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

one sec I'll do it real quick

- anonymous

maybe not quickly

- anonymous

:D take ur time:)

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## More answers

- anonymous

ooooo right bad at addition, woops

- anonymous

u can attach a picture of ur paper with solution

- anonymous

check to see if these work correctly:
A = |-4 -23|
|16 -5|

- anonymous

B = |3 26|
|-7 8|

- anonymous

may u send me ur steps also

- anonymous

nono B = |3 16|
|-7 8|

- anonymous

umm it's fairly illegible

- anonymous

just attach a picture with ur steps and btw its perfectly right:)

- anonymous

the camera on my phone is pretty terrible, can't read my writing in pencil

- 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

which are:
2a11 + 2b11 = 1
2a12 + 3b12 = 2
2a21 + 3b21 = -1
2a22+ 3b22 = -2

- 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

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- anonymous

thanks alot:)

- anonymous

yep, matricies are fun!

- anonymous

wants alot of them?:D

- anonymous

hit me with another one, I'm tired of my linear algebra studying anyway

- 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

check that alsoo s

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- anonymous

hmm, i konw it has to do with the cofactor expansion, let me chew on that a sec

- anonymous

err, cofactor matrix

- anonymous

??

- anonymous

thinking on it, I should really know this.

- anonymous

ok :) take ur time

- 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

and \[[C _{ij}]^T = adj(A)\]

- anonymous

might need an Assume the \[\det(A) \ne 0\]

- 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

and about the attachment did u check it? may u please check it:)

- anonymous

couldn't load it

- anonymous

http://en.wikipedia.org/wiki/Cramer's_rule

- anonymous

thanks:) i will send u the attachment again

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- anonymous

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- anonymous

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- anonymous

you there?

- anonymous

yeah sorry lost the connection

- anonymous

thats the Gauss Jordan algorithm metod whats about the adjugates

- anonymous

are u there?

- anonymous

wouldn't let me type, look at both

- anonymous

perfect:)

- anonymous

want more?

- anonymous

sure

- anonymous

start a new thread though

- anonymous

##### 1 Attachment

- anonymous

ohh sorry ok:)

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