10.1. Find all numbers $a$ and $b$ for which the equality $|a x+b y|+|b x+a y|=|x|+|y|$ holds for all values of the variables $x$ and $\cdot y$.
Answer. $a= \pm 1, b=0$ or $a=0, b= \pm 1$, a total of four pairs of values.

Solution. Substitute $x=1, y=0$ into the formula from the condition, we get $|a|+|b|=1$. Then substitute $x=1, y=1$, from which $|a+b|=1$. Finally, by setting $x=1, y=-1$, we have $|a-b|=1$. From the last two equations, we get $a+b= \pm(a-b)$, from which either $a=0$ or $b=0$. From the first equation, it then follows that $a= \pm 1, b=0$ or $a=0, b= \pm 1$.

Let's verify that the found pairs of $a$ and $b$ satisfy the condition of the problem. For example, if $a=1, b=0$, then $|a x+b y|+|b x+a y|=|a x|+|a y|=|a|(|x|+|y|)=|x|+|y|$. The other cases are verified similarly.

Grading criteria. If the correct answer with verification is provided: 2 points. Only the answer: 0 points. If there is no verification that the found candidates for the answer $a= \pm 1, b=0$ or $a=0, b= \pm 1$ satisfy the condition of the problem: deduct 2 points.