What property of determinants does this use?
I did the second part by adding together the first two determinants and then multiplied them by 3 since it looked like adding together the two matrices and then multiplying by 3 would give me a second column identical to the last matrix. The answer seems to be right but I don't understand why it's right because this doesn't seem to be following any determinant properties i'm aware of
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$\begingroup$The determinant is not a linear mapping. In particular, we do not have $\det(A_1+A_2)=\det(A_1)+\det(A_2)$ in general. So you cannot add the two matrices and then add their determinants to obtain the determinant of their sum.
What you can do: if you multiply a row in a matrix by a constant, then the determinant is changed by this constant. That is, if $B$ is obtained from $A$ by multiplying one of the rows by $k$, then $\det(B)=k\det(A)$. The same applies for columns (i.e. replace "rows" with "columns" in the last couple sentences). From this: if you multiply every row (or column) by $k$, then $\det(B)=k^n\det(A)$, where $A$ and $B$ are $n\times n$ matrices.
Edit (for the second matrix): we can use linearity in each column of the determinant (same goes for rows). That is, if $A=[a_1\cdots a_n]$ where $a_i$ are the column vectors of $A$, and $a_j=v+w$ for some $1\leq j\leq n$, then $$\det(A)=\det([a_1\cdots a_{j-1}~ v+w ~a_{j+1}\cdots a_n])=\det([a_1\cdots a_{j-1}~v~a_{j+1}\cdots a_n])+\det([a_1\cdots a_{j-1}~w~a_{j+1}\cdots a_n])$$
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