Maximum and Minimum of Determinant For a $n \times n$ matrix X, and let its row vectors $X_1,...,X_n$ where each vectors' norm is 1.
then, find the maximum and minimum of  $\det X$.
and using this, prove that for any $n \times n$ matrix A, $|\det A| \leq\prod_{i=1}^{n}{||A_{i}||}^2$ 
 A: The norm you're referring to is often called the Euclidean norm on $\mathbb{R}^n$ and is denoted $\|\cdot \|_2$.
First assume that $\|X_i\|_2 = 1$ for each row vector $X_i$.
To answer the max/min question: in what follows I will be working with the transpose, out of habit moreso than anything else. What does $|\det X|$ actually represent? Geometrically, it's the euclidean volume of the image of the unit n-thtant $[0,1]^N$ in $\mathbb{R}^N$.
The volume of this image, then, is zero when this image is contained in any proper subspace of $\mathbb{R}^N$. Convince yourself that this is the case if and only if the vectors $\{X_i\}$ have a nontrivial linear relation (i.e. are linearly dependent). Therefore your minimum is zero.
For the maximum value, I'll show you an argument for $N = 2$; the general case is not much different. Let $X = (X_1 X_2)$ be the two column vectors of $X$. We have that $X_2 = X_2^{\perp} + \alpha X_1$, decomposed into a sum of a piece orthogonal to $X_1$, and a piece parallel to $X_1$. Notice that $0 < \|X_2^{\perp}\|_2 < 1$ when $X_1, X_2$ are linearly independent and not orthogonal. By multilinearity of the determinant,
\begin{align*}
\det X &= \det ( X_1 X_2) = \det(X_1 X_2^{\perp}) + \alpha \det(X_1 X_1) = \det(X_1 X_2^{\perp}) \\
& = \det\left(X_1 \frac{\|X_2^{\perp}\|_2}{\|X_2^{\perp}\|_2} X_2^{\perp}\right) 
= \|X_2^{\perp}\|_2 \det\left(X_1 \frac{X_2^{\perp}}{\|X_2^{\perp}\|_2}\right)
\end{align*}
The matrix inside $\det$ on the right hand side of the bottom line is an orthogonal matrix, as its columns are orthogonal and of unit length. Orthogonal matrices represent rigid rotations, and moreover are isometries of $\mathbb{R}^N$- for us, they send unit volumes to unit volumes, and so the determinant equals $\pm 1$.
We have shown, then, that $|\det X| = \|X_2^{\perp}\|_2 < 1$, and so it is now clear that $|\det X|$ attains its maximum value at $1$, in particular only when the column vectors are orthogonal.
Second, to show the general inequality, apply the multilinearity of the determinant to factor out the norms of the column vectors, and then apply the maximum we just proved to this matrix whose column vectors have unit length. Correct me if I'm wrong, but the inequality should read
$$
|\det A| \leq \prod_{i = 1}^N \|A_i\|_2
$$
where the $A_i$ are the column vectors of $A$.
