Answer
$-76$
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 column $2$ (because of the zeros).
The first position, (i,j)=(1,2), 1+2=3,
defines the sequence of alternating signs starting with a $-$ (minus).
$\left|\begin{array}{lll}
2 & -4 & 2\\
-1 & 0 & 5\\
3 & 0 & 4
\end{array}\right|=$
$=-(-4)\cdot\left|\begin{array}{ll}
-1 & 5\\
3 & 4
\end{array}\right|+0\cdot\left|\begin{array}{ll}
2 & 2\\
3 & 4
\end{array}\right|-0\cdot\left|\begin{array}{ll}
2 & 2\\
-1 & 5
\end{array}\right|$
$=4[-1(4)-5(3)]$
$=4(-19)$
$= -76$