TITLE: If $A$ is a $2\times2$ matrix, what is $\det(4A)$ in terms of $\det(A)$?
QUESTION [11 upvotes]: If $A$ is a $2\times2$ matrix, what is $\det(4A)$ in terms of $\det(A)$?

This seems trivial, but I'm not sure exactly what they are asking. I'm guessing I have some matrix $A = \begin{bmatrix}a&b\\c&d\end{bmatrix}$ where I know $\det(A) = ad - bc$. So if they want to know what is $\det(4A)$ wouldn't it just be $4A = 4\begin{bmatrix}a&b\\c&d\end{bmatrix} = \begin{bmatrix}4a&4b\\4c&4d\end{bmatrix} = 
\det(A) = 16ad - 16bc$?

REPLY [2 votes]: Another view of the multilinear aspect of the determinant of a square matrix $A$ is the "recursive form" involving minors. If the matrix is $n\times n$, one can compute determinant with minors stemming from $(n-1)\times (n-1)$ matrices:
$$|A| = \sum (-1)^{i+j} a_{ij} M_{ij}\,.$$
The determinant of a   $1\times 1$ matrix (a scalar) grows linearly with the factor $\lambda$ applied to the (unique) matrix element. Hence the determinant of a $2\times 2$ matrix grows twice faster, since you get one $\lambda$ for the $a_{ij}$ terms (for instance $a$ and $c$ for you), and one for the  $M_{ij}$ terms, which derive  from dimension $1\times 1$ matrices (for instance $b$ and $d$ for you), as you wrote with $\lambda=4$. 
For a  $3\times 3$ matrix, you get one $\lambda$ for the $a_{ij}$ terms, and $\lambda^2$ for the $M_{ij}$ terms, coming from dimension $2\times 2$ objects. More generally i, you get one $\lambda$ for the $a_{ij}$ terms, and $\lambda^{n-1}$ for the $M_{ij}$ terms, which are computed from dimension $(n-1)\times (n-1)$ matrices, hence $\lambda^n $ total.