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QJEEM22Macfb1: JEE Main 2022 Mathematics - Class 12 JEEM22Macfb1 Solution: Let A = [ 0 − 2 2 0 ] . If M and N are two matrices given by…

Let A=[0−220]. If M and N are two matrices given by M=∑k=110A2kM = \sum\limits_{k = 1}^{10} {{A^{2k}}} and N=∑k=110A2k−1N = \sum\limits_{k = 1}^{10} {{A^{2k - 1}}} then MN2 is :

Multiple Choice Options

  • A a non-identity symmetric matrix
  • B a skew-symmetric matrix
  • C neither symmetric nor skew-symmetric matrix
  • D an identity matrix

Step-by-Step Verified Solution

A=[0−220]

A2=[0−220][0−220]=[−400−4]=−4I

M=A2+A4+A6+  .....  +  A20M = {A^2} + {A^4} + {A^6} + \,\,.....\,\, + \,\,{A^{20}}

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

=−I= - I [ 4−16+644 - 16 + 64 .... + upto 10 terms]

=−I . 4[(−4)10−1−4−1]=45(220−1)I= - I\,.\,4\left[ {{{{{( - 4)}^{10}} - 1} \over { - 4 - 1}}} \right] = {4 \over 5}({2^{20}} - 1)I

N=A1+A3+A5+  ....  +  A19N = {A^1} + {A^3} + {A^5} + \,\,....\,\, + \,\,{A^{19}}

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

=A((−4)10−1−4−1)=−(220−15)A= A\left( {{{{{( - 4)}^{10}} - 1} \over { - 4 - 1}}} \right) = - \left( {{{{2^{20}} - 1} \over 5}} \right)A

N2=(220−1)225A2=−424(220−1)2I{N^2} = {{{{({2^{20}} - 1)}^2}} \over {{2^5}}}{A^2} = {{ - 4} \over {24}}{({2^{20}} - 1)^2}I

MN2=−16125(220−1)3I=KI     (K≠± 1)M{N^2} = {{ - 16} \over {125}}{({2^{20}} - 1)^3}I = KI\,\,\,\,\,(K \ne \pm \,1)

(MN2)T=(KI)T=KI{(M{N^2})^T} = {(KI)^T} = KI

∴\therefore A is correct

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