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Question Number 41321 by rahul 19 last updated on 05/Aug/18

$$\mathrm{If}\mathrm{A}\mathrm{is}\mathrm{a}\mathrm{square}\mathrm{matrix}\mathrm{of}\mathrm{order}3,\mathrm{then}$$$$\mid {(\mathrm{A}-{\mathrm{A}}^{\mathrm{T}})}^{2011}\mid =?$$

Commented by rahul 19 last updated on 05/Aug/18

$$\mathrm{I}{}^{\prime }\mathrm{m}\mathrm{getting}\mathrm{A}-{\mathrm{A}}^{\mathrm{T}}\mathrm{a}\mathrm{skew}\mathrm{symmetric}$$$$\mathrm{of}\mathrm{order}3.$$$$\mathrm{I}\mathrm{wanted}\mathrm{to}\mathrm{know}\mathrm{why}{(\mathrm{A}-{\mathrm{A}}^{\mathrm{T}})}^{2011}\mathrm{will}$$$$\mathrm{also}\mathrm{be}\mathrm{a}\mathrm{skew}\mathrm{synmetric}\mathrm{of}\mathrm{order}3!$$

Answered by MJS last updated on 05/Aug/18

$$\left(\begin{array}{ccc}a& b& c\\ d& e& f\\ g& h& i\end{array}\right)-\left(\begin{array}{ccc}a& d& g\\ b& e& h\\ c& f& i\end{array}\right)=\left(\begin{array}{ccc}0& p& q\\ -p& 0& r\\ -q& -r& 0\end{array}\right)$$$${\left(\begin{array}{ccc}0& p& q\\ -p& 0& r\\ -q& -r& 0\end{array}\right)}^{2n-1}=$$$$=\left(\begin{array}{ccc}0& {(-1)}^{n-1}p{(p^2+q^2+r^2)}^{n-1}& {(-1)}^{n-1}q{(p^2+q^2+r^2)}^{n-1}\\ {(-1)}^{n}p{(p^2+q^2+r^2)}^{n-1}& 0& {(-1)}^{n-1}r{(p^2+q^2+r^2)}^{n-1}\\ {(-1)}^{n}q{(p^2+q^2+r^2)}^{n-1}& {(-1)}^{n}r{(p^2+q^2+r^2)}^{n-1}& 0\end{array}\right)$$

Commented by rahul 19 last updated on 06/Aug/18

thanks sir ����

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