anonymous
  • anonymous
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
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
one sec I'll do it real quick
anonymous
  • anonymous
maybe not quickly
anonymous
  • anonymous
:D take ur time:)

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anonymous
  • anonymous
ooooo right bad at addition, woops
anonymous
  • anonymous
u can attach a picture of ur paper with solution
anonymous
  • anonymous
check to see if these work correctly: A = |-4 -23| |16 -5|
anonymous
  • anonymous
B = |3 26| |-7 8|
anonymous
  • anonymous
may u send me ur steps also
anonymous
  • anonymous
nono B = |3 16| |-7 8|
anonymous
  • anonymous
umm it's fairly illegible
anonymous
  • anonymous
just attach a picture with ur steps and btw its perfectly right:)
anonymous
  • anonymous
the camera on my phone is pretty terrible, can't read my writing in pencil
anonymous
  • anonymous
steps 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
  • anonymous
which are: 2a11 + 2b11 = 1 2a12 + 3b12 = 2 2a21 + 3b21 = -1 2a22+ 3b22 = -2
anonymous
  • anonymous
which 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
  • anonymous
anonymous
  • anonymous
thanks alot:)
anonymous
  • anonymous
yep, matricies are fun!
anonymous
  • anonymous
wants alot of them?:D
anonymous
  • anonymous
hit me with another one, I'm tired of my linear algebra studying anyway
anonymous
  • anonymous
Let 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
  • anonymous
check that alsoo s
1 Attachment
anonymous
  • anonymous
hmm, i konw it has to do with the cofactor expansion, let me chew on that a sec
anonymous
  • anonymous
err, cofactor matrix
anonymous
  • anonymous
??
anonymous
  • anonymous
thinking on it, I should really know this.
anonymous
  • anonymous
ok :) take ur time
anonymous
  • anonymous
from 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
  • anonymous
and \[[C _{ij}]^T = adj(A)\]
anonymous
  • anonymous
might need an Assume the \[\det(A) \ne 0\]
anonymous
  • anonymous
??
anonymous
  • anonymous
i 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
  • anonymous
and about the attachment did u check it? may u please check it:)
anonymous
  • anonymous
couldn't load it
anonymous
  • anonymous
http://en.wikipedia.org/wiki/Cramer's_rule
anonymous
  • anonymous
thanks:) i will send u the attachment again
1 Attachment
anonymous
  • anonymous
anonymous
  • anonymous
anonymous
  • anonymous
you there?
anonymous
  • anonymous
yeah sorry lost the connection
anonymous
  • anonymous
thats the Gauss Jordan algorithm metod whats about the adjugates
anonymous
  • anonymous
are u there?
anonymous
  • anonymous
wouldn't let me type, look at both
anonymous
  • anonymous
perfect:)
anonymous
  • anonymous
want more?
anonymous
  • anonymous
sure
anonymous
  • anonymous
start a new thread though
anonymous
  • anonymous
1 Attachment
anonymous
  • anonymous
ohh sorry ok:)

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