linear equation cant solve

rcd = [-1 - 1  2] ;
ucd = rcd/norm(rcd);
syms Ax Ay Az Bx Bz T ;
eqn1 = [Ax Ay Az]+[0 0 -2*50*9.80665]+[Bx 0 Bz]+ T*ucd == [0 0 0] ;
eqn2 = cross([0 -2 0],[Ax Ay Az])+cross([2 0 0],[0 0 -50*9.80665])+cross([1 1 0],T*ucd)+cross([0 2 0],[Bx 0 Bz]) == [0,0,0] ;
solve = solve(eqn1,eqn2,[Ax Ay Az Bx Bz T])

//here the error message

Error in  +  (line 7)
X = privBinaryOp(A, B, 'symobj::zipWithImplicitExpansion', '_plus');

Error in myfisttest (line 4)
eqn1 = [Ax Ay Az]+[0 0 -2*50*9.80665]+[Bx 0 Bz]+ T*ucd == [0 0 0] ;
 

Answer 1

[Alan Stevens]

Like so

rcd = [-1, - 1,  2] ;
ucd = rcd/norm(rcd);
syms Ax Ay Az Bx Bz T ;
eqn1 = [Ax Ay Az]+[0 0 -2*50*9.80665]+[Bx 0 Bz]+ T*ucd == [0 0 0] ;
eqn2 = cross([0 -2 0],[Ax Ay Az])+cross([2 0 0],[0 0 -50*9.80665])+cross([1 1 0],T*ucd)+cross([0 2 0],[Bx 0 Bz]) == [0,0,0] ;
soln = solve(eqn1,eqn2,[Ax Ay Az Bx Bz T])

[Waritwong Sukprasongdee]

i cant find the difference here but your code works

[Alan Stevens]

Your rcd was interpreted as a 1x2 vector! Also, MATLAB didn't like solve = solve(...), so I used soln = solve(...).

[Waritwong Sukprasongdee]

thx but why does matlab interpret it as a 1 x 2 vector not 1 x 3 vector

[John D'Errico]

Be careful when you write something like this:

rcd = [-1 - 1  2]

Separating the minus sign from the number can introduce a bug into your code. (Which it did.)

>> rcd = [-1 - 1  2]
rcd =
    -2     2
    
>> rcd = [-1 -1  2]
rcd =
    -1    -1     2

Do you see the difference above? Note that it is perfectly valid to use spaces instead of a comma to separate terms in a vector, yet we get two different sized vectors as a result.

As it turns out, that the cause of your problem. What does this do:

rcd = [-1 - 1  2] ;

MATLAB sees the minus there, when distinct from the number as telling it to perform a SUBTRACTION. So it first subtracts 1 from -1, getting -2 as the result. Then it sees the number 2, and decides it needs to create a vector of length 2 because there is no operator between the elements.

Alan fixed it by putting in commas to delineate the numbers in the vector, though I would argue that explaining the problem is more important, regardless of whether a comma or a space is used to delineate the vector elements.

Answer 2

[Alan Stevens]

"... though I would argue that explaining the problem is more important, ..."

I agree. I didn't notice the extra space, just that the result was 1x2, so I didn't realise the explanation!

[John D'Errico]

(Please use comments, not answers for comments.)

Anyway, that is why I added my comment, to explain why your solution worked.

The minus sign "feature" in MATLAB is a dangerous thing, because it can be so often misunderstood what is happening. Here is another case where you need to be incredibly careful with minus signs.

>> -1^4
ans =
    -1

Yeah, right. Like raising a negative number to an even integer power can ever result in a negative number? But it does. And that is because of operator precedence rules.

https://www.mathworks.com/help/matlab/matlab_prog/operator-precedence.html

Concatenation operators fall at the very bottom of that list, apparently so low they fell completely off the bottom. :)

Anyway, in the example I just gave, power appears on a higher rung even than does negation (unary minus). So MATLAB essentially sees - (1)^4, which is -1. Of course, this gives a different result, the expected value of 1.

>> (-1)^4
ans =
     1

In the original case of

1 - 1

MATLAB parses this as using the dyadic minus operator, because it sees a number on either side of an operator. But when we write

1, - 1

here MATLAB has no choice but to see the - as a unary (or monadic) minus, so it negates the 1, and only then performs a concatenation.

As I said, you need to be very careful when playing with minus signs in MATLAB. But I'd not want to appear as negative. :)

[Walter Roberson]

There are other computer languages in which -1^4 would be interpreted as (-1)^4 because in those languages the unary operations are considered to bind very tightly. There are also languages where 10^-4 would be considered to be an error because the - is interpreted as a subtraction instead of a unary minus.