For the matrix $A = \left( {\begin{array}{*{20}{c}}
  1&5 \\
  6&7
\end{array}} \right)$,verify that
(i) $\left( {A + A'} \right)$is a symmetric matrix.
(ii) $\left( {A - A'} \right)$is a skew symmetric matrix.

Hint: Use the property of symmetric and skew symmetric matrices directly on the given matrix expression.


Given the matrix $A = \left( {\begin{array}{*{20}{c}}
  1&5 \\
  6&7
\end{array}} \right)$

We know that the transpose of a matrix is obtained by switching the rows with its columns
$ \Rightarrow A' = \left( {\begin{array}{*{20}{c}}
  1&6 \\
  5&7
\end{array}} \right)$

A Symmetric matrix is the one in which the matrix is equal to the transpose of itself.
\[ \Rightarrow if{\text{ }}A = {\left[ {{a_{ij}}} \right]_{n \times m}}{\text{ }}and{\text{ }}A' = {\left[ {{a_{ij}}} \right]_{m \times n}}{\text{ ,}}then{\text{ }}A = A'\]

We need to prove $\left( {A + A'} \right)$is a symmetric matrix.
$A + A' = \left( {\begin{array}{*{20}{c}}
  1&5 \\
  6&7
\end{array}} \right) + \left( {\begin{array}{*{20}{c}}
  1&6 \\
  5&7
\end{array}} \right) = \left( {\begin{array}{*{20}{c}}
  2&{11} \\
  {11}&{14}
\end{array}} \right){\text{ (1)}}$
Also,${\left( {A + A'} \right)^\prime } = \left( {\begin{array}{*{20}{c}}
  2&{11} \\
  {11}&{14}
\end{array}} \right){\text{ (2)}}$

From equations $(1)$and$(2)$ , we get ${\left( {A + A'} \right)^\prime } = \left( {A + A'} \right)$, which satisfies the above-mentioned condition of symmetric matrices.
Hence $\left( {A + A'} \right)$is a symmetric matrix.

A Skew symmetric matrix is the one in which the negative of the matrix is equal to the transpose of itself.

\[ \Rightarrow if{\text{ }}A = {\left[ {{a_{ij}}} \right]_{n \times m}}{\text{ }}and{\text{ }}A' = {\left[ {{a_{ij}}} \right]_{m \times n}}{\text{ ,}}then{\text{ }} - A = A'\]

We need to prove $\left( {A - A'} \right)$is a skew symmetric matrix.
$A - A' = \left( {\begin{array}{*{20}{c}}
  1&5 \\
  6&7
\end{array}} \right) + \left( {\begin{array}{*{20}{c}}
  1&6 \\
  5&7
\end{array}} \right) = \left( {\begin{array}{*{20}{c}}
  0&{ - 1} \\
  1&0
\end{array}} \right){\text{ (3)}}$
Also,${\left( {A - A'} \right)^\prime } = \left( {\begin{array}{*{20}{c}}
  0&1 \\
  { - 1}&0
\end{array}} \right){\text{ (4)}}$

From equations $(3)$and$(4)$ , we get ${\left( {A - A'} \right)^\prime } = - \left( {A - A'} \right)$, which satisfies the above-mentioned condition of skew symmetric matrices.
Hence $\left( {A - A'} \right)$is a skew symmetric matrix verified.

Note: The above-mentioned results are true for all square matrices. Similarly using above results, it can be proved that a square matrix can be expressed as the sum of a symmetric and skew symmetric matrix.