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NDA 2025 Maths · Mathematics · Matrices · medium

The skew-symmetric part (A-A^T/(2) of A=beginpmatrix1&3\\5&2endpmatrix is

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The skew-symmetric part A−AT2\dfrac{A-A^{T}}{2} of A=(1352)A=\begin{pmatrix}1&3\\5&2\end{pmatrix} is
  1. A. (01−10)\begin{pmatrix}0&1\\-1&0\end{pmatrix}
  2. B. (0−110)\begin{pmatrix}0&-1\\1&0\end{pmatrix}
  3. C. (1442)\begin{pmatrix}1&4\\4&2\end{pmatrix}
  4. D. (0−220)\begin{pmatrix}0&-2\\2&0\end{pmatrix}
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Answer: (B) (0−110)\begin{pmatrix}0&-1\\1&0\end{pmatrix}

AT=(1532)A^T=\begin{pmatrix}1&5\\3&2\end{pmatrix}, so A−AT=(0−220)A-A^T=\begin{pmatrix}0&-2\\2&0\end{pmatrix} and half of it is (0−110)\begin{pmatrix}0&-1\\1&0\end{pmatrix}. The matrix (1442)\begin{pmatrix}1&4\\4&2\end{pmatrix} is the symmetric part A+AT2\frac{A+A^T}2.
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