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3 and 4 .Determinants and Matrices
normal
If $A$ is skew symmetric matrix of order $3$ and $X$ be another matrix of same order, then $|XA + AX^T|$ is (where $|P|$ denotes determinant of matrix $P$ )
A
$|X + X^T|$
B
$|A + X|$
C
$|A -X|$
D
$0$
Solution
$A^{T}=-A$
$\left(\mathrm{XA}+\mathrm{AX}^{\mathrm{T}}\right)^{\mathrm{T}}=\mathrm{A}^{\mathrm{T}} \mathrm{X}^{\mathrm{T}}+\mathrm{XA}^{\mathrm{T}}$
$=-\mathrm{AX}^{\mathrm{T}}-\mathrm{XA}$
$=-\left(\mathrm{AX}^{\mathrm{T}}+\mathrm{XA}\right)$
$\left|\left(\mathrm{XA}+\mathrm{AX}^{\mathrm{T}}\right)^{\mathrm{T}}\right|=\left|-\left(\mathrm{XA}+\mathrm{AX}^{\mathrm{T}}\right)\right|$
$\left|\mathrm{XA}+\mathrm{AX}^{\mathrm{T}}\right|=-\left|\mathrm{XA}+\mathrm{AX}^{\mathrm{T}}\right|$
$\left|\mathrm{XA}+\mathrm{AX}^{\mathrm{T}}\right|=0$
Standard 12
Mathematics
