2. Eight numbers $a_{1}, a_{2}, a_{3}, a_{4}$ and $b_{1}, b_{2}, b_{3}, b_{4}$ satisfy the relations

$$
\left\{\begin{array}{c}
a_{1} b_{1}+a_{2} b_{3}=1 \\
a_{1} b_{2}+a_{2} b_{4}=0 \\
a_{3} b_{1}+a_{4} b_{3}=0 \\
a_{3} b_{2}+a_{4} b_{4}=1
\end{array}\right.
$$

It is known that $a_{2} b_{3}=7$. Find $a_{4} b_{4}$.
Solution. We will prove that ${ }^{1}$

$$
a_{2} b_{3}=a_{3} b_{2}
$$

Multiply equation (a) of the original system by $b_{2}$ and subtract from it equation (b) multiplied by $b_{1}$. The result is

$$
a_{2} \cdot \Delta=b_{2}
$$

Here $\Delta=b_{2} b_{3}-b_{1} b_{4}$. Similarly, from (c) and (d) we find that

$$
a_{3} \cdot \Delta=b_{3}
$$

Note that $\Delta \neq 0$, otherwise from (3) it would follow that $b_{3}=0$, and thus $a_{2} b_{3}=0$, which contradicts the problem's condition. It remains to express $a_{2}$ and $a_{3}$ from (2) and (3) and substitute the obtained expressions into (1). The validity of relation (1) will thereby be proven.

Further, from equation (d) and equality (1), it follows that $a_{4} b_{4}=1-a_{3} b_{2}=1-a_{2} b_{3}=-6$.

Answer: $a_{4} b_{4}=-6$.

Comment. ${ }^{1}$ The system of equations in the problem is the component-wise recording of the matrix equality

$$
\mathbf{A} \cdot \mathbf{B}=\left(\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right) \text {, where } \mathbf{A}=\left(\begin{array}{ll}
a_{1} & a_{2} \\
a_{3} & a_{4}
\end{array}\right) \text { and } \mathbf{B}=\left(\begin{array}{ll}
b_{1} & b_{2} \\
b_{3} & b_{4}
\end{array}\right)
$$

It is well known that if the product of two matrices is the identity matrix, then these matrices commute, which means the system of equations in the problem will remain valid if all $a_{i}$ are replaced by $b_{i}$ and vice versa. From this observation, equality (1) follows immediately.