1.9. Using the property that the determinant of the matrix $C=A \cdot B$, which represents the product of square matrices $A$ and $B$ of the same order, is equal to the product of the determinants of matrices $A$ and $B$, i.e., $\operatorname{det} C=\operatorname{det} A \cdot \operatorname{det} B$, calculate the determinant of the matrix $\operatorname{det} C$.

$$
A=\left(\begin{array}{lll}
3 & 2 & 5 \\
0 & 2 & 8 \\
4 & 1 & 7
\end{array}\right), B=\left(\begin{array}{ccc}
-2 & 3 & 4 \\
-1 & -3 & 5 \\
0 & 4 & 3
\end{array}\right)
$$
Solution. We will find the determinants of matrices $A$ and $B$ and then multiply them. We have

$$
\operatorname{det} A=\left|\begin{array}{lll}
3 & 2 & 5 \\
0 & 2 & 8 \\
4 & 1 & 7
\end{array}\right|=\left|\begin{array}{ccc}
3 & 2 & -3 \\
0 & 2 & 0 \\
4 & 1 & 3
\end{array}\right|=2 \cdot(-1)^{2+2}\left|\begin{array}{cc}
3 & -3 \\
4 & 3
\end{array}\right|=2 \cdot(9+12)=42
$$

Here we multiplied the second column of the determinant by $(-4)$, added it to the third column, and then expanded the resulting determinant along the second row.

Similarly,

$$
\operatorname{det} B=\left|\begin{array}{ccc}
-2 & 3 & 4 \\
-1 & -3 & 5 \\
0 & 4 & 3
\end{array}\right|=\left|\begin{array}{ccc}
0 & 9 & -6 \\
-1 & -3 & 5 \\
0 & 4 & 3
\end{array}\right|=(-1) \cdot(-1)^{1+2}\left|\begin{array}{cc}
9 & -6 \\
4 & 3
\end{array}\right|=51
$$

Here we multiplied the second row by $(-2)$ and added it to the first row, and then expanded the resulting determinant along the first column. Multiplying the obtained values of the determinants, we find that $\operatorname{det} C=51 \cdot 42=2142$. It is easy to verify that we get the same result if we first find the matrix $C$ by multiplying matrices $A$ and $B$, in which case the matrix $C$ has the form:

$$
C=\left(\begin{array}{lll}
-8 & 23 & 37 \\
-2 & 26 & 34 \\
-9 & 37 & 42
\end{array}\right)
$$

and then compute $\operatorname{det} C$.

In addition to the method of reducing the order of the determinant (Example 1.7) and the method of reducing to a triangular form (Example 1.8), the method of recurrence relations is used when calculating determinants. The essence of the method is that the initial determinant $D_{n}$ of order $n$ is expressed in terms of determinants of the same type but of a lower order, i.e., by a recurrence formula of the form: $D_{n}=f\left(D_{n-1}, D_{n-2}, \ldots, D_{n-k}\right)$, valid for all natural numbers $n$ greater than $k$. From this relation, using the method of mathematical induction, a formula is obtained that expresses the determinant $D_{n}$ in terms of the determinants $D_{n-1}, D_{n-2}, \ldots, D_{n-k}$.

As an example of applying the specified method, we will compute the Vandermonde determinant:

$$
D_{n}=\left|\begin{array}{ccccc}
1 & 1 & 1 & \ldots & 1 \\
a_{1} & a_{2} & a_{3} & \ldots & a_{n} \\
a_{1}^{2} & a_{2}^{2} & a_{3}^{2} & \ldots & a_{n}^{2} \\
\ldots & \ldots & \ldots & \ldots & \ldots \\
a_{1}^{n-1} & a_{2}^{n-1} & a_{3}^{n-1} & \ldots & a_{n}^{n-1}
\end{array}\right|
$$

We will show that for any $n \geqslant 2$ the Vandermonde determinant is equal to the product of all possible differences $a_{i}-a_{j}$, where $1 \leqslant j<i \leqslant n$. Indeed, for $n=2$ we have

$$
D_{2}=\left|\begin{array}{cc}
1 & 1 \\
a_{1} & a_{2}
\end{array}\right|=a_{2}-a_{1}
$$

Suppose the statement is proved for Vandermonde determinants of order $n-1$:

$$
D_{n-1}=\prod_{1 \leqslant j<i \leqslant n-1}\left(a_{i}-a_{j}\right)
$$

where the symbol

$$
\prod_{i=1}^{n} a_{i}
$$

denotes the product of the elements $a_{i}$, i.e.,

$$
\prod_{i=1}^{n} a_{i}=a_{1} \cdot a_{2} \cdot \ldots \cdot a_{n}
$$

We will prove that this formula is also valid for $D_{n}$. For this, we subtract the $(n-1)$-th row from the last $n$-th row, multiplied by $a_{1}$, and then sequentially subtract the $(k-1)$-th row from the $k$-th row, multiplied by $a_{1}$. As a result, we get

$$
D_{n}=\left|\begin{array}{ccccc}
1 & 1 & 1 & \ldots & 1 \\
0 & a_{2}-a_{1} & a_{3}-a_{1} & \ldots & a_{n}-a_{1} \\
0 & a_{2}^{2}-a_{2} a_{1} & a_{3}^{2}-a_{3} a_{1} & \ldots & a_{n}^{2}-a_{n} a_{1} \\
\ldots & \ldots & \ldots & \ldots & \ldots \\
0 & a_{2}^{n-1}-a_{2}^{n-2} a_{1} & a_{3}^{n-1}-a_{3}^{n-2} a_{1} & \ldots & a_{n}^{n-1}-a_{n}^{n-2} a_{1}
\end{array}\right|
$$

Next, we expand this determinant along the first column and factor out the common factors from all columns. As a result, we obtain the recurrence relation:

$$
\begin{aligned}
D_{n}=\left(a_{2}-a_{1}\right) & \cdot\left(a_{3}-a_{1}\right) \ldots\left(a_{n}-a_{1}\right) \cdot\left|\begin{array}{cccc}
1 & 1 & & 1 \\
a_{2} & a_{3} & & a_{n} \\
a_{2}^{2} & a_{3}^{2} & & a_{n}^{2} \\
\ldots & \ldots & & \ldots \\
a_{2}^{n-2} & a_{3}^{n-2} & \ldots & a_{n}^{n-2}
\end{array}\right|= \\
& =\left(a_{2}-a_{1}\right) \cdot\left(a_{3}-a_{1}\right) \ldots\left(a_{n}-a_{1}\right) \times \\
& \times \prod_{2 \leqslant j<i \leqslant n}\left(a_{i}-a_{j}\right)=\prod_{1 \leqslant j<i \leqslant n}\left(a_{i}-a_{j}\right) .
\end{aligned}
$$