Answer
0
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 row 1, ($1$s are easy to work with).
The first position, (i,j)=(1,1), 1+1=2,
defines the sequence of alternating signs starting with a plus.
$\left|\begin{array}{lll}
1 & 1 & 1\\
2 & 2 & 2\\
-3 & 4 & -5
\end{array}\right|=$
$=+1\cdot\left|\begin{array}{ll}
2 & 2\\
4 & -5
\end{array}\right|-1\cdot\left|\begin{array}{ll}
2 & 2\\
-3 & -5
\end{array}\right|+1\cdot\left|\begin{array}{ll}
2 & 2\\
-3 & 4
\end{array}\right|$
$=(-10-8)-(-10+6)+(8+6)$
$=-18+4+14$
$=0$
