Question

What is the symmetric part of matrix $A = \begin{bmatrix}1&2&4\\6&8&2\\2&{ - 2}&7\end{bmatrix}$?
A. $\begin{bmatrix}0&{ - 2}&{ - 1}\\{ - 2}&0&{ - 2}\\{ - 1}&{ - 2}&0\end{bmatrix}$
B. $\begin{bmatrix}1&4&3\\4&8&0\\3&0&7\end{bmatrix}$
C. $\begin{bmatrix}1&4&{ - 3}\\2&8&0\\3&0&7\end{bmatrix}$
D. $\begin{bmatrix}0&{ - 2}&1\\2&0&2\\{ - 1}&2&0\end{bmatrix}$

Answer 1

Hint: Using the fact that any matrix A can be written as the sum of a symmetric and a skew symmetric matrix we can solve the given problem.

Formula Used:
If $A$ be a matrix and $A'$ be its transpose then $A$ can be decomposed in two parts as
$A = \dfrac{1}{2}\left( {A + A'} \right)\, + \dfrac{1}{2}\left( {A - A'} \right)$
Where,
Symmetric part of matrix = $\dfrac{1}{2}\left( {A + A'} \right)$
Skew symmetric part of matrix =$\dfrac{1}{2}\left( {A - A'} \right)$

Complete step by step solution:
Given -$A = \begin{bmatrix}1&2&4\\6&8&2\\2&{ - 2}&7\end{bmatrix}$
We know that if $A$ be a matrix and$A'$be its transpose then the symmetric part of matrix is given by Symmetric part of matrix = $\dfrac{1}{2}\left( {A + A'} \right)$
Since $A = \begin{bmatrix}1&2&4\\6&8&2\\2&{ - 2}&7\end{bmatrix}$
So, $A' = \begin{bmatrix}1&6&2\\2&8&{ - 2}\\4&2&7\end{bmatrix}$
Now,
$\dfrac{1}{2}\left( {A + A'} \right) = \dfrac{1}{2}\begin{bmatrix}1&2&4\\6&8&2\\2&{ - 2}&7\end{bmatrix} + \begin{bmatrix}1&6&2\\2&8&{ - 2}\\4&2&7\end{bmatrix}$
$ \Rightarrow \dfrac{1}{2}\left( {A + A'} \right) = \dfrac{1}{2}\begin{bmatrix}2&8&6\\8&{16}&0\\6&0&{14}\end{bmatrix}$
$\Rightarrow \dfrac{1}{2}\left( {A + A'} \right) = \begin{bmatrix}1&4&3\\4&8&0\\3&0&7\end{bmatrix}$
∴ Symmetric part =$\begin{bmatrix}1&4&3\\4&8&0\\3&0&7\end{bmatrix}$

Option ‘B’ is correct

Note: Students may get confused in formula and wrong application of formula will give wrong answer. Correct formulas are –
Symmetric part = $\dfrac{1}{2}\left( {A + A'} \right)$
Skew symmetric =$\dfrac{1}{2}\left( {A - A'} \right)$