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 $4A = 4\begin{bmatrix}a&b\\c&d\end{bmatrix} = \begin{bmatrix}4a&4b\\4c&4d\end{bmatrix} = \det(A) = 16ad - 16bc$?

Solution 1:

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.

Solution 2:

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.