44. Determinant

$$
D=\left|\begin{array}{rrr}
3 & 1 & 2 \\
-1 & 2 & 5 \\
0 & -4 & 2
\end{array}\right|
$$

expand: a) by the elements of the 1st row; b) by the elements of the 2nd column.
Solution.

a) $D=3\left|\begin{array}{rr}2 & 5 \\ -4 & 2\end{array}\right|-1 \cdot\left|\begin{array}{rr}-1 & 5 \\ 0 & 2\end{array}\right|+2\left|\begin{array}{rr}-1 & 2 \\ 0 & -4\end{array}\right|=3(4+20)-$ $-1(-2-0)+2(4-0)=72+2+8=82$
b) $D=-1 \cdot\left|\begin{array}{rr}-1 & 5 \\ 0 & 2\end{array}\right|+2\left|\begin{array}{ll}3 & 2 \\ 0 & 2\end{array}\right|-(-4)\left|\begin{array}{rr}3 & 2 \\ -1 & 5\end{array}\right|=$ $=(-1)(-2)+2 \cdot 6-(-4) 17=2+12+68=82$.

If the determinant has a fourth or higher order, it can also be expanded along the elements of a row or column, and then reduce the order of the algebraic complements.