 Solution

Expansion down the last column gives the characteristic polynomialThe quadratic polynomial has roots $(1\pm \sqrt{5})/2$, which are not rational numbers. For example, the rational root theorem restricts any rational root of this monic integer polynomial to $1$ or $-1$, and neither works. If $A$ had a diagonalization over $\mathbb{Q}$, its characteristic polynomial would be a product of linear factors with rational numbers as their roots. It is not, so $A$ is not diagonalizable over $\mathbb{Q}$.