A) null matrix
B) identity matrix
C) symmetric
D) skew-symmetric
Correct Answer: D
Solution :
Given, \[A=\left[ \begin{matrix} 3 & -4 \\ 1 & -1 \\ \end{matrix} \right]\] Then, \[A'=\left[ \begin{matrix} 3 & 1 \\ -4 & -1 \\ \end{matrix} \right]\] Now, \[A-A'=\left[ \begin{matrix} 3 & -4 \\ 1 & -1 \\ \end{matrix} \right]-\left[ \begin{matrix} 3 & 1 \\ -4 & -1 \\ \end{matrix} \right]\] \[\Rightarrow \] \[A-A'=\left[ \begin{matrix} 0 & -5 \\ 5 & 0 \\ \end{matrix} \right]\] ?. (i) Now, we have \[A'-A=\left[ \begin{matrix} 3 & 1 \\ -4 & -1 \\ \end{matrix} \right]-\left[ \begin{matrix} 3 & -4 \\ 1 & -1 \\ \end{matrix} \right]=\left[ \begin{matrix} 0 & 5 \\ -5 & 0 \\ \end{matrix} \right]\] \[\Rightarrow \] \[(A'-A)'=\left[ \begin{matrix} 0 & -5 \\ 5 & 0 \\ \end{matrix} \right]=(A-A')\] [From Eq. (i)] which represent that \[(A-A')\] is skew-symmetric matrix.
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