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Division matrixHello, I need to solve an exercise follows, first calculate the inverse matrix and then multiply the first matrix.
I await help. Thank you. follows the code below incomplete. m = [[1,2,3],[4,5,6],[7,8,9]] x = [] for i in [0,1,2]: y = [] for linha in m: y.append(linha[i]) x.append(y) print x [[1, 4, 7], [2, 5, 8], [3, 6, 9]] def ProdMatrix(x,b): tamL = len(x) tamC = len(x[0]) c = nullMatrix(tamL,tamC) for i in range(tamL): for j in range(tamC): val = 0 for k in range(len(b)): val = val + x[i][l]*b[k][j] c[i][j] return c |

Re: Division matrixOn Tue, 13 Nov 2012 10:19:43 -0200, Cleuson Alves <cleuson.o@gmail.com>
declaimed the following in gmane.comp.python.general: > Thanks, I'm starting to plan now, so I'm still confused with the production > code, but what I need is to divide array 2x2 or 3x3. > I still can not! Divide it by what? A scalar... Another square matrix of the same size... Or a square matrix of a different size (is that even possible?)... Based http://en.wikipedia.org/wiki/Divisio...on_of_matrices upon, I can understand where the need for the inverse comes from -- and multiplication by the inverse gives the "division". Next up, http://en.wikipedia.org/wiki/Inverti...3.972_matrices gives direct formulations for 2x2 and 3x3 matrices. -- Wulfraed Dennis Lee Bieber AF6VN wlfraed@ix.netcom.com HTTP://wlfraed.home.netcom.com/ |

Re: Division matrixOn Tue, Nov 13, 2012 at 1:00 AM, Cleuson Alves <cleuson.o@gmail.com> wrote:
> Hello, I need to solve an exercise follows, first calculate the inverse matrix and then multiply the first matrix. I would just point out that in most numerical applications, you rarely need to calculate the intermediate of the matrix inverse directly. See, e.g., http://www.johndcook.com/blog/2010/0...t-that-matrix/ Of course, if this hasn't been said yet: NumPy. Michael |

Re: Division matrixLe mardi 13 novembre 2012 02:00:28 UTC+1, Cleuson Alves a écrit*:
> Hello, I need to solve an exercise follows, first calculate the inverse matrix and then multiply the first matrix. > > I await help. > > Thank you. > > follows the code below incomplete. > > > > m = [[1,2,3],[4,5,6],[7,8,9]] > > x = [] > > for i in [0,1,2]: > > y = [] > > for linha in m: > > y.append(linha[i]) > > x.append(y) > > > > print x > > [[1, 4, 7], [2, 5, 8], [3, 6, 9]] > > > > def ProdMatrix(x,b): > > tamL = len(x) > > tamC = len(x[0]) > > c = nullMatrix(tamL,tamC) > > for i in range(tamL): > > for j in range(tamC): > > val = 0 > > for k in range(len(b)): > > val = val + x[i][l]*b[k][j] > > c[i][j] > > return c ------ Pedagogical hint: Before blindly calculating the inverse matrix, it may be a good idea to know if the inverse matrix exists. jmf |

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