So if alpha was to equal something other than nine, you could keep row reducing until you have a trivial solution? I'm a little confused because it seems that you can have infinite solutions even if alpha was not nine.
I think given the math in the video, the only way to get a free variable is when alpha = 9. That was the point of performing the row reduction. If you can find another value that works, please let me know. However, given the manipulations that result in a row of all zeros except for the alpha-9 term, I'm pretty sure alpha = 9 is the only way this will occur. Hope that helps, Adam
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For what values of alpha are the vectors linearly independent? I am working on a similar problem, first found the value of alpha where vectors are dependent, but not sure how to show for which values of alpha the vectors are linearly independent. Your videos are very helpful! Thank you:)
+Lindsay Anderson The easiest way to tell if a set of three vectors of length 3 is linearly independent is to compute the determinant of the matrix where each column of the matrix corresponds to one of the vectors. Specifically, let A = [2 3 -1; -4 2 -6; 5 -4 a]. Then, compute det(A). I used the Wolfram 3x3 determinant calculator (see this link: www.wolframalpha.com/widgets/view.jsp?id=7fcb0a2c0f0f41d9f4454ac2d8ed7ad6) and found that det(A) = 16a--144 As long as det(A) is NOT equal to zero, this means that the set of vectors are linearly independent. So, there are quite a few values for a that you can choose that yield det(A) not equal to zero. Hope that helps, thanks for watching! Adam
We're trying to find a solution to Ax = 0. One obvious answer is when x is equal to the all zero vector. Since this is always a solution to Ax = 0, we call this the "trivial solution". The work in the video shows that when alpha = 9, we end up with a system of equations with x3 being a free variable. This means there are an INFINITE number of solutions to Ax = 0, and almost all of these values for x are vectors that AREN'T the all-zero vector. So, we say that we've found a non-trivial solution. Hope that helps! Adam
Yes, that's correct. The value of alpha = 9 is the only value that will yield linearly dependent vectors. For all other values of alpha, the vectors will be independent.
that Netflix ad reminded me of BREAKING BAD and then I was wondering can WE do something like them (they using chemistry) with using this linear algebra......????
