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Not sure if this is a bug in WolframAlpha or not

Posted 11 years ago
 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+dependentIt seems as if I'm phrasing the query correctly.  Am I missing something obvious?Thanks,Evan
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Posted 11 years ago
 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) // ExpandOut[30]= 0
Posted 11 years ago
 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 11 years ago
 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 , will correct now.
Posted 11 years ago
 You need to find the coefficents c's first. As inClearAll[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 givesc1 + 3 c2 + 3 c3 + (3 c1 + 10 c2 + 11 c3) x + (4 c1 + 15 c2 + 18 c3) x^2In 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.