Answer
72
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.
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We select the first row (because of the zeros).
The first position, (i,j)=(1,1), 1+1=2,
defines the sequence of alternating signs starting with a +.
$\left|\begin{array}{lll}
3 & 0 & 0\\
2 & 1 & -5\\
2 & 5 & -1
\end{array}\right|=$
$=+3\cdot\left|\begin{array}{ll}
1 & -5\\
5 & -1
\end{array}\right|-0\cdot\left|\begin{array}{ll}
2 & -5\\
2 & -1
\end{array}\right|+0\cdot\left|\begin{array}{ll}
2 & 1\\
2 & 5
\end{array}\right|$
$=3[1(-1)-5(5)]$
$=3(24)$
$= 72$
