Division matrix


C

Cleuson Alves

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)
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[l]*b[k][j]
c[j]
return c
 
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D

Dennis Lee Bieber

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/Division_(mathematics)#Division_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/Invertible_matrix#Inversion_of_2.C3.972_matrices
gives direct formulations for 2x2 and 3x3 matrices.
 
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W

wxjmfauth

Le 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)

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[l]*b[k][j]

c[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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