$A(A^{17}-cI)=0$, what is $c$?
Solution 1:
Since $A$ is invertible we obtain $A^{17}=cI_3$. Taking determinants we have $$ 6^{17}=\det(A)^{17}=\det(A^{17})=\det(cI_3)=c^3. $$
$A(A^{17}-cI)=0$, what is $c$?
Solution 1:
Since $A$ is invertible we obtain $A^{17}=cI_3$. Taking determinants we have $$ 6^{17}=\det(A)^{17}=\det(A^{17})=\det(cI_3)=c^3. $$