I had write the function for find the difference of maximum and minimum from each martix row and express it as matrix .

function [mmr mmm]= minimax (A)
mmr=[abs(max(A(1:end,:))-min(A(1:end,:)))]
%mmr=[abs((max(A(1,:))-min(A(1,:)))),abs((max(A(2,:))-min(A(2,:)))),abs((max(A(3,:))-min(A(3,:))))]
mmm=abs(max(A(:))-min(A(:)))
end

code call

A=rand(100,3,4)

It return the difference between maximum and minimum value of colum but i want row.And it display two times ans

Answer 1

[Rik]

If you want a column instead of a row, why not transpose? And if you want both outputs, you will have to use a syntax with two outputs. If you want to suppress the output in your function you should look at the orange line underneath your code: mlint is warning you that you might have forgotten a semicolon.

And why are you taking the absolute value of when computing the maximum value?

%Note: this only fixes the main issues in the function you posted, it doesn't give the correct result for your assignment
A=rand(100,3,4);
[mmr,mmm]= minimax (A)
function [mmr,mmm]= minimax (A)
mmr=(abs(max(A(1:end,:))-min(A(1:end,:)))).';
mmm=abs(max(A(:))-min(A(:))).';
end

[YOGESHWARI PATEL]

when i call the code

A=[1:3;4:7;8:11;12:15]

minimax(A')

This gives the answer.

Can I make directly change in function to get answer.

[YOGESHWARI PATEL]

for row .This code give answer for column.

[Rik]

Sure, just do the transpose inside your function:

function [mmr,mmm]= minimax (A)
A=A.';

[YOGESHWARI PATEL]

Thank you.It works wow

[Image Analyst]

I don't see why abs() is needed. The difference will always be 0 or positive for real numbers, right? Can you give me one example where the max minus the min will be negative?

[YOGESHWARI PATEL]

I made the correct.

[Rik]

The difference will always be 0 or positive for real numbers, right?

Since max and min use the magnitude, this even holds for complex values.

[YOGESHWARI PATEL]

thanks for solving the problem