How do you analyze this matrix to find its determinant?

Question

How do I apply the theorems regarding the determinant of matrix when that matrix is multiplied by a scalar?

Answer 1

Rewrite your first matrix $A = (A_{i,j})_{1\leqslant i,j \leqslant 3}$ and your second matrix $B = (B_{i,j})_{1\leqslant i,j \leqslant 3}$.

$\operatorname{det}(B) = \sum_{\sigma \in \mathcal{S}_3} sgn(\sigma) \prod_{i=1}^3 B_{i,\sigma(i)}$, where $\mathcal{S}_3$ is the group of all permutations of $\{1;2;3\}$ and $sgn(\sigma)$ is the signature of $\sigma$.
For all $1\leqslant j \leqslant 3$, we have that $B_{1,j} = 3 A_{1,j}$, $B_{2,j} = -2 A_{2,j}$ and $B_{3,j} = 6 A_{3,j}$. Hence, for any $\sigma \in \mathcal{S}_3$, we get $$\prod_{i=1}^3 B_{i,\sigma(i)} = 3\cdot(-2)\cdot 6 \cdot \prod_{i=1}^3 A_{i,\sigma(i)} = -36 \cdot \prod_{i=1}^3 A_{i,\sigma(i)} $$ All in all, we get $\boxed{\operatorname{det}(B) = -36\cdot \operatorname{det}(A)}$.