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Suppose $A$ is a $3\times 3$ matrix with columns $v_{1}, v_{2}, v_{3}$.

If $b = 2v_{1}- v_{3}$, then $Ax = b$ has one or more solutions.

Is this true or false?

Is this false, since $Ax= b$ has solutions iff $A$ has a nonzero determinant, so the fact that $b = 2v_{1}-v_{3}$ is independent of the fact that $Ax=b$ has solutions?

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what you're probably forgetting is that when $A$ is a matrix and $x$ is a column vector, $Ax$ is a linear combination of the columns of $A$. – symplectomorphic Jul 18 '14 at 11:18
A non-zero determinant is required only for a unique solution. – James Jul 18 '14 at 14:52
up vote 4 down vote accepted

$Ax = b$ has solutions whenever $b$ lies in the image (column space) of $A$. Since $b$ given in your question is a linear combination of the columns of $A$, it is by definition in the image of $A$. Hence, a solution exists.

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Hint. No that's wrong. What can you say about $A \cdot (2, 0, -1)^t$ in terms of the $v_i$? $A$ being invertible gives solutions for all $b$, otherwise the solvability depends on $b$.

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+1 Just ahead of me – drhab Jul 18 '14 at 8:49

When $A$ has a non-zero determinante, i.e. $A$ is invertible, you can put whatever you want for $b$ and calculate $x=A^{-1}b$...and with the given $b$ and assuming orthogonality between the $v_k$'s you can even calculate $x$ by hand...

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For a $n\times n$ matrix $A$ composed of columns $\{v_1,v_2,\dots,v_n\}$ and a vector $x=[x_1,x_2,\dots,x_n]^\top\in \mathbb R^n$, we have that

$$Ax=x_1 v_1+x_2v_2+\dots x_n v_n$$

In your case, can you now find such an $x$ that $Ax=2v_1-v_3$?

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