Express the following matrices as the sum of a symmetric and a skew symmetric matrix :

(i) $\left[ {\begin{array}{cccccccccccccccccccc}3&5\\1&{ - 1}\end{array}} \right]$

(ii) $\left[ {\begin{array}{cccccccccccccccccccc}6&{ - 2}&2\\{ - 2}&3&{ - 1}\\2&{ - 1}&3\end{array}} \right]$

(iii)$\left[ {\begin{array}{cccccccccccccccccccc}3&3&{ - 1}\\{ - 2}&{ - 2}&1\\{ - 4}&{ - 5}&2\end{array}} \right]$

(iv)$\left[ {\begin{array}{cccccccccccccccccccc}1&5\\{ - 1}&2\end{array}} \right]$

.

:(i) Here, A =$\left[ {\begin{array}{cccccccccccccccccccc}3&5\\1&{ - 1}\end{array}} \right]$ $\Rightarrow$ $A' = \left[ {\begin{array}{cccccccccccccccccccc}3&1\\5&{ - 1}\end{array}} \right]$

Let $P = \cfrac{1}{2}(A + A')$

Now, $A + A' = \left[ {\begin{array}{cccccccccccccccccccc}3&5\\1&{ - 1}\end{array}} \right] + \left[ {\begin{array}{cccccccccccccccccccc}3&1\\5&{ - 1}\end{array}} \right] = \left[ {\begin{array}{cccccccccccccccccccc}{3 + 3}&{5 + 1}\\{1 + 5}&{ - 1 - 1}\end{array}} \right] = \left[ {\begin{array}{cccccccccccccccccccc}6&6\\6&{ - 2}\end{array}} \right]$
$\Rightarrow$ $P = \cfrac{1}{2}(A + A')$

$= \cfrac{1}{2}\left[ {\begin{array}{cccccccccccccccccccc}6&6\\6&{ - 2}\end{array}} \right] = \left[ {\begin{array}{cccccccccccccccccccc}3&3\\3&{ - 1}\end{array}} \right]$

$\Rightarrow$ $P' = \left[ {\begin{array}{cccccccccccccccccccc}3&3\\3&{ - 1}\end{array}} \right] = P$

Thus, P = $\cfrac{1}{2}$ (A+ A’) is a symmetric matrix.
Also, let Q= $\cfrac{1}{2}$(A$-$A’)

Now, A$-$A’$= \left[ {\begin{array}{cccccccccccccccccccc}3&5\\1&{ - 1}\end{array}} \right] - \left[ {\begin{array}{cccccccccccccccccccc}3&1\\5&{ - 1}\end{array}} \right]$
$= \left[ {\begin{array}{cccccccccccccccccccc}{3 - 3}&{5 - 1}\\{1 - 5}&{ - 1 + 1}\end{array}} \right] = \left[ {\begin{array}{cccccccccccccccccccc}0&4\\{ - 4}&0\end{array}} \right]$

$\Rightarrow$ $Q = \cfrac{1}{2}(A - A') = \cfrac{1}{2}\left[ {\begin{array}{cccccccccccccccccccc}0&4\\{ - 4}&0\end{array}} \right] = \left[ {\begin{array}{cccccccccccccccccccc}0&2\\{ - 2}&0\end{array}} \right]$

$\Rightarrow$ $Q' = \left[ {\begin{array}{cccccccccccccccccccc}0&{ - 2}\\2&0\end{array}} \right] = - \left[ {\begin{array}{cccccccccccccccccccc}0&2\\{ - 2}&0\end{array}} \right] = - Q$

Thus, $Q = \cfrac{1}{2}$= ( A$-$ A’ ) is a skew$-$symmetric matrix.
$\therefore$ $P + Q = A = \left[ {\begin{array}{cccccccccccccccccccc}3&3\\3&{ - 1}\end{array}} \right] + \left[ {\begin{array}{cccccccccccccccccccc}0&2\\{ - 2}&0\end{array}} \right]$

