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

Hint: 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:
\[A' + B' = \left[ {\begin{array}{*{20}{c}}
  2&5 \\
  3&{ - 1}
\end{array}} \right]\] ………. (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 $A = \dfrac{1}{2}\left( {A + A'} \right) + \dfrac{1}{2}\left( {A - A'} \right)$