: \[\Omega \times B = A ^{T}\times AB\] If you remember, this "Omega" matrix is the 5x5 square matrix we obtained only by multiplying A by it's transposed, so we know how it looks like and it means we can get it's inverse, after we get it, multiply it on both sides of the equation: \[\ B = I_{5} \times B = \Omega ^{1}\times \Omega \times B = \Omega ^{1}\times A ^{T}\times AB\]Where "ifive" is the 5x5 identity matrix. Now you can easily calculate it using matlab, since you got all the matrices on the right side. My answer was 8.2188 0.3281 20.1758 3.5195 11.7969 16.7813 5.0508 19.4258 13.2344 18.4297 0.2500 0.6875 1.8438 1.3438 4.5000 0.8438 0.6719 4.9727 0.8711 0.2344 0.0208 0.0186 0.0391 0.0034 1.9796 Altough I can be wrong. Hope it helped and I'm sorry for any mess I made.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0This answer has an important mistake. The 5 x 5 matrix Ohmega has rank four, so it is not invertible. You could use the Moore Penrose generalized inverse as a pseudoinverse. Then the solution would be A^+(AB) = B. For your numbers, A^+ is 0.2899 0.3750 0.1319 0.0122 0.1016 0.1821 0.0369 0.0643 0.0988 0.1343 0.1124 0.2119 0.2515 0.1404 0.2568 0.0441 0.0159 0.1034 0.3350 0.1480 Then Bhat= A^+AB is the best guess for B. It is 1.7188 0.3160 0.7344 0.0816 0.0035 0.6319 0.2900 0.6254 0.1833 0.0078 0.2396 0.2692 2.5596 0.0695 0.0030 0.1632 0.1833 0.0681 0.9527 0.0020 0.0035 0.0039 0.1078 0.0010 2.0000 For sure ABhat = AA^+AB = AB.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Perhaps I'm being simplistic here, but it seems that all of the column vectors in AB are scalar multiples of the column vectors in A except for the entry in row 2 column 3 of AB. If that entry was a 3, the answer for B would be 2 0 0 0 0 0 1 0 0 0 0 0 3 0 0 0 0 0 1 0 0 0 0 0 2 Is it possible that there was a copying mistake?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0It is not possible to inverse the nonsquare matrix. To inverse matrix the two candition have to accomplished  if u want to inverse A in K  body and n  dimension  1) \[A,B \in M _{n \times n}\left( K \right)\] 2) \[A \times B = I\]
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