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Let A be a 5 x 3 matrix. If $$b = a_1 + a_2 = a_2 + a_3$$

then what can you conclude about the number of solutions of the linear system Ax = b? Explain.

I'm not sure about this question. All I know is that if b can be written as a combination of column vectors a, then the linear system is consistent. I am not sure what this says about the number of solutions, however.

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    So, wait a second - are you using $a_j$ as notation for Column $j$ of the matrix $A$? If so, can you edit that information into the question? Without *some* explanation of what these $a_i$ are, the question makes no sense.2012-09-30
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    The question doesn't specify, but because this is a problem about linear combination, I assumed that the a's are column vectors. I wrote the question exactly as it is written in the book.2012-09-30
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    What book? What page? Can you put up a scan somewhere, and link to it?2012-09-30

2 Answers 2

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Since $A$ maps $(1, 0, 0)$ to $a_1$, $(0, 1, 0)$ to $a_2$ and $(0, 0, 1)$ to $a_3$, $(1, 1, 0)$ and $(0, 1, 1)$ are both solutions to $Ax = b$. Linear transformations give only one, zero or infinite solutions, thus there are infinite solutions. Alternatively, $m(1, 1, 0) + n(0, 1, 1)$ is a solution for any $m + n = 1$.

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    Thanks. That makes sense. So, if I had a similar problem b = a1 + a2 + a3 + a4, the correct answer would be that the linear system has one solution (1,1,1,1)?2012-09-30
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    Oh and it's a 3x4 matrix I forgot to mention2012-09-30
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Yeah, old question, but I came across it so someone else might, as well. More clearly, since b is a combination of vectors a1, a2, and a3 (either [1,1,0] or [0,1,1]), the system MUST be consistent and have at least one solution. Because the system is overdetermined (5x3), it could be exactly one solution OR infinitely many solutions, depending on whether the system is linearly independent.

MH234's system is underdetermined (3x4) so one of the variables is necessarily free and provides for infinitely many solutions.