= Solution
{c}
Expansion down the last column gives the <characteristic polynomial>
$$
\chi_A(\lambda)=\det(\lambda I-A)=(\lambda-1)(\lambda^2-\lambda-1).
$$
The quadratic <polynomial> has <roots> $(1\pm\sqrt5)/2$, 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 \b[$A$ is not diagonalizable over $\mathbb Q$].
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