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Not sure if this is a bug in WolframAlpha or not

Posted 10 years ago
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Hi,

This is my first post here, so please excuse it if it's in the wrong place—I couldn't seem to find anywhere related to possible bugs.  This is related to Linear Algebra; specifically, whether polynomials in P2 are linearly dependent or independent.  For p1 = 1 + 3x + 4x^2, p2 = 3 + 10x + 15x^2, p3 = 3 + 11x + 18x^2 I believe the answer should be that these polynomials are linearly dependent, as the determinant of the coefficient matrix is 0 and it yields nontrivial solutions (and my course textbook affirms this).  However, WA says it's linearly independent when this query is performed:

http://www.wolframalpha.com/input/?i=is+%7B%7B1%2B3x%2B4x%5E2%7D%2C%7B3%2B10x%2B15x%5E2%7D%2C%7B3%2B11x%2B18x%5E2%7D%7D+linearly+dependent

It seems as if I'm phrasing the query correctly.  Am I missing something obvious?

Thanks,
Evan
POSTED BY: Evan Roth
4 Replies
So there is no solution. Since det is zero.
So the polynomials are L.I.

Another easier way to see this:
Solve[eqs, {c1, c2, c3}]
Will say there is no solution.
 But why would Solve say there is no solution? The polynomials are not linearly independent.
In[30]:= 3 (1 + 3 x + 4 x^2) - 2 (3 + 10 x + 15 x^2) + (3 + 11 x + 18 x^2) // Expand

Out[30]= 0
POSTED BY: Ilian Gachevski
So there is a solution. Since det is not zero.
So the polynomials are L.I.
 
It should be {1, 3, 3} instead of {1, 3, 4} and Det[ mat] is indeed zero.
POSTED BY: Ilian Gachevski
It should be {1, 3, 3} instead of {1, 3, 4} and Det[ mat] is indeed zero.

Opps, thanks.  I typed it wrong. (I should have programmed this part, so I do not have to type it in, but was lazy emoticon, will correct now.
POSTED BY: Nasser M. Abbasi
You need to find the coefficents c's first. As in
ClearAll[c1, c2, c3, x]
p1 = 1 + 3 x + 4 x^2; p2 = 3 + 10 x + 15 x^2; p3 = 3 + 11 x + 18 x^2;
Collect[c1 p1 + c2 p2 + c3 p3, x]
which gives
c1 + 3 c2 + 3 c3 + (3 c1 + 10 c2 + 11 c3) x + (4 c1 + 15 c2 + 18 c3) x^2
In the above, since the basis vectors of the polynominals are {1,x,x^2}, then see if there are c's, not all zero which makes the above sum zero. If you can find at least one c which makes the sum zero, then the polynomials are L.D. Else they are L.I.
So, write
eqs = {c1 + 3 c2 + 3 c3 == 0,
  3 c1 + 10 c2 + 11 c3 == 0,
  4 c1 + 15 c2 + 18 c3 == 0}
mat = {{1, 3, 3}, {3, 10, 11}, {4, 15, 18}}
Det[mat]
(* 0 *)

So there is no solution. Since det is zero.
So the polynomials are L.I.

Another easier way to see this:
Solve[eqs, {c1, c2, c3}]
Will say there is no solution.
POSTED BY: Nasser M. Abbasi
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