If $A$ is a $2\times2$ matrix, what is $\det(4A)$ in terms of $\det(A)$? 
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 $4A = 4\begin{bmatrix}a&b\\c&d\end{bmatrix} = \begin{bmatrix}4a&4b\\4c&4d\end{bmatrix} = 
\det(A) = 16ad - 16bc$?
 A: In particular, $\det(4A) = 4^2 \det(A)$.  
Another way to think about it is to remember that the determinant of an $n \times n$ matrix gives the (signed) volume of the $n$-dimensional box enclosed by its column vectors.  Increasing each side length by a factor of $k$ will increase the volume by a factor of $k^n$.
A: determinant is a multilinear form so for all matrix $A\in M_n(\mathbb{R})$  and for all $\alpha\in \mathbb{R}$ 
$$
\det(\alpha A)=\alpha^n \det(A) 
$$
A: It is indeed trivial. With $A$ as you have it,  $\det(A)=ad-bc$.
Now, $\det(4A)=\det(\begin{bmatrix}
4a & 4b\\ 
4c & 4d
\end{bmatrix})$=$16ad-16bc=16(ad-bc)=16*\det(A)$.
Alternatively, just use the formula: if $A$ is a real valued $n \times n$ matrix, then $\det(cA) = c^n\det(A)$  $\forall$ $c\in \mathbb{R}$. 
A: Seeing the determinant as the volume of the unit cube $Q$ under the map $A$, one sees that the composition with the multiplication $x\mapsto kx$ will dialate the image $AQ$ by a factor $k$. 
But as the volume scales like $k^n$ in $n-$dimensional space, you get that 
$$\det(kA)=vol(kA(Q))=k^n vol(A(Q))=k^n\det(A).$$
A: Let's consider your equation $$4A = 4\begin{bmatrix}a&b\\c&d\end{bmatrix} = \begin{bmatrix}4a&4b\\4c&4d\end{bmatrix} = 
\det(A) = 16ad - 16bc$$
The first three objects in this are $2\times 2$ matrices while the last 2 are numbers.  So clearly these can't all be equal.  What you really want to say here is
$$\color{red}{\det(}4A\color{red}{)} = \color{red}{\det\left(\color{black}{4\begin{bmatrix}a&b\\c&d\end{bmatrix}}\right)} = \color{red}{\begin{vmatrix}\color{black}{4a}&\color{black}{4b}\\\color{black}{4c}&\color{black}{4d}\end{vmatrix}} = 
16ad - 16bc$$
Then just complete the logic with a final $$=16\det(A)$$ and you're done.
A: More generally, if $A$ is $n \times n$, $\det(cA) = c^n\det(A)$ for any scalar $c$. This is because the determinant is a multilinear function of its columns.
In fact, one can define the determinant as the unique function from $n \times n$ matrices to scalars that is $n$-linear alternating in the columns, and takes the value $1$ for the identity matrix.
A: 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.
