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DribleDaVaca19
 one year ago
Best ResponseYou've already chosen the best response.0dw:1370965723891:dw

amistre64
 one year ago
Best ResponseYou've already chosen the best response.1Ax = b x = A^(1) b might be one of the options

amistre64
 one year ago
Best ResponseYou've already chosen the best response.1they was nice enough to give you a determinant of 1 ... how sweet

DribleDaVaca19
 one year ago
Best ResponseYou've already chosen the best response.0i have 4 answers

amistre64
 one year ago
Best ResponseYou've already chosen the best response.1the answers might help us determine a suitable soluton process

DribleDaVaca19
 one year ago
Best ResponseYou've already chosen the best response.0dw:1370966096557:dw

amistre64
 one year ago
Best ResponseYou've already chosen the best response.1ax + by = n cx + cy = m matrix A = a b c d vector b = n m so yeah, they are looking for an inverse setup

DribleDaVaca19
 one year ago
Best ResponseYou've already chosen the best response.0dw:1370966191827:dw

amistre64
 one year ago
Best ResponseYou've already chosen the best response.1do you recall how to find the inverse of a 2x2 matrix?

amistre64
 one year ago
Best ResponseYou've already chosen the best response.1the shorcut involves a swap and some negations, youll also need a determinant

DribleDaVaca19
 one year ago
Best ResponseYou've already chosen the best response.0never mind I found it out, thx anyways

amistre64
 one year ago
Best ResponseYou've already chosen the best response.1\[\begin{pmatrix}a&b\\c&d\end{pmatrix}^{1}=\begin{pmatrix}d/A&b/A\\c/A&a/A\end{pmatrix}\]
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