Find $2\det ( \frac{1}{2} A )$ given that $A$ is $3\times 3$ and $\det(A)= -2$ Here is a question that should be done today:

If $A$ is $3\times 3$ and $\det(A)= -2$, find $2\det(\frac{1}{2}A)$.

I solved this problem but I am not sure because the way I used is not accurate!
 A: The question is how do you find $\det(cA)$ given $\det(A)$ for any scalar $c$.
The answer is that $\det(cA)$ = $c^n\det(A)$ for an $n\times n$ matrix. For this reason, $$2|\frac{1}{2}A| = 2\left(\frac{1}{2}\right)^3|A| = \boxed{-\dfrac{1}{2}}$$
To demonstrate this identity for $n = 3$, consider that the determinant of a $3\times3$ matrix can be represented as the signed area of a parallelepiped made from the column (or row) vectors of the matrix. If each of the column vectors are multiplied by $c$, the length of the sides of the parallelepiped will be multiplied by $c$ and the volume will increase by a factor of $c^3$.
If you are interested, the volume of a parallelepiped is given by:
$$V = a b c \sqrt{1+2\cos(\alpha)\cos(\beta)\cos(\gamma)-\cos^2(\alpha)-\cos^2(\beta)-\cos^2(\gamma)}$$
where $a$, $b$, and $c$ are side lengths and $\alpha$, $\beta$, and $\gamma$ are internal angles. Internal angles are preserved when the length of the sides change so it is clear that multiplying each of the side lengths by a constant factor will increase the volume by the cube of that factor.
A: If $A$ is composed of column vectors $v_i$, then for some $c \in \mathbb{R}$:
$$\det(v_1, v_2, ..., cv_i, ..., v_n) = c \cdot \det(v_1, v_2, ..., v_i, ..., v_n)$$
So if an entire $n \times n$ matrix $A$ is multiplied by some $c \in \mathbb{R}$, then that $c$ can be pulled from each column.  And so $\det(cA) = c^n\det(A)$.
