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Joint Entrance Examination
Mathematics
Matrices and Determinants
Question 406
Question 406 of 472
Medium
MCQ
Let \[A = \left[ {
(
0
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2
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\begin{pmatrix} 0 & { - 2} \\ 2 & 0 \\ \end{pmatrix}
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} \right]\]. 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 :
A
a non-identity symmetric matrix
B
a skew-symmetric matrix
C
neither symmetric nor skew-symmetric matrix
D
an identity matrix
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Solutions · 1
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Official solution
\[A = \left[ {
(
0
−
2
2
0
)
\begin{pmatrix} 0 & { - 2} \\ 2 & 0 \\ \end{pmatrix}
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)
} \right]\] \[{A^2} = \left[ {
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\begin{pmatrix} 0 & { - 2} \\ 2 & 0 \\ \end{pmatrix}
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} \right]\left[ {
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\begin{pmatrix} 0 & { - 2} \\ 2 & 0 \\ \end{pmatrix}
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} \right] = \left[ {
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4
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4
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\begin{pmatrix} { - 4} & 0 \\ 0 & { - 4} \\ \end{pmatrix}
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4
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} \right] = - 4I\] \[M = {A^2} + {A^4} + {A^6} + \,\,.....\,\, + \,\,{A^{20}}\] \[ = - 4I + 16I - 64I\,\, + \] ..... upto 10 terms
=
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I
= - I
=
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I
[
4
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16
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64
4 - 16 + 64
4
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16
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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 \[ = A\left( {{{{{( - 4)}^{10}} - 1} \over { - 4 - 1}}} \right) = - \left( {{{{2^{20}} - 1} \over 5}} \right)A\] \[{N^2} = {{{{({2^{20}} - 1)}^2}} \over {{2^5}}}{A^2} = {{ - 4} \over {24}}{({2^{20}} - 1)^2}I\] \[M{N^2} = {{ - 16} \over {125}}{({2^{20}} - 1)^3}I = KI\,\,\,\,\,(K \ne \pm \,1)\]
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N
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T
=
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I
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T
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I
{(M{N^2})^T} = {(KI)^T} = KI
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T
=
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I
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T
=
K
I
\[\therefore\] A is correct
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