Answer
$-75$
Work Step by Step
To evaluate an nth-order determinant, where $n > 2$,
1. Select a row or column about which to expand.
2. Multiply each element $a_{ij}$ in the row or column by $(-1)^{i+j}$ times the determinant obtained by deleting the ith row and the $j\mathrm{t}\mathrm{h}$ column in the given array of numbers.
3. The value of the determinant is the sum of the products found in step 2.
-----
We select the column 3 (because of the zeros).
The first position, (i,j)=(1,3), 1+3=4,
defines the sequence of alternating signs starting with a +.
$\left|\begin{array}{lll}
3 & 1 & 0\\
-3 & 4 & 0\\
-1 & 3 & -5
\end{array}\right|=$
$=+0\cdot\left|\begin{array}{ll}
-3 & 4\\
-1 & 3
\end{array}\right|-0\cdot\left|\begin{array}{ll}
3 & 1\\
-1 & 3
\end{array}\right|+(-5)\cdot\left|\begin{array}{ll}
3 & 1\\
-3 & 4
\end{array}\right|$
$=-5[3(4)-1(-3)]$
$= -5(15)=-75$
