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

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

It is known that $a_{2} b_{3}=7$. Find $a_{4} b_{4}$.

Interregional Olympiad for Schoolchildren based on Departmental Educational Organizations
Solution. We will prove that ${ }^{1}$

Multiply the equation (a) of the original system

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

$$
\begin{cases}a_{1} b_{1}+a_{2} b_{3}=1 & (\mathrm{a}) \\ a_{1} b_{2}+a_{2} b_{4}=0 & \text { (b) } \\ a_{3} b_{1}+a_{4} b_{3}=0 & \text { (c) } \\ a_{3} b_{2}+a_{4} b_{4}=1 & \text { (d) }\end{cases}
$$

by $b_{2}$ and subtract from it the equation (b), multiplied by $b_{1}$. As a result, we get

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

Here $\boldsymbol{\Delta}=\mathbf{b}_{\mathbf{2}} \boldsymbol{b}_{\mathbf{3}}-\boldsymbol{b}_{\mathbf{1}} \boldsymbol{b}_{\mathbf{4}}$. 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 condition of the problem. It remains to express $a_{2}$ and $a_{3}$ from (2) and (3) and substitute the obtained expressions into (1). The validity of the 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 $\mathbf{A} \cdot \mathbf{B}=\left(\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right)$, where $\mathbf{A}=\left(\begin{array}{ll}a_{1} & a_{2} \\ a_{3} & a_{4}\end{array}\right)$ 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 equal to the identity matrix, then such matrices commute, which means that 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.