backslash(\) is through error in 2013a but not to 2018b

the below syntax is working in matlab 2018b

s = [1 2 3 4; 5 6 7 9; 1 1 1 1 ; 1 1 1 1];
b=[1;1;1;1];
d = s.\b
%%result
d =

1.0000    0.5000    0.3333    0.2500
0.2000    0.1667    0.1429    0.1111
1.0000    1.0000    1.0000    1.0000
1.0000    1.0000    1.0000    1.0000

but in 2013a is throug "Matrix dimensions must agree."

may i know why?

Thanks in advance

Murugan

Answer 1

[Walter Roberson]

You did not use the \ operator: you used the .\ operator.

A.\B is the same as B./A -- element-wise division but with the order of operands reversed from the usual element-wise division.

The .\ operator is obscure, and I have never found a reason to use it other than for testing.

The reason the operator works on newer versions but fails on older versions is that you are doing element-by-element division between two arrays the same size. R2015b added the feature of "implicit expansion": for the element-by-element binary operators between two arrays, if the two arrays are the same size except that one might have size 1 in a dimension that the other is not 1, then the array with size 1 for the dimension will be automatically replicated to be the same size as the other.

Now, if you had used d = s\b with the \ operator instead of the .\ operator, then you would have had proper array sizes for the matrix-left-divide operator even back in R2013 time frame.

[Murugan C]

If I use s\b then my output will 'NaN'. this is not expected

[Walter Roberson]

What calculation are you trying to perform?

The last two rows of s are the same, so s cannot possibly be full rank, so if you were hoping to do least-squared fitting or trying to solve s*x = b then you will not be able to do so.

[Murugan C]

i got your point. thanks.