Given a matrix $A\in M_3(Z_{11})$, calculate $\det(3A)^{-1}$ given $\det A=4$

Question

$(3A)^{-1} = \frac{1}{3}A^{-1}$

So that, $\det(3A)^{-1} = (\frac{1}{3})^3\det(A^{-1}) = \frac{1}{27}*\frac{1}{4} = \frac{1}{9}$

Am I correct?

Answer 1

As you work in $\mathbf Z/11\mathbf Z$, you cannot really use fractions, which live i, $\mathbf Q$, and you have to replace $\frac13$ and $\frac14$ with $3^{-1}=4$ and $4^{-1}=3$ repectively. Other than that, the formulæ are correct, if they are interpreted properly. This $$\det\bigl((3A)^{-1}\bigr)=\bigl(\det(3A)\bigr)^{-1}=(3^3\det A)^{-1}=(5\cdot4)^{-1}=9^{-1}=5. $$