substitute each element of a vector into a matrix without using loop
3 views (last 30 days)
Show older comments
AllKindsofMath AllKinds
on 19 Nov 2013
Answered: Alfonso Nieto-Castanon
on 20 Nov 2013
Hi I want to substitute each element of vector1 into 'x' in matrix1 and store each matrix in an array without using a loop. Please tell me how.
vector1=[1:1:10];
matrix1=[4*x 5*x ; 4*x 2*x];
Thanks in advance.
0 Comments
Accepted Answer
Jan
on 19 Nov 2013
Like this?
vector = 1:1:10;
vector = reshape( vector, [1, 1, numel(vector)] );
one_vector = ones( 1, 1, numel(vector) );
matrix = [4 * vector, 5 * one_vector; 4 * one_vector, 2 * vector ];
5 Comments
Jan
on 20 Nov 2013
Not as far as I can see. You will probably have to iterate over the third dimension.
More Answers (3)
Sean de Wolski
on 19 Nov 2013
So:
vector1=[1:1:10];
x = vector1;
matrix1=[4*x 5*x ; 4*x 2*x];
Or is x symbolic?
clear x;
syms x
matrix1=[4*x 5*x ; 4*x 2*x];
matrix1 = subs(matrix1,x,vector1)
5 Comments
Jan
on 19 Nov 2013
Maybe something like the following?
matrix = [4, 5; 4, 2];
[p, q] = size( matrix );
vector = 1:1:10;
matrix = repmat( matrix(:), 1, numel( vector ) );
matrix = matrix .* repmat( vector, p*q, 1 );
matrix = reshape( matrix, p, q, numel( vector ) );
This gives you a 3d matrix, where each layer contains the specified matrix, mulitplied by one entry in vector
Alfonso Nieto-Castanon
on 20 Nov 2013
perhaps something like:
f = @(x)[4*x 5 ; 4 2*x]; % Matrix in functional form
vector = 1:10; % Your vector of values for 'x'
matrix = arrayfun(f,vector,'uni',0); % A cell array of matrices
values = cellfun(@det,matrix); % Determinant of each of those matrices
0 Comments
See Also
Categories
Find more on Multidimensional Arrays in Help Center and File Exchange
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!