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I'm asked to find out for what values of $a$ and $b$ is the vector $(-a,1,b)$ a linear combination of the set $$(a,1,2a),(0,1,2a),(2a,-1,a+2)$$ I tried to solve it by expressing the linear combination in matrix form, but I'm not being able to reduce it and solve it.
Is there a better way to do this?

Lourenco Entrudo
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    Please show how you "tried to solve it by expressing the linear combination in matrix form". I'd put the vectors in rows and try to row reduce the resulting matrix. There's a tutorial on using MathJax in Meta Math.SE, and I can help with details. – hardmath Jan 16 '22 at 03:34

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Consider the matrix$$M=\begin{bmatrix}a&0&2a\\1&1&-1\\2a&2a&a+2\end{bmatrix},$$whose columns are those three vectors. If its determinant is not $0$, then those vectors span $\Bbb R^3$, and therefore $(-a,1,b)$ is a linear combination of them. And $\det M=3a^2+2a$, which is $0$ if and only if $a=0$ or $a=-\frac23$. So, you only need to deal with these two cases.

If $a=0$, then your vectors are $(0,1,0)$, $(0,1,0)$, and $(0,-1,2)$, and $(-a,1,b)$ is $(0,1,b)$. Clearly, it is a linear combination of the three vectors for any value of $b$:$$(0,1,b)=\left(1+\frac b2\right)(0,1,0)+\frac b2(0,-1,2).$$

Can you deal with the case $a=-\frac23$ now?

José Carlos Santos
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