2. Let $(a, b)$ be real numbers. Then the minimum value of $a^{2}+a b+b^{2}-a-2 b$ is . $\qquad$
Hint: Use the method of completing the square. The original expression can be transformed into
$$
\left(a+\frac{1}{2} b-\frac{1}{2}\right)^{2}+\frac{3}{4}(b-1)^{2}-1 \geqslant-1 \text {. }
$$

When $a+\frac{1}{2} b-\frac{1}{2}=0, b-1=0$, i.e., $a=0$, $b=1$, the original expression can achieve the minimum value of -1.