Error in comparing equality of two matrices

Hi, I am trying to add a conditional statement which compares whether two matrices are equal and generates a result. But it fails everytime. Please find my code below:

alpha = input("Enter value of alpha: ")
beta = input("Enter the value of beta: ")
gamma = input("Enter the value of gamma: ")


R_BA = rotz(alpha*pi/180)*roty(beta*pi/180)*rotx(gamma*pi/180)

X = R_BA(:,1)
Y = R_BA(:,2)
Z = R_BA(:,3)

mod_X = X(1).^2 + X(2).^2 + X(3).^2
mod_Y = Y(1).^2 + Y(2).^2 + Y(3).^2
mod_Z = Z(1).^2 + Z(2).^2 + Z(3).^2

if int64(mod_X)==int64(mod_Y)==int64(mod_Z)
    disp("X, Y and Z vectors are unit vectors")
else
    disp("Unitary condition not satisfied")
end

if dot(X,Y) == dot(Y,Z) == dot(Z,X)
    disp("Orthonormal condition satisfied")
else
    disp("Orthonormal condition not satisfied")
end

inv_RBA = inv(R_BA) %inverse of rotation matrix

transpose_RBA = transpose(R_BA)

if isequal(inv_RBA,transpose_RBA)
    disp("Property satisfied")
else
    disp("Fail")
end
    

The code at the end checks if inv_RBA is equal to transpose_RBA. The two matrices have the same exact elements but I do not get a positive output.

Highly appreciate your help here.

Thanks

Answer 1

[Chunru]

Use "isequal" for matrix comparison and "&&" for logic and.

% if int64(mod_X)==int64(mod_Y)==int64(mod_Z)
if isequal(int64(mod_X), int64(mod_Y) && isequal(int64(mod_Y), int64(mod_Z))
    ...
end

The same applies to "if dot(X,Y) == dot(Y,Z) == dot(Z,X)" as well.

[Stephen23]

Note that ISEQUAL accepts any number of input arguments (not just two), so this simplifies to:

isequal(int64(mod_X), int64(mod_Y, int64(mod_Z))

Note that this is unlikely to be useful as it does not take into account the binary floating point error of those calculations.

It is not clear what hypothesis converting to INT64 (which rounds the values) is expected to confirm or reject.

[Walter Roberson]

I suspect that they were thinking of typecast() to uint64 and then comparing those underlying representations. But that does not give any more utility than comparing double precision.

Answer 2

[N/A]

I had a problem using the last "if" statement in my code where I was not able to compare inv_RBA and transpose_RBA. I figured out this way to compare them both:

k = 0;
for m = 1:size(inv_RBA)
    for n = 1:size(transpose_RBA)
        
        if isequal(i,j)
            continue
        else
            disp("Fail")
            k = 1
        end
    end

end

if k==0
    disp("Property satisfied")
end

[Walter Roberson]

@Sampath Vinay Pilla

That code does not define i or j so they will either be whatever was in the workspace, or else they will be the default values for those two names, which are functions that return sqrt(-1) .

Caution: size() of a variable, when no option is provided in the size() call, always returns a vector, even if the variable is a scalar. When you use 1:size() then you are using 1 colon a vector . MATLAB does define that: it takes the first element of the vector. So 1:size(inv_RBA) is going to be the same as 1:size(inv_RBA,1) which might not be what you want;.

[N/A]

Hi, it was not entirely clear to me as I'm a beginner, but thanks for your feedback. Would you be able to recommend a better way to compare each element of these matrices and generate a result which says both are equal?

[Walter Roberson]

all( abs(inv_RBA - transpose_RBA) < TOLERANCE )

where TOLERANCE is some value that you feel is "close enough" to equal. For example if the values are near 2, then you might feel that 1e-13 was "close enough"