backslash(\) is through error in 2013a but not to 2018b
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Murugan C
am 19 Apr. 2023
Kommentiert: Murugan C
am 20 Apr. 2023
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
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Walter Roberson
am 19 Apr. 2023
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.
3 Kommentare
Walter Roberson
am 19 Apr. 2023
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.
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