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JEE Advanced 2015Paper 1
Exam
JEE Advanced
Year
2015
Paper
Paper 1
Subject
Mathematics
Chapter
Matrices

JEE Advanced 2015 · Paper 1 · Mathematics · Matrices

Let XandY be two arbitrary, 3 × 3 , non - zero, skew - symmetric...

Let XandYXandY be two arbitrary, 3×33 \times 3 , non - zero, skew - symmetric matrices and Z be an arbitrary 3×33 \times 3 , non - zero, symmetric matrix. Then which of the following matrices is (are) skew symmetric ?

Options

  1. AY3Z4−Z4Y3{Y}^{3}{Z}^{4} - {Z}^{4}{Y}^{3}
  2. BX44+Y44{X}^{44} + {Y}^{44}
  3. CX4Z3−Z3X4{X}^{4}{Z}^{3} - {Z}^{3}{X}^{4}
  4. DX23+Y23{X}^{23} + {Y}^{23}

Correct answer

C. X4Z3−Z3X4{X}^{4}{Z}^{3} - {Z}^{3}{X}^{4}

Step-by-step solution

XT=−X,YT=−Y{X}^{T} = - X, {Y}^{T} = - Y and ZT=Z{Z}^{T} = Z
P=Y3Z4−Z4Y3P = {Y}^{3}{Z}^{4} - {Z}^{4}{Y}^{3}
PT=(Z4){P}^{T} = {\left({Z}^{4}\right)}^{T}{\left({Y}^{3}\right)}^{T} - {\left({Y}^{3}\right)}^{T}{\left({Z}^{4}\right)}^{T}
=(ZT)= {\left({Z}^{T}\right)}^{4}{\left({Y}^{T}\right)}^{3} - {\left({Y}^{T}\right)}^{3}{\left({Z}^{T}\right)}^{4}
=Z4= {Z}^{4}\left(- {Y}^{3}\right) - (−Y3)Z4\left(- {Y}^{3}\right){Z}^{4}
=Y3Z4–Z4Y3=P(symmetric)= {Y}^{3}{Z}^{4} – {Z}^{4}{Y}^{3} = P ( symmetric )
Q=X44+Y44Q = {X}^{44} + {Y}^{44}
QT=(XT){Q}^{T} = {\left({X}^{T}\right)}^{44} + {\left({Y}^{T}\right)}^{44} = {X}^{44} + {Y}^{44} = Q(symmetric)
RT=(Z3)T(X4)T−(X4)T(Z3)T{R}^{T} = {\left({Z}^{3}\right)}^{T}{\left({X}^{4}\right)}^{T} - {\left({X}^{4}\right)}^{T}{\left({Z}^{3}\right)}^{T}
=(ZT)= {\left({Z}^{T}\right)}^{3}{\left({X}^{T}\right)}^{4} - {\left({X}^{T}\right)}^{4}{\left({Z}^{T}\right)}^{3}
=Z3(−X)= {Z}^{3}{\left(- X\right)}^{4} - {\left(- X\right)}^{4}{\left(Z\right)}^{3}
=Z3X4−X4Z3=−R= {Z}^{3}{X}^{4} - {X}^{4}{Z}^{3} = - R (Skew Symmetric)
S=X23+Y23S = {X}^{23} + {Y}^{23}
ST=(XT){S}^{T} = {\left({X}^{T}\right)}^{23} + {\left({Y}^{T}\right)}^{23} = {\left(- X\right)}^{23} + {\left(- Y\right)}^{23}
=−X23−Y23=−S= - {X}^{23} - {Y}^{23} = - S (Skew symmetric)
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