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Mathematics Question on Matrices

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

A

Y3Z4−Z4Y3Y^3Z^4 - Z^4Y^3

B

X44+Y44X^{44} + Y^{44}

C

X4Z3−Z3X4X^4Z^3 - Z^3X^4

D

X23+Y23X^{23} + Y^{23}

Answer

X23+Y23X^{23} + Y^{23}

Explanation

Solution

XT=−XYT=−YZT=ZX^{T} = -X Y^{T} = -Y Z^{T} = Z
(A)(Y3Z4−Z4Y3)T=(Y3Z4)T−(Z4Y3)T\left(A\right) \left(Y^{3} Z^{4} - Z^{4} Y^{3}\right)^{T} = \left(Y^{3} Z^{4}\right)^{T }- \left(Z^{4} Y^{3}\right)^{T}
=(Z4)T(Y3)T−(Y3)T(Z4)T= \left(Z^{4}\right)^{T} \left(Y^{3}\right)^{T }- \left(Y^{3}\right)^{T} \left(Z^{4}\right)^{T}
=(ZT)4(YT)3−(YT)3(ZT)4= \left(Z^{T}\right)^{4} \left(Y^{T}\right)^{3} - \left(Y^{T}\right)^{3} \left(Z^{T}\right)^{4}
=Z4(−Y)3−(−Y)3(Z)4= Z^{4 }\left(-Y\right)^{3} - \left(-Y\right)^{3} \left(Z\right)^{4}
=−Z4Y3+Y3Z4= -Z^{4}Y^{3} + Y^{3} Z^{4}
=Y3Z4−Z4Y3= Y^{3}Z^{4} - Z^{4}Y^{3}
Hence it is symmetric matrix.
(B)(X44+Y44)T=(XT)44+(YT)44\left(B\right) \left(X^{44} + Y^{44}\right)^{T} = \left(X^{T}\right)^{44} + \left(Y^{T}\right)^{44}
=X44+Y44= X^{44} + Y^{44}
Hence it is symmetric matrix.
(C)(X4Z3−Z3X4)T=(X4Z3)T−(Z3X4)T\left(C\right) \left(X^{4} Z^{3} - Z^{3} X^{4}\right)^{T} = \left(X^{4} Z^{3}\right)^{T} - \left(Z^{3} X^{4}\right)^{T}
=(ZT)3(XT)4−(XT)4(ZT)3= \left(Z^{T}\right)^{3} \left(X^{T}\right)^{4} - \left(X^{T}\right)^{4} \left(Z^{T}\right)^{3}
=Z3X4−X4Z3= Z^{3} X^{4} - X^{4 }Z^{3}
=−(X4Z3−Z3X4)= - \left(X^{4 }Z^{3} - Z^{3} X^{4}\right)
Hence it is skew symmetric matrix.
(D)(X23+Y23)T=(XT)23+(YT)23\left(D\right) \left(X^{23} + Y^{23}\right)^{T} = \left(X^{T}\right)^{23} + \left(Y^{T}\right)^{23}
=−(X23+Y23)= - \left(X^{23} + Y^{23}\right)
Hence it is skew symmetric matrix.