Question

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

विकल्प

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

• \[\begin{bmatrix}2&-2&-4\\-1&3&4\\1&-2&-3\end{bmatrix}\]

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

• \[\begin{bmatrix}4&-3&-3\\-3&6&2\\-3&2&-6\end{bmatrix}\]

Answer 1

First, \[B+B^T=\begin{bmatrix}4&-3&-3\\-3&6&2\\-3&2&-6\end{bmatrix}\]. Multiplying every entry by \[\frac{1}{2}\] gives \[P\], and its reflected entries are equal.