(ii) Here, $A = \left[ {\begin{array}{cccccccccccccccccccc}6&{ - 2}&2\\{ - 2}&3&{ - 1}\\2&{ - 1}&3\end{array}} \right]$ $\Rightarrow$ $A' = \left[ {\begin{array}{cccccccccccccccccccc}6&{ - 2}&2\\{ - 2}&3&{ - 1}\\2&{ - 1}&3\end{array}} \right]$

Let $P = \cfrac{1}{2}(A + A')$
$\therefore$ $A + A' = \left[ {\begin{array}{cccccccccccccccccccc}6&{ - 2}&2\\{ - 2}&3&{ - 1}\\2&{ - 1}&3\end{array}} \right] + \left[ {\begin{array}{cccccccccccccccccccc}6&{ - 2}&2\\{ - 2}&3&{ - 1}\\2&{ - 1}&3\end{array}} \right]$

$= \left[ {\begin{array}{cccccccccccccccccccc}{6 + 6}&{ - 2 - 2}&{2 + 2}\\{ - 2 + ( - 2)}&{3 + 3}&{ - 1 + ( - 1)}\\{2 + 2}&{ - 1 + ( - 1)}&{3 + 3}\end{array}} \right] = \left[ {\begin{array}{cccccccccccccccccccc}{12}&{ - 4}&4\\{ - 4}&6&{ - 2}\\4&{ - 2}&6\end{array}} \right]$

$\Rightarrow$ $P = \cfrac{1}{2}(A + A') = \cfrac{1}{2}\left[ {\begin{array}{cccccccccccccccccccc}{12}&{ - 4}&4\\{ - 4}&6&{ - 2}\\4&{ - 2}&6\end{array}} \right] = \left[ {\begin{array}{cccccccccccccccccccc}6&{ - 2}&2\\{ - 2}&3&{ - 1}\\2&{ - 1}&3\end{array}} \right]$

$\Rightarrow$ $P' = \left[ {\begin{array}{cccccccccccccccccccc}6&{ - 2}&2\\{ - 2}&3&{ - 1}\\2&{ - 1}&3\end{array}} \right] = \left[ {\begin{array}{cccccccccccccccccccc}6&{ - 2}&2\\{ - 2}&3&{ - 1}\\2&{ - 1}&3\end{array}} \right] = P$

Thus, $P = \cfrac{1}{2}(A + A')$is a symmetric matrix.

Also, let $Q = \cfrac{1}{2}(A - A')$
Now, A$-$A’$= \left[ {\begin{array}{cccccccccccccccccccc}6&{ - 2}&2\\{ - 2}&3&{ - 1}\\2&{ - 1}&3\end{array}} \right] - \left[ {\begin{array}{cccccccccccccccccccc}6&{ - 2}&2\\{ - 2}&3&{ - 1}\\2&{ - 1}&3\end{array}} \right]$

$= \left[ {\begin{array}{cccccccccccccccccccc}{6 - 6}&{ - 2 + 2}&{2 - 2}\\{ - 2 + 2}&{3 - 3}&{ - 1 + 1}\\{2 - 2}&{ - 1 + 1}&{3 - 3}\end{array}} \right] = \left[ {\begin{array}{cccccccccccccccccccc}0&0&0\\0&0&0\\0&0&0\end{array}} \right]$

$\Rightarrow$ $Q = \cfrac{1}{2}(A - A') = \cfrac{1}{2}\left[ {\begin{array}{cccccccccccccccccccc}0&0&0\\0&0&0\\0&0&0\end{array}} \right] = \left[ {\begin{array}{cccccccccccccccccccc}0&0&0\\0&0&0\\0&0&0\end{array}} \right]$and $Q' = - Q$

So, Q = $\cfrac{1}{2}$(A$-$A’) is a skew symmetric matrix.

