Question

If $\Delta =\left|\begin{array}{ccc}{a}_{11}& a_{12}& a_{13}\\ a_{21}& a_{22}& a_{23}\\ a_{31}& a_{32}& a_{33}\end{array}\right|$ and $c_{ij}={(-1)}^{i+j}$ (determinant obtained by deleting ith row and jth column), then $\left|\begin{array}{ccc}{c}_{11}& c_{12}& c_{13}\\ c_{21}& c_{22}& c_{23}\\ c_{31}& c_{32}& c_{33}\end{array}\right|={\Delta }^2$

If $\left|\begin{array}{ccc}1& x& x^2\\ x& x^2& 1\\ x^2& 1& x\end{array}\right|=7$ and $\Delta =\left|\begin{array}{ccc}{x}^3-1& 0& x-x^4\\ 0& x-x^4& x^3-1\\ x-x^4& x^3-1& 0\end{array}\right|$, then

• A. $\Delta =7$
• B. $\Delta =343$
• C. $\Delta =-49$
• D. $\Delta =49$

Answer 1

The correct option is D $\Delta =49$

For $\left|\begin{array}{ccc}1& x& x^2\\ x& x^2& 1\\ x^2& 1& x\end{array}\right|=7$
$\left|\begin{array}{ccc}{c}_{11}& c_{12}& c_{13}\\ c_{21}& c_{22}& c_{23}\\ c_{31}& c_{32}& c_{33}\end{array}\right|=\left|\begin{array}{ccc}{x}^3-1& 0& x-x^4\\ 0& x-x^4& x^3-1\\ x-x^4& x^3-1& 0\end{array}\right|$
$\Delta =7^2=49$