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For example, given a 3d point p(x,y,z), with (x,y,z) as its coordinates. Then it is transformed by rotation and translation, as 

p'=R(p)*p+t(p), where R(p) is a 3x3 rotation matrix that is a matrix of functions of p, and t(p) is a 3x1 vector function of p. 

My question is how to derive dp'/d(as a 3x3 matrix) using maple? 

To make it clear,I want to do it in a way that dp'/dp = ∂p'/∂p + ∂p'/∂R*∂R/∂p +  ∂p'/∂t*∂t/∂p

And I'd like to know each intermediate quantity, such as p'/∂R, R/∂p.


Anyone can help?

Thanks a lot. 

Hi
I have my question makes any sense. I am from Denmark and not used to write math in english.

I have an characteristic matrix with an variable λ that takes on differen values.

How do I write λ in the matrix so Maple knows that when I call out a row with the variable λ in it and asssign

λ to a specific value, Maple changes λ to the specific value.

 

Example (I was thinking something like this):

A:=Matrix(2,2,[(1-λ_i),2,3,(4-λ_i)])

λ_1:=2

λ_2:=4

A[1],λ_1               (1-2) 2

A[1],λ_2               (1-4) 2

A[2],λ_1               3 (4-2)

A[2],λ_2               3 (4-4)

I want to generate all 4x4 matrices of rank-1 over the field with two elements, F_2. I don't want to use actual arrays though, I just want to use symbols representing the unit matrices. So for example Eij represents the 4x4 matrix with a 1 in row i & column j, and 0's elsewhere. 

We have 16 unit matrices: E11, E12, ..., E14, ... E41, E42, E43, E44. Then we have to look at all possible linear combinations that result in a rank-1 matrix. So for example: E11+E12,  E11+E21,  E11+E12+E13,  E33+E43, etc. The total number of rank-1 4x4 matrices over F_2 is 225. How can I find all of these quickly?

Hi

I need sort data of an imported matrix to ordered pairs.

'M' is a matrix with 2 columns and unknown numbers of row.

 

with(LinearAlgebra):

M := RandomMatrix(12, 2);

pairs := [M[1], M[2], ... , M[numelems(DeleteColumn(M, [1]))]];

with(CurveFitting):

PolynomialInterpolation(pairs, x);

Hey guys, I'm new to the forum, so please tell if I need to set up the question in a different way :) I've tried to find an answer for this, but have struggled since our learning book is in danish, so the used terms may not be technically correct,

 

Anyways, how do you solve this problem in maple? 

 

Find the complete real solution for the differential equation system:

 

 

For the homogenous part I've found. (Is this correct?)

  , c1,c2 € R

 

 

I've tried to find an answer for the inhomogenous part but I get a really complicated result, so I doubt it's correct.

 

Thanks for the help :)

-Alex 

 

I have a markov matrix that is 500x500. I need to take the standard deviation of the top row of this matrix after I raise it to high powers, like 10^17 by 10^12, (from a do loop.) I would also like ot make a histogram of this data too. 

 

Any ideas what I can do? 

I've got a vector x=[0.36,1.3279,1.6882] (1*3) obtained as x:=pseudoInverse(A)*b where A=[<2,4,1>|<1,-1,1>|<3,3,2>] where <2,4,1>,<1,-1,1>,<3,3,2> are the columns of A and b is [8,5,4](1*3). Now, when I find the rowspace of A using RowSpace(A) I get the row vectors <1,0,1> and <0,1,1>, neither of which are equivalent to x. How do I arrive at my result that x is in the rowspace?

where A^T is the transpose of A and it's given that (A^T).A is not invertible.

I am stuck as to how to arrive at the solution for x in this case. I initially thought I could multiply both sides by the inverse of A^T reducing it to Ax=b but that was obviously wrong since A^T is itself not invertible(it is singular).

Hello people in Mapleprimes,

 

I wish you to teach me about this.

From a matrix, I want to make a string.

 

pt3 := Matrix(4, 2, [1, 3, 4, 5, 6, 8, 5, 10]);

 

string1 := cat(

  "(", convert(pt3[1, 1], string), ",", convert(pt3[1, 2], string), ",", ")",

  "==",

  "(", convert(pt3[2, 1], string), ",", convert(pt3[2, 2], string), ",", ")",

  "==",

  "(", convert(pt3[3, 1], string), ",", convert(pt3[3, 2], string), ",", ")",

  "==",

  "(", convert(pt3[4, 1], string), ",", convert(pt3[4, 2], string), ")"

)

 

The above code makes 

string1 := "(1,3,)==(4,5,)==(6,8,)==(5,10)"

 

What I want to know is how to write a code of programming without writing each row of the above code.

Actually, though I wrote 4 row matrix above, the number of rows of matrix I want to deal with might be more than 100.

 

Best wishes.

 

taro

 

M := Matrix([[0,1,1,1,1],[1,0,1,1,1],[1,1,0,1,1],[1,1,1,0,1],[1,1,1,1,0]]);

how to multiply M[1] which is [0,1,1,1,1] with Matrix([[x0], [x1], [x2], [x3], [x4]])
to become [0,x1,x2,x3,x4]

may not use Matrix([[x0], [x1], [x2], [x3], [x4]])

just want to use first element as x0 when it is not 0

second element as x1, etc

LM:=proc(n)
local L;
uses combinat;
L:=permute([1$(n*(n-1)/2), 0$(n*(n-1)/2)], n*(n-1)/2);
[seq(Matrix(n,{seq(seq((i+1,j)=L[k][(i-1)*i/2+j], j=1..i), i=1..n-1)}, shape=symmetric), k=1..nops(L))];
end proc:
M := LM(5);
N := nops(M);
append("E:\\mm.txt");
for i from 1 to N do
ExportMatrix( "E://mm.txt", M[i]);
od:

 

hope to export to a text file

and show

for example

matrix([[1,1,1],[1,1,0],[0,0,0]]);


matrix([[1,1,1],[1,1,0],[0,0,0]]);


matrix([[1,1,1],[1,1,0],[0,0,0]]);

...

etc

 

how to solve this eq by maple:

P:=Matrix([[ 0 , .5 , .5 , 0 , 0 , 0 ], [ 1/3 , 0 , 0 , 1/3 , 1/3 , 0 ], [ 1/3 , 0 , 0 , 0 , 1/3 , 1/3 ], [ 0 , 1 , 0 , 0 , 0 , 0 ], [ 0 , .5 , .5 , 0 , 0 , 0 ], [ 0 , 0 , 1 , 0 , 0 , 0 ]]);

 

pii:=Vector[row]([ a , b , c , d , e , f ])

 

how to find

pii.P=pii

 

Silaws

let m3 = [[0; 1; 0]; [1; 0; 1]; [0; 1; 0]]

1. Firstly, express this matrix into sequence function expression

2. how to express this matrix in terms of forloop code

3. for complicated case such as 1 is not in easy pattern, can it intelligently express the matrix in terms of for loop code

 

is there exist extra tools to express matrix in terms of for loop code or sequence function code?

how do minor generate 2*2 matrix from 5*5 matrix or n*n matrix which n >= 3

first case is the symmetric or nonsymmetric ladder matrix which fill full upper right and lower left corner

{0,1,1,1,1}
{1,0,1,1,1}
{1,0,0,0,1}
{1,1,1,0,0}
{1,1,1,1,0}

second case is overlap two different first case , one is 1 and another is 0 which is for display partial fill the upper right corner and lower right corner

{0,1,0,0,0}
{1,0,1,1,0}
{1,0,0,0,1}
{0,1,1,0,0}
{0,0,0,1,0}

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