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anonymous
 5 years ago
et A and B be 2*2 matrices such that
2A+3B= 1 2
1 2
Atranspose +B transpose=1 3
7 6
Compute A and B
anonymous
 5 years ago
et A and B be 2*2 matrices such that 2A+3B= 1 2 1 2 Atranspose +B transpose=1 3 7 6 Compute A and B

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0one sec I'll do it real quick

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ooooo right bad at addition, woops

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0u can attach a picture of ur paper with solution

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0check to see if these work correctly: A = 4 23 16 5

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0may u send me ur steps also

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0nono B = 3 16 7 8

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0umm it's fairly illegible

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0just attach a picture with ur steps and btw its perfectly right:)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0the camera on my phone is pretty terrible, can't read my writing in pencil

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0steps I used: A^T + B^T = (A+B)^T thus: a11 + b11 = 1 a22+ b22 = 6 a12 + b12 = 7 a21 + b21 = 3 and i solved for the a values and plugged them into the other set of equations

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0which are: 2a11 + 2b11 = 1 2a12 + 3b12 = 2 2a21 + 3b21 = 1 2a22+ 3b22 = 2

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0which gave me the values for B matrix then I plugged those values back into the first set of equations to get my A matrix

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0yep, matricies are fun!

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0wants alot of them?:D

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0hit me with another one, I'm tired of my linear algebra studying anyway

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Let A be a 2*2 matrix Prove that if det(a)not equal zero then the inverse is given by A^1=(1/det(A))adj(A)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0hmm, i konw it has to do with the cofactor expansion, let me chew on that a sec

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0thinking on it, I should really know this.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0from cramer's rule we get:\[x_{j} = \det(A _{i}(e _{j})) / \det(A)\] thus \[x _{ij} = (1/\det(A))C _{ji}\] which gives us:\[A^{1} = X = (1/\det(A))[C _{ji}] = (1/\det(A))[C _{ij}]^T\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0and \[[C _{ij}]^T = adj(A)\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0might need an Assume the \[\det(A) \ne 0\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i pieced together a few statements, I'm pretty sure that is the right aproach but can't tell you how to do it exactly

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0and about the attachment did u check it? may u please check it:)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0thanks:) i will send u the attachment again

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0yeah sorry lost the connection

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0thats the Gauss Jordan algorithm metod whats about the adjugates

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0wouldn't let me type, look at both

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0start a new thread though
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