Question

If $\Delta =\left|\begin{array}{ccc}1& a& a^2\\ a& a^2& 1\\ a^2& 1& a\end{array}\right|=-4$ then find the value of $\left|\begin{array}{ccc}{a}^3-1& 0& a-a^4\\ 0& a-a^4& a^3-1\\ a-a^4& a^3-1& 0\end{array}\right|.$

Answer 1

Given that $\Delta =\left|\begin{array}{ccc}1& a& a^2\\ a& a^2& 1\\ a^2& 1& a\end{array}\right|=-4.$
Consider $C_{ij}$ be the cofactor of element $a_{ij}$.
Then $C_{11}=a^3-1,C_{12}=0,C_{13}=a-a^4;C_{21}=0,C_{22}=a-a^4,C_{23}=a^3-1;C_{31}=a-a^4,C_{32}=1-a^3,C_{33}=0.$
So determinant formed by using the cofactors of $\Delta$ is $\left|\begin{array}{ccc}{a}^3-1& 0& a-a^4\\ 0& a-a^4& a^3-1\\ a-a^4& a^3-1& 0\end{array}\right|={\Delta }_1$ say.
As we know that $=\Delta =\left|\begin{array}{ccc}{C}_{11}& C_{12}& C_{13}\\ C_{21}& C_{22}& C_{23}\\ C_{31}& C_{32}& C_{33}\end{array}\right|={\Delta }^{3-1}={\Delta }^2$
(Here we've used $|adj.A|=|A{|}^{n-1},$ where n is order of A; also $|A|=|A^T|.$)
Hence ${\Delta }_1=(-4{)}^2=16.$