!
First part was very simple, second part though I've been wrking on it and am still confused.
Okay now let me start by saying that I know all the rules for determinant algebra (I think). The thing is that I see a fundamental issue here in that I think I need to add the matrices together but from what I know, there isn't a good formula for finding a determinant through matrix addition. The other option is to multiple rows but then the variables would have coefficients
I have decided that I could always just use one of the matrices and pull out the algebra for finding its determinant, and then use the relation of a b c d e f to find the determinant of the final matrix. But I think that would be missing the point of the problem no?
So I think I know how to get the answer but I think it'd be the wrong way of getting it, if that makes sense.
Thanks.
Determinants are multilinear functions of their column vectors (and its row vectors, too, BTW, but you need to consider column vectors here). So say you have $\det(\mathbf a, \mathbf b, \mathbf c)=u$ and $\det(\mathbf a, \mathbf d, \mathbf c)=v$ and you'd like to find $\det(\mathbf a, \mathbf e, \mathbf c)$, where $\mathbf e=\beta \mathbf b + \delta \mathbf d$. Then $\det(\mathbf a, \mathbf e, \mathbf c) = \beta\det(\mathbf a, \mathbf b, \mathbf c)+\delta \det(\mathbf a, \mathbf d, \mathbf c)=\beta u + \delta v$ by the linearity of the 2nd argument of the $\det$ function.
So you just need to check if $(1,-1,-3) \in \operatorname{span}\left((1,1,1),(1,2,3)\right)$ -- it is -- and then find the linear combination of $(1,1,1)$ and $(1,2,3)$ which gives you $(1,-1,-3)$.