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anonymous
 one year ago
matrix equation help?
{{12,7},{5,3}}x={{2,1},{3,2}}
anonymous
 one year ago
matrix equation help? {{12,7},{5,3}}x={{2,1},{3,2}}

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amistre64
 one year ago
Best ResponseYou've already chosen the best response.0an augment might be beneficial

amistre64
 one year ago
Best ResponseYou've already chosen the best response.0unless you know how to work an inverse? Ax = b A'Ax = A'b Ix = A'b x = A'b

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0how can i plug in matrices?

amistre64
 one year ago
Best ResponseYou've already chosen the best response.0what are you working with? ti83?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0No. I meant how can I plug in matrices on Openstudy

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0So I can show the equation correctly

amistre64
 one year ago
Best ResponseYou've already chosen the best response.0its labor intensive, but i did make an app for it, if you are using a laptop or desktop that is

amistre64
 one year ago
Best ResponseYou've already chosen the best response.0\[\left[ \begin{array}c 12 & 7\\5 & 3\\\end{array} \right] x= \left[ \begin{array}c 2 & 1\\3 & 2\\\end{array} \right]\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[\[\left[ \begin{array}c 12 & 7\\5 & 3\\\end{array} \right]\] X = \[\left[ \begin{array}c 2 & 1\\3 & 2\\\end{array} \right]\]\]

amistre64
 one year ago
Best ResponseYou've already chosen the best response.0it was designed to post one matrix hence the oddity of trying multiples, but knowing the latex sorts it out

amistre64
 one year ago
Best ResponseYou've already chosen the best response.0so, how do we approach this? inverse are augment? either way is fine. inverse may be simpler in this case tho

amistre64
 one year ago
Best ResponseYou've already chosen the best response.0we can augment it as this, and row reduce. \[\left[ \begin{array}c 12 & 7 &  & 2 & 1\\5 & 3 &  & 3 & 2\\\end{array} \right]\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0how do you reduce it

amistre64
 one year ago
Best ResponseYou've already chosen the best response.0elementary row operations for course, get the left side worked into the identity matrix. and the right side works itself into x

amistre64
 one year ago
Best ResponseYou've already chosen the best response.0divide row1 by 12 and multiply by 5 and add to row2 12,7,2,1 1,7/12,2/12,1/12 5,35/12,10/12,5/12 5, 3, 3, 2 5,35/12,10/12, 5/12 5, 36/12, 36/12,24/12  0, 1/12, 26/12, 29/12 multiply by 12 0, 1, 26, 29 new setup 1, 7/12, 2/12, 1/12 0, 1 , 26 , 29 multiply row 2 by 7/12 and add it to row1 1, 7/12, 2/12, 1/12 0, 7/12 , 26(7)/12 , 29(7)/12  1, 0 , 90/6 , 204/12 1, 0 , 15 , 17 1, 0, 15, 17 0, 1, 26, 29 I x

amistre64
 one year ago
Best ResponseYou've already chosen the best response.0http://www.wolframalpha.com/input/?i=rref%7B%7B12%2C7%2C2%2C1%7D%2C%7B5%2C3%2C3%2C2%7D%7D

amistre64
 one year ago
Best ResponseYou've already chosen the best response.0did you try for an inverse?

amistre64
 one year ago
Best ResponseYou've already chosen the best response.035 12 7 5 3 36 det = 1 by some basic shortcut, A' has a pattern swap corners and negate the other 2 A' = 3 7 5 12
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