Hence, $P + Q = A =$ $\left[ {\begin{array}{cccccccccccccccccccc}6&{ - 2}&2\\{ - 2}&3&{ - 1}\\2&{ - 1}&3\end{array}} \right]$+ $\left[ {\begin{array}{cccccccccccccccccccc}0&0&0\\0&0&0\\0&0&0\end{array}} \right]$

(iii) Here, $A = \left[ {\begin{array}{cccccccccccccccccccc}3&3&{ - 1}\\{ - 2}&{ - 2}&1\\{ - 4}&{ - 5}&2\end{array}} \right]$ $\Rightarrow$ $A' = \left[ {\begin{array}{cccccccccccccccccccc}3&{ - 2}&{ - 4}\\3&{ - 2}&{ - 5}\\{ - 1}&1&2\end{array}} \right]$

Let $P = \cfrac{1}{2}(A + A')$

Now, $A + A' = \left[ {\begin{array}{cccccccccccccccccccc}3&3&{ - 1}\\{ - 2}&{ - 2}&1\\{ - 4}&{ - 5}&2\end{array}} \right] + \left[ {\begin{array}{cccccccccccccccccccc}3&{ - 2}&{ - 4}\\3&{ - 2}&{ - 5}\\{ - 1}&1&2\end{array}} \right]$

$= \left[ {\begin{array}{cccccccccccccccccccc}{3 + 3}&{3 - 2}&{ - 1 - 4}\\{ - 2 + 3}&{ - 2 - 2}&{1 - 5}\\{ - 4 - 1}&{ - 5 + 1}&{2 + 2}\end{array}} \right] = \left[ {\begin{array}{cccccccccccccccccccc}6&1&{ - 5}\\1&{ - 4}&{ - 4}\\{ - 5}&{ - 4}&4\end{array}} \right]$

So, $P = \cfrac{1}{2}(A + A') = \cfrac{1}{2}\left[ {\begin{array}{cccccccccccccccccccc}6&1&{ - 5}\\1&{ - 4}&{ - 4}\\{ - 5}&{ - 4}&4\end{array}} \right]$

$= \left[ {\begin{array}{cccccccccccccccccccc}3&{1/2}&{ - 5/2}\\{1/2}&{ - 2}&{ - 2}\\{ - 5/2}&{ - 2}&2\end{array}} \right]$and $P' = \left[ {\begin{array}{cccccccccccccccccccc}3&{1/2}&{ - 5/2}\\{1/2}&{ - 2}&{ - 2}\\{ - 5/2}&{ - 2}&2\end{array}} \right] = P$

Thus, P = $\cfrac{1}{2}(A + A')$is a symmetric matrix.
Also, let Q = $\cfrac{1}{2}(A - A')$

Again, $A - A' = \left[ {\begin{array}{cccccccccccccccccccc}3&3&{ - 1}\\{ - 2}&{ - 2}&1\\{ - 4}&{ - 5}&2\end{array}} \right] - \left[ {\begin{array}{cccccccccccccccccccc}3&{ - 2}&{ - 4}\\3&{ - 2}&{ - 5}\\{ - 1}&1&2\end{array}} \right]$

$= \left[ {\begin{array}{cccccccccccccccccccc}{3 - 3}&{3 + 2}&{ - 1 + 4}\\{ - 2 - 3}&{ - 2 + 2}&{1 + 5}\\{ - 4 + 1}&{ - 5 - 1}&{2 - 2}\end{array}} \right] = \left[ {\begin{array}{cccccccccccccccccccc}0&5&3\\{ - 5}&0&6\\{ - 3}&{ - 6}&0\end{array}} \right]$

$\therefore$ $\cfrac{1}{2}(A - A') = \cfrac{1}{2}\left[ {\begin{array}{cccccccccccccccccccc}0&5&3\\{ - 5}&0&6\\{ - 3}&{ - 6}&0\end{array}} \right] = \left[ {\begin{array}{cccccccccccccccccccc}0&{5/2}&{3/2}\\{ - 5/2}&0&3\\{ - 3/2}&{ - 3}&0\end{array}} \right]$

$\Rightarrow$ $Q' = \left[ {\begin{array}{cccccccccccccccccccc}0&{ - 5/2}&{ - 3/2}\\{5/2}&0&{ - 3}\\{3/2}&{ - 3}&0\end{array}} \right] = - \left[ {\begin{array}{cccccccccccccccccccc}0&{5/2}&{3/2}\\{ - 5/2}&0&3\\{ - 3/2}&{ - 3}&0\end{array}} \right] = - Q$.

