1.7. Calculate the determinant

$$
\Delta=\left|\begin{array}{cccc}
3 & 1 & -1 & 2 \\
-3 & 1 & 4 & -5 \\
2 & 0 & 1 & -1 \\
3 & -5 & 4 & -4
\end{array}\right|
$$
S o l u t i o n. Notice that the second column of the determinant already contains one zero element. Add to the elements of the second row the elements of the first row multiplied by -1, and to the elements of the fourth row - the elements of the first row multiplied by 5. We get:

$$
\Delta=\left|\begin{array}{cccc}
3 & 1 & -1 & 2 \\
(-3-3) & (1-1) & (1+4) & (-2-5) \\
2 & 0 & 1 & -1 \\
(15+3) & (5-5) & (-5+4) & (10-4)
\end{array}\right|=\left|\begin{array}{cccc}
3 & 1 & -1 & 2 \\
-6 & 0 & 5 & -7 \\
2 & 0 & 1 & -1 \\
18 & 0 & -1 & 6
\end{array}\right| .
$$

Expanding the obtained determinant along the second row, we have:

$$
\Delta=(-1)^{1+2} \cdot 1 \cdot\left|\begin{array}{ccc}
-6 & 5 & -7 \\
2 & 1 & -1 \\
18 & -1 & 6
\end{array}\right|=-2 \cdot\left|\begin{array}{ccc}
-3 & 5 & -7 \\
1 & 1 & -1 \\
9 & -1 & 6
\end{array}\right|
$$

(Then we factored out the multiplier 2 from the first column based on property 4.) Next, add to the elements of the first and second columns the elements of the third column of the determinant. We get:

$$
\Delta=-2 \cdot\left|\begin{array}{ccc}
-10 & -2 & -7 \\
0 & 0 & -1 \\
15 & 5 & 6
\end{array}\right|=-2 \cdot 5 \cdot\left|\begin{array}{ccc}
-2 & -2 & -7 \\
0 & 0 & -1 \\
3 & 5 & 6
\end{array}\right|=10 \cdot\left|\begin{array}{ccc}
2 & 2 & 7 \\
0 & 0 & -1 \\
3 & 5 & 6
\end{array}\right|
$$

Here we factored out the multiplier in the first column, and then the common multiplier $(-1)$ in the first row. Expanding the resulting third-order determinant along the elements of the second row, we get:

$$
\Delta=10 \cdot(-1) \cdot(-1)^{2+3}\left|\begin{array}{cc}
2 & 2 \\
3 & 5
\end{array}\right|=10 \cdot(10-6)=40
$$

Here the second-order determinant is calculated according to its definition, by the formula

$$
\left|\begin{array}{ll}
a_{11} & a_{12} \\
a_{21} & a_{22}
\end{array}\right|=a_{11} \cdot a_{12}-a_{21} \cdot a_{22}
$$