If a square matrix of coefficients is non-singular there will be a unique solution.
A non-singular matrix will have a non-zero determinant.
This matrix is fairly sparse so computing the determinant via minors shouldn't be that difficult.
Of course you can just dump it into Wolfram as well. It should be
Det[{{1,0,3,0},{0,1,0,-3},{0,-2,3,2},{3,0,0,7}}]
on the command line
I don't know what you mean by "determine at face value". If you mean "just look at it an immediately know", I can't help you. I am just not that good. I would have to actually do the calculations!
The system of equation is
$\displaystyle x_1+ 0x_2+ 3x_3+ 0x_4= 2$
$\displaystyle 0x_1+ x_2+ 0x_3- 3x_4= 2$
$\displaystyle 0x_1- 2x_2+ 3x_3+ 2x_4= 1$
$\displaystyle 3x_1+ 0x_2+ 0x_3+ 3x_4= -5$.
As romek said, that is fairly sparce- there are a whole lot of 0s in there.
We can get even more 0s by:
1) Subtract 3 times the first equation from the fourth and add twice the second equation to the third:
$\displaystyle x_1+ 0x_2+ 3x_3+ 0x_4= 2$
$\displaystyle 0x_1+ x_2+ 0x_3- 3x_4= 3$
$\displaystyle 0x_1+ 0x_2+ x_3- 4x_4= 7$
$\displaystyle 0x_1+ 0x_2- 9x_3+ 7x_4= -11$.
That is almost triangular.
2) Just add three times the third of those equations to the fourth to get
$\displaystyle x_1+ 0x_2+ 3x_3+ 0x_4= 2$
$\displaystyle 0x_1+ x_2+ 0x_3- 3x_4= 3$
$\displaystyle 0x_1+ 0x_2+ x_3- 4x_4= 7$
$\displaystyle 0x_1+ 0x_2+ 0x_3- 5x_4= 10$.
Which is triangular and it is easy to see that is a consistent set of equations. In fact, it is easy to solve the system by "back-substituting".