MCQ
If $A = \left[ {\begin{array}{*{20}{c}}1&3\\2&1\end{array}} \right]$, then determinant of ${A^2} - 2A$ is
- A$5$
- ✓$25$
- C$-5$
- D$-25$
${A^2} = \left[ {\begin{array}{*{20}{c}}1&3\\2&1\end{array}} \right]\,\left[ {\begin{array}{*{20}{c}}1&3\\2&1\end{array}} \right]\, = \,\left[ {\begin{array}{*{20}{c}}7&6\\4&7\end{array}} \right]$
and ${A^2} - 2A = \left[ {\begin{array}{*{20}{c}}5&0\\0&5\end{array}} \right]\,,{\rm{det }}({A^2} - 2A) = \left| {\,\begin{array}{*{20}{c}}5&0\\0&5\end{array}\,} \right| = 25$.
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$z^5=1$ then value of $\left| {\begin{array}{*{20}{c}}
{{e^\alpha }}&{{e^{2\alpha }}}&{{e^{3\alpha + 1}}}&{ - {e^{ - \delta }}} \\
{{e^\beta }}&{{e^{2\beta }}}&{{e^{3\beta + 1}}}&{ - {e^{ - \delta }}} \\
{{e^\gamma }}&{{e^{2\gamma }}}&{{e^{3\gamma + 1}}}&{ - {e^{ - \delta }}}
\end{array}} \right|$