Question

For \[B=\begin{bmatrix}2&-2&-4\\-1&3&4\\1&-2&-3\end{bmatrix}\], what is its skew-symmetric part \[Q=\frac{1}{2}(B-B^T)\]?

Options

• \[\begin{bmatrix}0&-\frac{1}{2}&-\frac{5}{2}\\\frac{1}{2}&0&3\\\frac{5}{2}&-3&0\end{bmatrix}\]

• \[\begin{bmatrix}0&\frac{1}{2}&\frac{5}{2}\\-\frac{1}{2}&0&-3\\-\frac{5}{2}&3&0\end{bmatrix}\]

• \[\begin{bmatrix}2&-\frac{3}{2}&-\frac{3}{2}\\-\frac{3}{2}&3&1\\-\frac{3}{2}&1&-3\end{bmatrix}\]

• \[\begin{bmatrix}0&-1&-5\\1&0&6\\5&-6&0\end{bmatrix}\]

Answer 1

Subtracting gives \[B-B^T=\begin{bmatrix}0&-1&-5\\1&0&6\\5&-6&0\end{bmatrix}\]. Dividing by \[2\] gives \[Q\], whose diagonal entries are zero and whose reflected entries are negatives.