Thus, Q =$\cfrac{1}{2}$(A$-$A’) is a skew symmetric matrix.

Hence, $P + Q = A = \left[ {\begin{array}{cccccccccccccccccccc}3&{\cfrac{1}{2}}&{\cfrac{{ - 5}}{2}}\\{\cfrac{1}{2}}&{ - 2}&{ - 2}\\{\cfrac{{ - 5}}{2}}&{ - 2}&2\end{array}} \right] + \left[ {\begin{array}{cccccccccccccccccccc}0&{\cfrac{5}{2}}&{\cfrac{3}{2}}\\{\cfrac{{ - 5}}{2}}&0&3\\{\cfrac{{ - 3}}{2}}&{ - 3}&0\end{array}} \right]$

(iv) Here, $A = \left[ {\begin{array}{cccccccccccccccccccc}1&5\\{ - 1}&2\end{array}} \right] \Rightarrow A' = \left[ {\begin{array}{cccccccccccccccccccc}1&{ - 1}\\5&2\end{array}} \right]$

Let $P = \cfrac{1}{2}(A + A')$
Now, $A + A' = \left[ {\begin{array}{cccccccccccccccccccc}1&5\\{ - 1}&5\end{array}} \right] + \left[ {\begin{array}{cccccccccccccccccccc}1&{ - 1}\\5&2\end{array}} \right] = \left[ {\begin{array}{cccccccccccccccccccc}{1 + 1}&{5 - 1}\\{ - 1 + 5}&{2 + 2}\end{array}} \right] = \left[ {\begin{array}{cccccccccccccccccccc}2&4\\4&4\end{array}} \right]$

$\Rightarrow$ $P = \cfrac{1}{2}(A + A')$

$= \cfrac{1}{2}\left[ {\begin{array}{cccccccccccccccccccc}2&4\\4&4\end{array}} \right] = \left[ {\begin{array}{cccccccccccccccccccc}1&2\\2&2\end{array}} \right] \Rightarrow P' = \left[ {\begin{array}{cccccccccccccccccccc}1&2\\2&2\end{array}} \right] = P$

Thus, P = $\cfrac{1}{2}$(A+ A’) is a symmetric matrix.

Also, let Q = $\cfrac{1}{2}$(A$-$A’)

Now, A$-$A’$= \left[ {\begin{array}{cccccccccccccccccccc}1&5\\{ - 1}&2\end{array}} \right] - \left[ {\begin{array}{cccccccccccccccccccc}1&{ - 1}\\5&2\end{array}} \right] = \left[ {\begin{array}{cccccccccccccccccccc}{1 - 1}&{5 + 1}\\{ - 1 - 5}&{2 - 2}\end{array}} \right] = \left[ {\begin{array}{cccccccccccccccccccc}0&6\\{ - 6}&0\end{array}} \right]$

$\Rightarrow$ $Q = \cfrac{1}{2}(A - A') = \cfrac{1}{2}\left[ {\begin{array}{cccccccccccccccccccc}0&6\\{ - 6}&0\end{array}} \right] = \left[ {\begin{array}{cccccccccccccccccccc}0&3\\3&0\end{array}} \right]$

$\Rightarrow$ $Q' = \left[ {\begin{array}{cccccccccccccccccccc}0&{ - 3}\\3&0\end{array}} \right] = - \left[ {\begin{array}{cccccccccccccccccccc}0&3\\{ - 3}&0\end{array}} \right] = - Q$.

Thus, Q =$\cfrac{1}{2}$(A$-$A’) is a skew symmetric matrix.

Hence, P + Q = A$= \left[ {\begin{array}{cccccccccccccccccccc}1&2\\2&2\end{array}} \right] + \left[ {\begin{array}{cccccccccccccccccccc}0&3\\{ - 3}&0\end{array}} \right]$.