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I'm working on a problem that asks for all the real number vectors $b_1$,$b_2$ such that

$$ \begin{pmatrix} 4 & 1\\ 12 & 3\\ \end{pmatrix} * \begin{pmatrix} x \\ y \\ \end{pmatrix} = \begin{pmatrix} b_1\\ b_2\\ \end{pmatrix} $$

So far, I have reduced the matrix to $$ \begin{pmatrix} 4 & 1 & b_1 \\ 0 & 0 & -3b_1 + b_2 \\ \end{pmatrix} $$

How do I solve from here?

Thank you all!

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    Hint: what do the equations your last matrix is representing tell you about the last row? What about how $b_1$ and $b_2$ are related? – 2017-02-24
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    $-3b_1 + b_2$ must equal zero. – 2017-02-24
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    So is there some relationship between $b_1$ and $b_2$ that you could use to simplify your matrix? – 2017-02-24
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    For the equation to be consistent, the last row of the reduced matrix must be zero. – 2017-02-24

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We have: $$\begin{bmatrix} 4&1&b_1\\ 0&0&-3b_1+b_2 \end{bmatrix} $$ This really encodes the system of equations $$ 4x+y=b_1$$ $$ 0=-3b_1+b_2.$$ It's easy to see, then, that $3b_1=b_2$. So, we have that $b_1=\frac{1}{3}b_2.$ Thus, we have that $4x+y=b_1$, and so the solution set for this matrix equation is the line $y=-4x+b_1$. With these data, we can see that the range is of the set of vectors of the form $$ \begin{pmatrix} b_1\\ 3b_1 \end{pmatrix}$$

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    My professor has the answer as $ \begin{matrix} b_1 \\ 3b_1 \\ \end{matrix} $ What does this mean? – 2017-02-24
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    Based on our relation $3b_1=b_2$, we can see that anything in the range must be of the form $(b_1,3b_1)$. Perhaps I should add this to the answer. – 2017-02-24