45. Calculate the determinant

$$
D=\left|\begin{array}{rrrr}
3 & 0 & 2 & 0 \\
2 & 3 & -1 & 4 \\
0 & 4 & -2 & 3 \\
5 & 2 & 0 & 1
\end{array}\right|
$$
Solution. We will expand the determinant along the elements of the 1st row (since it contains two zero elements):

$$
\begin{gathered}
D=\left|\begin{array}{rrrr}
3 & 0 & 2 & 0 \\
2 & 3 & -1 & 4 \\
0 & 4 & -2 & 3 \\
5 & 2 & 0 & 1
\end{array}\right|=3\left|\begin{array}{rrr}
3 & -1 & 4 \\
4 & -2 & 3 \\
2 & 0 & 1
\end{array}\right|-0 \cdot\left|\begin{array}{rrr}
2 & -1 & 4 \\
0 & -2 & 3 \\
5 & 0 & 1
\end{array}\right|+ \\
\\
+2\left|\begin{array}{lll}
2 & 3 & 4 \\
0 & 4 & 3 \\
5 & 2 & 1
\end{array}\right|-0 \cdot\left|\begin{array}{ccc}
2 & 3 & -1 \\
0 & 4 & -2 \\
5 & 2 & 0
\end{array}\right| .
\end{gathered}
$$

Since the second and fourth terms of the expansion are zero, we have

$$
\begin{aligned}
& D=3\left[3\left|\begin{array}{rr}
-2 & 3 \\
0 & 1
\end{array}\right|-(-1)\left|\begin{array}{ll}
4 & 3 \\
2 & 1
\end{array}\right|+4\left|\begin{array}{rr}
4 & -2 \\
2 & 0
\end{array}\right|\right]+2\left[2\left|\begin{array}{ll}
4 & 3 \\
2 & 1
\end{array}\right|-\right. \\
& \left.-3\left|\begin{array}{ll}
0 & 3 \\
5 & 1
\end{array}\right|+4\left|\begin{array}{ll}
0 & 4 \\
5 & 2
\end{array}\right|\right]=3[3(-2)-(-1)(4-6)+4 \cdot 4]+ \\
& +2[2(4-6)-3(-15)+4(-20)]=3(-6-2+16)+ \\
& +2(-4+45-80)=24-78=-54
\end{aligned}
$$

46-48. Calculate the third-order determinants: