"Let $A = \left[ {\matrix{ 0 & { - 2} \cr 2 & 0 \cr } } \right]$. If M and N are two matrices given by $M = \sum\limits_{k = 1}^{10} {{A^{2k}}}$ and $N = \sum\limits_{k = 1}^{10} {{A^{2k - 1}}}$ then MN2 is :"

Question

Accepted Answer

"

$A = \left[ {\matrix{ 0 & { - 2} \cr 2 & 0 \cr } } \right]$

$${A^2} = \left[ {\matrix{ 0 & { - 2} \cr 2 & 0 \cr } } \right]\left[ {\matrix{ 0 & { - 2} \cr 2 & 0 \cr } } \right] = \left[ {\matrix{ { - 4} & 0 \cr 0 & { - 4} \cr } } \right] = - 4I$$

$M = {A^2} + {A^4} + {A^6} + \,\,.....\,\, + \,\,{A^{20}}$

$= - 4I + 16I - 64I\,\, +$ ..... upto 10 terms

$= - I$ [$4 - 16 + 64$ .... + upto 10 terms]

$$ = - I\,.\,4\left[ {{{{{( - 4)}^{10}} - 1} \over { - 4 - 1}}} \right] = {4 \over 5}({2^{20}} - 1)I$$

$N = {A^1} + {A^3} + {A^5} + \,\,....\,\, + \,\,{A^{19}}$

$= A - 4A + 16A\,\, +$ ..... upto 10 terms

$$ = A\left( {{{{{( - 4)}^{10}} - 1} \over { - 4 - 1}}} \right) = - \left( {{{{2^{20}} - 1} \over 5}} \right)A$$

$${N^2} = {{{{({2^{20}} - 1)}^2}} \over {{2^5}}}{A^2} = {{ - 4} \over {24}}{({2^{20}} - 1)^2}I$$

$M{N^2} = {{ - 16} \over {125}}{({2^{20}} - 1)^3}I = KI\,\,\,\,\,(K \ne \pm \,1)$

${(M{N^2})^T} = {(KI)^T} = KI$

$\therefore$ A is correct

"

Let $A = \left[ {\matrix{ 0 & { - 2} \cr 2 & 0 \cr } } \right]$. If M and N are two matrices given by $M = \sum\limits_{k = 1}^{10} {{A^{2k}}}$ and $N = \sum\limits_{k = 1}^{10} {{A^{2k - 1}}}$ then MN² is :

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Let $A = \left[ {\matrix{ 0 & { - 2} \cr 2 & 0 \cr } } \right]$. If M and N are two matrices given by $M = \sum\limits_{k = 1}^{10} {{A^{2k}}}$ and $N = \sum\limits_{k = 1}^{10} {{A^{2k - 1}}}$ then MN2 is :

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Let $A = \left[ {\matrix{ 0 & { - 2} \cr 2 & 0 \cr } } \right]$. If M and N are two matrices given by $M = \sum\limits_{k = 1}^{10} {{A^{2k}}}$ and $N = \sum\limits_{k = 1}^{10} {{A^{2k - 1}}}$ then MN² is :