As suggested in the comments, Gauss elimination is usually the way to go, and the fastest in this case, too:
$$\det A=
\det\begin{pmatrix}
2 & 3 & 1 & 0 \\
4 & -2 & 0 & -3\\
8 & -1 & 2 & 1\\
1 & 0 & 3 & 4\\
\end{pmatrix}
=
\det\begin{pmatrix}
0 & 3 & -5 & -8 \\
0 & -2 & -12 & -19\\
0 & -1 & -22 & -31\\
1 & 0 & 3 & 4\\
\end{pmatrix}
=
(-1)^{4+1}\cdot 1\cdot\det\begin{pmatrix}
3 & -5 & -8 \\
-2 & -12 & -19\\
-1 & -22 & -31\\
\end{pmatrix}
=
-\det\begin{pmatrix}
0 & -71 & -101 \\
0 & 32 & 43\\
-1 & -22 & -31\\
\end{pmatrix}
=
-1\cdot(-1)^{3+1}\cdot(-1)\cdot\det\begin{pmatrix}
-71 & -101 \\
32 & 43\\
\end{pmatrix}
= (-71)\cdot 43-(-101)\cdot 32=179
$$
(Wolfram Alpha-verified result; I never could remember the 3x3-formula, so I don't use it)
If you absolutely want an upper diagonal matrix, you can do this, but it's only a restriction of the normal algorithm:
$$\det A=
\det\begin{pmatrix}
2 & 3 & 1 & 0 \\
4 & -2 & 0 & -3\\
8 & -1 & 2 & 1\\
1 & 0 & 3 & 4\\
\end{pmatrix}
=
\det\begin{pmatrix}
2 & 3 & 1 & 0 \\
0 & -8 & -2 & -3\\
0 & -13 & -2 & 1\\
0 & -\frac12 & \frac52 & 4\\
\end{pmatrix}
=
\det\begin{pmatrix}
2 & 3 & 1 & 0 \\
0 & -8 & -2 & -3\\
0 & 0 & ? & ?\\
0 & 0 & ? & ?\\
\end{pmatrix}
=
\det\begin{pmatrix}
2 & 3 & 1 & 0 \\
0 & -8 & -2 & -3\\
0 & 0 & ? & ?\\
0 & 0 & 0 & ?\\
\end{pmatrix}
$$
(I'm too lazy to calculate the $?$ now, just continue with the Gaussian Elimination. The determinant will then be the product of the entries on the diagonal.)
For the eigenvalues: yes, you have to calculate the characteristic polynomial.
In your example, if you interchange two first colums you'll get
$$
\left[\matrix{
x_1 & x_2 & x_3 \\
2 & 1 & 3 \\
4 & 4 & 2 \\
1 & 1 & 4}\right]\quad\sim\quad
\left[\matrix{
x_2 & x_1 & x_3 \\
1 & 2 & 3 \\
4 & 4 & 2 \\
1 & 1 & 4}\right].
$$
What happened in the top most row is that the roles of $x_1$ and $x_2$ have been interchanged. When you get a solution to the modified system as $(2,-2,4)$, you must not forget to switch the variables back to the original ones.
Similar thing happens if you take, say, the first column minus the second column
$$
\left[\matrix{
x_1 & x_2 & x_3 \\
2 & 1 & 3 \\
4 & 4 & 2 \\
1 & 1 & 4}\right]\quad\sim\quad
\left[\matrix{
\tilde x_1 & \tilde x_2 & \tilde x_3 \\
1 & 1 & 3 \\
0 & 4 & 2 \\
0 & 1 & 4}\right].
$$
To see the relation between the old and the new variables, we write
$$
A_1x_1+A_2x_2+A_3x_3=(A_1-A_2)\tilde x_1+A_2\tilde x_2+A_3\tilde x_3=
A_1\tilde x_1+A_2(\tilde x_2-\tilde x_1)+A_3\tilde x_3.
$$
hence, $x_1=\tilde x_1$, $x_2=\tilde x_2-\tilde x_1$ and $x_3=\tilde x_3$, i.e. your variables has changed again. It means that
working with columns you are working with variables while working with rows you are working with equations.
In general, elementary row operations are equivalent to left multiplication by an elementary matrix transformation
$$
Ax=b\quad\implies\quad \underbrace{LA}_{\tilde A}x=\underbrace{Lb}_{\tilde b},
$$
while column operation corresponds to right multiplication by an elementary matrix. Because this multiplication has to be done for $A$ and matrix multiplication is not commuting, we have to compensate this action by the inverse matrix
$$
Ax=b\quad\implies\quad \underbrace{AR}_{\tilde A}\underbrace{R^{-1}x}_{\tilde x}=b.
$$
Best Answer
Given $$\begin{vmatrix} 4 & -7 & 9 & 1 \\ 6 & 2 & 7 & 0 \\ 3 & 6 & -3 & 3 \\ 0 & 7 & 4 & -1 \end{vmatrix}_{R_1<->R_2}$$ $$\begin{vmatrix} 6 & 2 & 7 & 0 \\ 4 & -7 & 9 & 1 \\ 3 & 6 & -3 & 3 \\ 0 & 7 & 4 & -1 \end{vmatrix}_{R_2\rightarrow R_2-\frac23 R_1}$$ $$\begin{vmatrix} 6 & 2 & 7 & 0 \\ 0 & -\frac { 25 }{ 3 } & \frac { 13 }{ 3 } & 1 \\ 3 & 6 & -3 & 3 \\ 0 & 7 & 4 & -1 \end{vmatrix}_{R_3\rightarrow R_3-\frac12R_1}$$ $$\begin{vmatrix} 6 & 2 & 7 & 0 \\ 0 & -\frac { 25 }{ 3 } & \frac { 13 }{ 3 } & 1 \\ 0 & 5 & -\frac { 13 }{ 2 } & 3 \\ 0 & 7 & 4 & -1 \end{vmatrix}_{R_3\rightarrow R_3+\frac35R_2}$$ $$\begin{vmatrix} 6 & 2 & 7 & 0 \\ 0 & -\frac { 25 }{ 3 } & \frac { 13 }{ 3 } & 1 \\ 0 & 0 & -\frac { 39 }{ 10 } & \frac { 18 }{ 5 } \\ 0 & 7 & 4 & -1 \end{vmatrix}$$ and so on.... till you get $0's$ in the first $3$ columns of the last row and then you will get the value of the determinant