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T or F

If $\mathbf{A}\vec{x} = \vec{b_{1}}$ has no solution, $\mathbf{A}\vec{x} = \vec{b_{2}}$ has many solutions, then $\mathbf{A}\vec{x} = \vec{b_{1}} + \vec{b_{2}}$ has many solutions.


This is false because the addition of the two $\vec{b}$ vectors doesn't mean the sum is a multiple of either $\vec{b_{1}}$ or $\vec{b_{2}}$ and will not necessarily fall in the same plane as either? Meaning it could result in either no solution or many solutions since $\mathbf A$ doesn't span the whole space?

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I don't quite understand your reasoning, so I'll just go on and disprove the statement :)

Choose $x_2$ to be the solution of $Ax_2 = b_2$ and $x_1$ to fulfil $x_1 + x_2 = x$ where $x$ is assumed to be any solution of $Ax = b_1 + b_2$. Now we have

$b_1 + b_2 = Ax = A (x_1 + x_2) = Ax_1 + Ax_2 = Ax_1 + b_2$

Subtracting $b_2$ from the first and the last term gives:

$Ax_1 = b_1$, which is a contradiction, because $x_1$ does not exist.

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    Pretty much so. The assumption which lead to the contradiction was that a solution of $Ax = b_1 + b_2$ exists.2012-09-19