What is the difference between backward slash vs forward slash in MATLAB?

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Jamie Al
Jamie Al el 19 de Jun. de 2022
Editada: Stephen23 el 20 de Jun. de 2022
I have a failry simple question in MATLAB. What s the difference between the backslash operator vs the forward slash operator. For example,
A=[4,12;6,8];
b=[6,12;14,8];
x1 = A/b;
1.1333 -0.2000
0.5333 0.2000
%which is different from:
x2 = A\b;
3.0000 0
-0.5000 1.0000
I am asking because I am trying to convert a simple line of code from MATLAB to c++ which it turns out there's no forward slash in c++ unfortunately.
  2 comentarios
Jamie Al
Jamie Al el 20 de Jun. de 2022
Well, I am talking about more like "matrix" division using Eigen library which seems like not working.

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Respuestas (3)

Steven Lord
Steven Lord el 20 de Jun. de 2022
A=[4,12;6,8];
b=[6,12;14,8];
From the documentation for mrdivide, /, if x = A/b then x*b should be close to A.
x1 = A/b
x1 = 2×2
1.1333 -0.2000 0.5333 0.2000
check = x1*b-A
check = 2×2
0 0 0 0
Similarly from the documentation for mldivide, \, if x = A\b then A*x should be close to b.
x2 = A\b
x2 = 2×2
3.0000 0 -0.5000 1.0000
check = A*x2 - b
check = 2×2
1.0e-15 * 0.8882 0 0 0

John D'Errico
John D'Errico el 20 de Jun. de 2022
BOTH of them are linear algebraic solutions. Where matrices are involved, they solve subtly different problems.
A\b solves the linear algebra problem A*X=b.
For these matrices...
A=[4,12;6,8];
b=[6,12;14,8];
X1 = A\b
X1 = 2×2
3.0000 0 -0.5000 1.0000
Did it work?
A*X1
ans = 2×2
6.0000 12.0000 14.0000 8.0000
Did it recover the matrix b? Yes.
What does forward slash do? Again, when matrices are involved, it solves a different problem. A/b is equivalent to solving the linear algebra problem X2*b=A.
X2 = A/b
X2 = 2×2
1.1333 -0.2000 0.5333 0.2000
Did it work?
X2*b
ans = 2×2
4 12 6 8
Essentially, the two are similar in philosophy. The difference is where the unknown matrix would be in the problem you are implicitly solving.

William Rose
William Rose el 20 de Jun. de 2022
X=A\B computes X=inv(A)*B.
Y=A/B computes Y=A*inv(B)
A=rand(2,2); B=rand(2,2);
X1=A\B
X1 = 2×2
8.5138 7.1476 -3.5676 -2.2030
X2=inv(A)*B
X2 = 2×2
8.5138 7.1476 -3.5676 -2.2030
Y1=A/B
Y1 = 2×2
0.2325 -0.2566 -0.0595 0.7033
Y2=A*inv(B)
Y2 = 2×2
0.2325 -0.2566 -0.0595 0.7033
When I look at A\B, I try to remember that the A looks like it is "under" the divide sign, which reminds me that A is the denominator in A\B. And it comes first, so inv(A) is before B in the (non-commutative) multiplication.
  7 comentarios
Jamie Al
Jamie Al el 20 de Jun. de 2022
@William Rose @John D'Errico Thanks all for all the help, I greatly appreciate it. For now inv seems to work for my square matrices but I will keep in mind what you discussed here as well.
Stephen23
Stephen23 el 20 de Jun. de 2022
Editada: Stephen23 el 20 de Jun. de 2022
" Matlab's left divide may not use the equation I gave above - @John D'Errico says it doesn't, and I trust him."
Even better is to read the MLDIVIDE() documentation yourself:
"The equation I gave in my comment (not my original answer) is standard in a statistics class when discussing linear regression."
But almost completely useless when doing numeric computations on numeric data.
In much the same way beginners use the determinent to test if a matrix is singular or not, because that is what they were taught in "statistics class", but in the real world of numeric computing: almost completely useless.
"But A'A is not necessarily invertible (although I have never encoutered a linear regression problem where it's not)."
Whether A'A is invertible is not really the problem here, this still avoids the numeric issue.
"So maybe Matlab has a way to deal with that possibility, by not using inv() when it does left matrix divide."
The MLDIVIDE() documentation explains what algorithms it uses. Read it.

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