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Given that A $\begin{bmatrix}1\\-2\\1\end{bmatrix}=\begin{bmatrix}2\\-3\\-5\end{bmatrix}$ and A $\begin{bmatrix}1\\4\\-2\end{bmatrix}=\begin{bmatrix}0\\2\\-1\end{bmatrix}$ find A $\begin{bmatrix}1 && 1\\-2 && 4\\1 && -2\end{bmatrix}$

Not really sure how to approach this question. Help would be much appreciated :)

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Hint: $A[v_1 v_2] = [Av_1 Av_2]$

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Oh wow, thank you very much. I have just started learning Matrix Algebra and this popped up under an extension question in my workbook... is there a rule or something that I can google to understand the intuition behind this? – user162048 Jul 7 '14 at 9:36
I think what you want to do is understand how you multiply two matrices. Once you understand that, the solution should be immediate. – Gerry Myerson Jul 7 '14 at 9:58
The wikipedia article about matrix multiplication ( provides a lot of intuitive ways to understand matrix multiplications. – Surb Jul 7 '14 at 9:59

To get more intuition for working with matrices, it might also help to consider a $n$ x $m$ matrix A as a representation of a linear transformation from an m-dimensional vector space $V$ to an n-dimensional vector space $W$, see:

A is not just a table of numbers with related operations; it produces elements of $W$ when fed with elements of $V$.

If you feed A with one $V$ element, it produces one $W$ element; if you feed it with two, it produces two. Also note that the linearity of the transformation implies (and is defined by) that for $v_1, v_2 \in V$ and $\lambda, \mu$ scalars, A fed with $\lambda v_1 + \mu v_2$ will produce $\lambda Av_1 + \mu Av_2$, thus: $A(\lambda v_1 + \mu v_2) = \lambda A(v_1) + \mu A(v_2)$

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