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 $a_{4} b_{4}$.
Solution. We will prove that ${ }^{1}$

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

![](https://cdn.mathpix.com/cropped/2024_05_06_3ff3c6fd678a38f38275g-1.jpg?height=63&width=780&top_left_y=2529&top_left_x=115)

Interregional Olympiad for Schoolchildren Based on Departmental Educational Organizations in Mathematics

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

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

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

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

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

Note that $\Delta \neq 0$, because 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 $\mathbf{A} \cdot \mathbf{B}=\left(\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right)$, 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 the identity matrix, then these matrices commute, which means the system of equations in the problem remains valid if all $a_{i}$ are replaced by $b_{i}$ and vice versa. From this observation, equality (1) follows immediately.