anonymous
  • anonymous
Is it possible to find an inverse of a non-square 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?
MIT 18.06 Linear Algebra, Spring 2010
  • Stacey Warren - Expert brainly.com
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schrodinger
  • schrodinger
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anonymous
  • anonymous
Here 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 "i-five" 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
  • anonymous
This 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 pseudo-inverse. 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
  • anonymous
Perhaps 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?

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anonymous
  • anonymous
It is not possible to inverse the non-square 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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