# Matrix operation connecting two Matrix

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Pretesh John on 17 Nov 2020
Commented: Pretesh John on 27 Nov 2020
I have two matrices A=[11;15;45;17;1] B=[2 4 0; 3 4 5; 4 0 0; 5 0 0; 0 0 0] Where B is a kind of refrence matrix which is showing the element number of A. For example B11 (=2) means second element of A (=15), B22 (=4) means forth element of A (=17) I want to create a mathematical expression like X=A+u(B-A) where 'u' is a constant number and 'X' is also and column matrix containing all the values of expression. Example: for first row of B X11=A11+u(B11-A11)=A11+u(A21-A11) X21=A11+u(B12-A11)=A11+u(A41-A11) Then X31=A21+u(B21-A21)=A21+u(A31-A21) and so on
How can I create a program. Thanks

CHENG QIAN LAI on 24 Nov 2020
Edited: CHENG QIAN LAI on 24 Nov 2020
For example:
B(1,1)=2 means second element of A (=15) --> A( B(1,1),1 ) = A( 2,1 ) = 15
B(2,2)=4 means forth element of A (=17) --> A( B(2,2),1 ) = A( 4,1 ) = 17
B(1,3)=0 --> A( B(1,3),1 ) = A( 0,1 ) -->Index in position 1 is invalid. Array indices must be positive integers or logical values.
if B(r,c)=0 ,set X(r,c)=NaN
Example:
let u=1;
for first row of B
X(1,1)=A(1,1)+u*( A(B(1,1),1) - A(1,1) )=A(1,1)+ u*( A(2,1) -A(1,1) )=15
X(1,2)=A(1,1)+u*( A(B(1,2),1) - A(1,1) )=A(1,1)+ u*( A(4,1) -A(1,1) )=17
X(1,3)=A(1,1)+u*( A(B(1,3),1) - A(1,1) )=A(1,1)+ u*( A(0,1) -A(1,1) ) -->NaN
for secon row of B
X(2,1)=A(2,1)+u*( A(B(2,1),1) - A(2,1) )=A(2,1) +u*( A(3,1) -A(2,1) )=45
X(2,2)=A(2,1)+u*( A(B(2,2),1) - A(2,1) )=A(2,1) +u*( A(4,1) -A(2,1) )=17
X(2,3)=A(2,1)+u*( A(B(2,3),1) - A(2,1) )=A(2,1) +u*( A(5,1) -A(2,1) )=1
for third row of B
X(3,1)=A(3,1)+u*( A(B(3,1),1) - A(3,1) )=A(3,1) +u*( A(4,1) -A(3,1) ) =17
X(3,2)=A(3,1)+u*( A(B(3,2),1) - A(3,1) )=A(3,1) +u*( A(0,1) -A(3,1) ) -->NaN
X(3,3)=A(3,1)+u*( A(B(3,3),1) - A(3,1) )=A(3,1) +u*( A(0,1) -A(3,1) ) -->NaN
row=4
X(4,1)=A(4,1)+u*( A(B(4,1),1) - A(4,1) )=A(4,1) +u*( A(5,1) -A(4,1) )=29
X(4,2)=A(4,1)+u*( A(B(4,2),1) - A(4,1) )=A(4,1) +u*( A(0,1) -A(4,1) ) -->NaN
X(4,3)=A(4,1)+u*( A(B(4,3),1) - A(4,1) )=A(4,1) +u*( A(0,1) -A(4,1) ) -->NaN
row=5
X(5,1)=A(3,1)+u*( A(B(5,1),1) - A(3,1) )=A(3,1) +u*( A(0,1) -A(5,1) ) -->NaN
X(5,2)=A(3,1)+u*( A(B(5,2),1) - A(3,1) )=A(3,1) +u*( A(0,1) -A(5,1) ) -->NaN
X(5,3)=A(3,1)+u*( A(B(5,3),1) - A(3,1) )=A(3,1) +u*( A(0,1) -A(5,1) ) -->NaN
u=1;
A=[11;15;45;17;1];
B=[2 4 0;
3 4 5;
4 0 0;
5 0 0;
0 0 0];
X=NaN(size(B)) % =NaN(5,3)
% X =
% NaN NaN NaN
% NaN NaN NaN
% NaN NaN NaN
% NaN NaN NaN
% NaN NaN NaN
a=repmat(A,1,3)
% a =
% 11 11 11
% 15 15 15
% 45 45 45
% 17 17 17
% 1 1 1
idx=B>0 & B <=numel(A) % or idx=find( B>0 & B <=numel(A) )
% idx =
% 5×3 logical array
% 1 1 0
% 1 1 1
% 1 0 0
% 1 0 0
% 0 0 0
X(idx)=a(idx) + u*( a(B(idx),1) - a(idx) )
% X =
% 15 17 NaN
% 45 17 1
% 17 NaN NaN
% 1 NaN NaN
% NaN NaN NaN
% or
Y=a(idx) + u*( a(B(idx),1) - a(idx) )
% Y =
% 15
% 45
% 17
% 1
% 17
% 17
% 1
Pretesh John on 27 Nov 2020
Thanks a lot