Find maximal possible determinant value given constraint Task is to find maximal possible determinant value for 2x2 and 3x3 matrices given following constraint: 
$$\sum_{i,j=1}^na_{ij}^2 \le 1$$
I was able to come up with solution, but I received the test result, where I scored zero for this task. I still hasn't got a chance to take a look at my examined paper, but for now I just want to find whether or not my solution was actually true.
Solution
First trivial idea is that optimum would be reached in the point of exact equality $\sum_{i,j=1}^na_{ij}^2 = 1$.
Then, terms which participate in det calculation with minus sign, must be negative or zero.
$$\left( \begin{matrix}  \frac 1 {\sqrt 2} &amp; 0 \\
0 &amp; \frac 1 {\sqrt 2} \end{matrix}\right)$$

$$\left( \begin{matrix} \frac 1 {\sqrt 3} &amp; 0 &amp; 0 \\
0 &amp; \frac 1 {\sqrt 3} &amp; 0 \\
0 &amp; 0 &amp; \frac 1 {\sqrt 3} \end{matrix}\right)$$
So, I came up with these matrices and respective det values are $\frac 1 2$ and $\frac 1 {3\sqrt 3}$.  

Is that correct?
Thanks!
Here is a geometric argument that I would give:

It is always possible to decompose any matrix into a product of an orthogonal matrix and an upper triangular matrix. Orthogonal transformation is isometry, so $\sum\limits_{i,j=1}^{n}a_{ij}^{2}$ is preserved on the upper triangular matrix. Orthogonal matrix have determinant 1 or -1, so no effects on absolute value of determinant. Hence we need to look only at determinant of upper triangular matrix with the same constraint, but determinant there are only affected by the diagonal.
It's not obvious to me that it would be impossible to make the determinant larger in some way by having larger terms subtracted, if this caused larger terms to be added at the same time. Can you clarify why you think the terms participating with a minus sign would need to be zero or negative? 

Edit:

Let the columns be denoted by $u, v, w$.

I agree about the maximum being attained on the surface where $|u|^2 + |v|^2 + |w|^2 = 1$.

The differential of $\det(u,v,w)$ is $(h,k,l) \mapsto \det(h,v,w) + \det(u,k,w) + \det(u,v,l) = (v \times w) \cdot h + (w \times u) \cdot k + (u \times v) \cdot l$.

The differential of $|u|^2 + |v|^2 + |w|^2$ is $(h,k,l) \mapsto 2u \cdot h + 2v \cdot k + 2w \cdot l$.

The Lagrange multiplier method implies that at a point that optimizes the determinant, these linear functionals need to be proportional. So the triple of vectors $(u,v,w)$ needs to be proportional to $(v \times w, w \times u, u\times v)$. This proves that $u$, $v$ and $w$ all need to be orthogonal, and after an (orthogonal) change of basis, you can assume your matrix is diagonal. From there I reach the same result as you.