substitute each element of a vector into a matrix without using loop
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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.
Accepted Answer
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
AllKindsofMath AllKinds
on 19 Nov 2013
Sean de Wolski
on 19 Nov 2013
Edited: Sean de Wolski
on 19 Nov 2013
i.e. the second half of what I have...
AllKindsofMath AllKinds
on 19 Nov 2013
Sean de Wolski
on 19 Nov 2013
subs(matrix1,x,vector1(4))
AllKindsofMath AllKinds
on 19 Nov 2013
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
1 Comment
AllKindsofMath AllKinds
on 19 Nov 2013
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
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