Question

The symmetric part of the matrix $A =\begin{bmatrix} {1}&{2} &{4}\\\ {6}&{8}& {2} \\\ {2}&{-2}&{7}\\\ \end{bmatrix}$ is

• A. $\begin{bmatrix} {1}&{4} &{3}\\\ {2}&{8}& {0} \\\ {3}&{0}&{7}\\\ \end{bmatrix}$
• B. $\begin{bmatrix} {1}&{4} &{3}\\\ {4}&{8}& {0} \\\ {3}&{0}&{7}\\\ \end{bmatrix}$
• C. $\begin{bmatrix} {0}&{-2} &{-1}\\\ {-2}&{0}& {-2} \\\ {-1}&{-2}&{0}\\\ \end{bmatrix}$
• D. $\begin{bmatrix} {0}&{-2} &{1}\\\ {2}&{0}& {2} \\\ {-1}&{2}&{0}\\\ \end{bmatrix}$

Answer 1

Answer: (B)

$\begin{bmatrix} {1}&{4} &{3}\\\ {4}&{8}& {0} \\\ {3}&{0}&{7}\\\ \end{bmatrix}$

Answer 2

Given, matrix $A= \begin{bmatrix}1 & 2 & 4 \\\ 6 & 8 & 2 \\\ 2 & -2 & 7\end{bmatrix}$ $\therefore$ Symmetric part of $A=\frac{1}{2}\left[A+A'\right]$ =\frac{1}{2}\left\\{\begin{bmatrix}1 & 2 & 4 \\\ 6 & 8 & 2 \\\ 2 & -2 & 7\end{bmatrix}+ \begin{bmatrix}1 & 6 & 2 \\\ 2 & 8 & -2 \\\ 4 & 2 & 7\end{bmatrix}\right\\} =\frac{1}{2}\left\\{\begin{bmatrix}2 & 8 & 6 \\\ 8 & 16 & 0 \\\ 6 & 0 & 14\end{bmatrix}\right\\}= \begin{bmatrix}1 & 4 & 3 \\\ 4 & 8 & 0 \\\ 3 & 0 & 7\end{bmatrix}