Sparse $3\times 3$ matrix

Expand along the first row, where a zero entry saves you one of the terms.

1. Along row $1$ the entries are $2$, $0$ and $1$, so the middle term vanishes.
2. Deleting row $1$ and column $1$ leaves a minor equal to $4\times 6-0\times 5=24$, which contributes $2\times 24=48$.
3. Deleting row $1$ and column $3$ leaves a minor equal to $3\times 5-4\times 0=15$; its sign in the checkerboard is positive, so it contributes $1\times 15=15$.
4. So $\det (A)=48+15=63$.

Answer: $63$