Question

If A is a symmetric matrix and B is a skew symmetric matrix such that A + B = \left[ {\begin{array}{*{20}{c}} 2&3 \\\ 5&{ - 1} \end{array}} \right], then AB is equal to?
A. \left[ {\begin{array}{*{20}{c}} { - 4}&2 \\\ 1&4 \end{array}} \right]
B. \left[ {\begin{array}{*{20}{c}} { - 4}&{ - 2} \\\ { - 1}&4 \end{array}} \right]
C. \left[ {\begin{array}{*{20}{c}} 4&{ - 2} \\\ { - 1}&{ - 4} \end{array}} \right]
D. \left[ {\begin{array}{*{20}{c}} 4&{ - 2} \\\ 1&{ - 4} \end{array}} \right]

Answer 1

To solve this question, we have to remember that a matrix A is a symmetric matrix if $A' = A$ and A is skew symmetric matrix if $A' = - A$, where $A'$ is the transpose of matrix A. If $A = {\left[ {{a_{ij}}} \right]_{m \times n}}$, then $A' = {\left[ {{a_{ji}}} \right]_{n \times m}}$

Complete step-by-step answer :
Given that,
A is a symmetric matrix, i.e. $A' = A$ ,
And,
B is a skew symmetric matrix, i.e. $B' = - B$
Such that,
A + B = \left[ {\begin{array}{*{20}{c}} 2&3 \\\ 5&{ - 1} \end{array}} \right] ………. (i)
We have to find AB.
So,
We know that,
${\left( {A + B} \right)^\prime } = A' + B'$ ……….. (ii)
We have,
Transposing equation (i), we will get
{\left( {A + B} \right)^\prime } = \left[ {\begin{array}{*{20}{c}} 2&5 \\\ 3&{ - 1} \end{array}} \right]
Using equation (ii), we can write this as:

2&5 \\\ 3&{ - 1} \end{array}} \right]$$ ………. (iii) Putting $A' = A$ and $B' = - B$ in equation (iii), we will get $$A - B = \left[ {\begin{array}{*{20}{c}} 2&5 \\\ 3&{ - 1} \end{array}} \right]$$ ……….. (iv) Adding equation (i) and (iv), we will get $$ \Rightarrow A - B + A + B = \left[ {\begin{array}{*{20}{c}} 2&5 \\\ 3&{ - 1} \end{array}} \right] + \left[ {\begin{array}{*{20}{c}} 2&3 \\\ 5&{ - 1} \end{array}} \right]$$ $$ \Rightarrow 2A = \left[ {\begin{array}{*{20}{c}} 4&8 \\\ 8&{ - 2} \end{array}} \right]$$ $$ \Rightarrow A = \dfrac{1}{2}\left[ {\begin{array}{*{20}{c}} 4&8 \\\ 8&{ - 2} \end{array}} \right]$$ $$ \Rightarrow A = \left[ {\begin{array}{*{20}{c}} 2&4 \\\ 4&{ - 1} \end{array}} \right]$$ Putting this in equation (i), we will get $$ \Rightarrow \left[ {\begin{array}{*{20}{c}} 2&4 \\\ 4&{ - 1} \end{array}} \right] + B = \left[ {\begin{array}{*{20}{c}} 2&3 \\\ 5&{ - 1} \end{array}} \right]$$ $$ \Rightarrow B = \left[ {\begin{array}{*{20}{c}} 2&3 \\\ 5&{ - 1} \end{array}} \right] - \left[ {\begin{array}{*{20}{c}} 2&4 \\\ 4&{ - 1} \end{array}} \right]$$ $$ \Rightarrow B = \left[ {\begin{array}{*{20}{c}} 0&{ - 1} \\\ 1&0 \end{array}} \right]$$ Now, we will find AB, So, $$ \Rightarrow AB = \left[ {\begin{array}{*{20}{c}} 2&4 \\\ 4&{ - 1} \end{array}} \right]\left[ {\begin{array}{*{20}{c}} 0&{ - 1} \\\ 1&0 \end{array}} \right]$$ $$ \Rightarrow AB = \left[ {\begin{array}{*{20}{c}} {0 + 4}&{ - 2 + 0} \\\ {0 - 1}&{ - 4 + 0} \end{array}} \right]$$ $$ \Rightarrow AB = \left[ {\begin{array}{*{20}{c}} 4&{ - 2} \\\ { - 1}&{ - 4} \end{array}} \right]$$ Here, we get $$AB = \left[ {\begin{array}{*{20}{c}} 4&{ - 2} \\\ { - 1}&{ - 4} \end{array}} \right]$$ **Hence, the correct answer is option (C).** **Note** : whenever we asked such type of questions, we should also remember that any square matrix can be represented as the sum of a symmetric and a skew symmetric matrix, i.e. A be a square matrix, then we can write $A = \dfrac{1}{2}\left( {A + A'} \right) + \dfrac{1}{2}\left( {A - A'} \right)$