Maximize $\det (A)$ subject to $\|A\|_{\text{F}} \le 1$ 
$A$ is an $n \times n$ matrix such that the sum of squares of its elements is less than or equal to 1.
What is $\max \det (A)$?
(a) $n = 2$
(b) $n = 3$

My partial solution
For $n = 2$, I think the answer is 0.5. Define the constrained maximisation problem as $$\max(a_{11}a_{22}-a_{21}a_{12})$$ such that $$a_{11}^2+a_{12}^2+a_{21}^2+a_{22}^2 \le 1$$
If $a_{21}=a_{12}=0$, the maximum $0.5$ is achieved at $a_{11}=a_{22}=1/\sqrt{2}$.
If $a_{21} \ne 0, a_{12}\ne0$, the maximum $0.5$ is achieved at $a_{11}=a_{22}=a_{21}=0.5$ and $a_{12}=-0.5$
But how to I prove that formally? And how to I tackle the more difficult case of $n=3$?
 A: Let $v_1, \ldots, v_n$ be the column vectors of $A$.
By Hadamard's inequality, we have
$$\det(A) = \prod_{i=1}^n \|v_i\|$$
We are told the sum of squares of $A$'s  elements $\le 1$. In terms of $v_i$, this
means
$$\sum_{i=1}^n \|v_i\|^2 = 1$$
By GM $\le$ AM, we have
$$\left(\prod_{i=1}^n \|v_i\|^2\right)^{1/n} \le \frac1n \sum_{i=1}^n \|v_i\|^2 = \frac1n
\quad\implies\quad\det(A) \le \frac{1}{n^{n/2}}$$
Notice when $A = \frac{1}{\sqrt{n}} I_n$, the equality on RHS is achieved. 
This means above inequality is an optimal one.
Update
People seem to be interested in an explicit proof for the $n = 3$ case.
For completeness, here is a proof of Hadamard's inequality at $n = 3$.
Notice


*

*For $3 \times 3$ matrix $A$ with columns $v_1,v_2,v_3$; we have  $\det A = v_1 \cdot (v_2 \times v_3)$.

*For any $u, v \in \mathbb{R}^3$, we have the vector identity $|u|^2 |v|^2 = (u\cdot v)^2 + |u \times v|^2$. 


We find
$$\begin{align}|\det A|^2 
&= |v_1 \cdot (v_2 \times v_3)|^2 = |v_1|^2|v_2 \times v_3|^2 -
|v_1 \times (v_2 \times v_3)|^2\\
&= |v_1|^2|v_2|^2|v_3|^2 - ( |v_1|^2 (v_2 \cdot v_3)^2 +
|v_1 \times (v_2 \times v_3)|^2)\\ &\le |v_1|^2|v_2|^2|v_3|^2
\end{align}
$$
A: Maybe one should look at $\det (A)$ as a volume of cube spanned by its rows. Then $\det (A)$ is maximal when the rows are perpendicular, and in this case $\det (A)$ is the product of the lengths of the rows. So, the problem reduces to the classical inequality between arithmetical and geometrical means.
A: For $n=2$, you can use the inequality $a_{11} a_{22} \le 0.5(a_{11}^2+a_{22}^2)$ and the same for $a_{21}$, $a_{12}$ to see that 
$$a_{11}a_{22}-a_{21}a_{12} \le \frac{a_{11}^2+a_{22}^2+a_{12}^2+a_{21}^2}{2} = \frac 12$$.
For the case $n=3$ look here:
Update: no, it was not a good idea.
