matrix column multiplication with rows of another matrix

this is my (9*9) matrix.

     1     1     1     1     0     1     1     1     1
     1     1     1     1     1     0     1     1     1
     1     1     1     1     1     1     0     1     1
     1     1     1     1     1     1     1     0     1
     1     1     1     1     1     1     1     1     0
     0     1     1     1     1     1     1     1     1
     1     0     1     1     1     1     1     1     1
     1     1     0     1     1     1     1     1     1
     1     1     1     0     1     1     1     1     1

I would like to multiply entire column 2, 3, and 4 etc. of this matrix with rows of the below column vector matrix one by one.

000
001
010
011
100
101
110
111

This is multiply 000 with column 2 , 3 and 4. which will result entire column 2, 3 and 4 as zero.

Then multiply 001 with column 2 , 3 and 4. which will result entire column 2, 3 as zero and column 4 will remain same. This process should continue for each row.

[CS Researcher]

For multiplication the number of elements in the row must be same as the number of elements in the column. In your data the rows have 3 elements each while the columns have 9 elements.

[Monika Kok]

I want to multiply 010 by column 2, 3 and 4 that means entire column 2 should be multiplied by zero.column 3 by 1 and column 4 by zero. I think we can multiply a column by a number in matalb

[Star Strider]

Does the small detail that the matrix you posted is (9x9), not (15x15) have any bearing on your Question?

[Monika Kok]

yeah that is (9*9) matrix... sorry for the wrong typing.

[Image Analyst]

The so-called 15*15 matrix is actually 9*9. Are the 000, etc. supposed to be binary numbers, or decimal?

Answer 1

[Image Analyst]

Try this:

m=[...
  1     1     1     1     0     1     1     1     1
  1     1     1     1     1     0     1     1     1
  1     1     1     1     1     1     0     1     1
  1     1     1     1     1     1     1     0     1
  1     1     1     1     1     1     1     1     0
  0     1     1     1     1     1     1     1     1
  1     0     1     1     1     1     1     1     1
  1     1     0     1     1     1     1     1     1
  1     1     1     0     1     1     1     1     1]

v = [...
  000
  001
  010
  011
  100
  101
  110
  111
  111]
vm = repmat(v, [1, size(m, 2)])

out = m .* repmat(v, [1, size(m, 2)])

Use bin2dec if your v is actually binary numbers.

Answer 2

[Roger Stafford]

Let M be your 15 x 15 matrix and A be your n x 3 matrix.

B = zeros(size(M,1),3,size(A,1)); % A three-dimensional result
for k = 1:size(A,1)
  B(:,:,k) = bsxfun(@times,A(k,:),M(:,2:4));
end

You have asked that a size(M,1) x 3 result be obtained for each row of A, which has forced me to give you a three-dimensional result.

[Roger Stafford]

Monika, you stated “This is multiply 000 with column 2 , 3 and 4. which will result entire column 2, 3 and 4 as zero.” This means, if I understand you correctly, that you first want the entire nine-element columns 2, 3, and 4 to be made into zeros by multiplication with [0 0 0]. That is a 9 by 3 group of zeros. Then you want the same repeated for each row of your eight rows in your second matrix. I conclude that there should be eight times nine times three results and that is why I produced a three dimensional array as the result. The question is why you have now expressed a preference for a different answer with just nine times three elements.