Question

Find the symmetric and skew-symmetric parts of the matrix $A = \left[ {\begin{array}{*{20}{c}}
  1&2&4 \\
  6&8&1 \\
  3&5&7
\end{array}} \right]$

Answer 1

Hint: We can find the transpose of the matrix by interchanging the rows and columns. We can obtain the symmetric matrix by adding the matrix and its transpose and dividing it with 2. Similarly, the skew symmetric matrix can be obtained by subtracting the transpose of the matrix from the matrix and diving it with 2. Then we will get the symmetric and skew-symmetric parts of the matrix.

Complete step-by-step answer:
We have the matrix
 $A = \left[ {\begin{array}{*{20}{c}}
  1&2&4 \\
  6&8&1 \\
  3&5&7
\end{array}} \right]$
We know that transpose of a matrix is obtained by interchanging the rows and columns of the matrix.
So $A'$ can be found out by interchanging the rows and columns of the matrix A.
 $ \Rightarrow A' = \left[ {\begin{array}{*{20}{c}}
  1&6&3 \\
  2&8&5 \\
  4&1&7
\end{array}} \right]$
We know that $A + A'$ is a symmetric matrix and $A - A'$ is a skew symmetric matrix.
We can add both these vectors,
 $ \Rightarrow \left( {A + A'} \right) + \left( {A - A'} \right) = 2A$
On dividing throughout with 2, we get.
 $ \Rightarrow A = \dfrac{1}{2}\left( {A + A'} \right) + \dfrac{1}{2}\left( {A - A'} \right)$
Now A is expressed as the sum a symmetric and a skew-symmetric matrix.
Now we can find $A + A'$ ,
 $ \Rightarrow A + A' = \left[ {\begin{array}{*{20}{c}}
  1&2&4 \\
  6&8&1 \\
  3&5&7
\end{array}} \right] + \left[ {\begin{array}{*{20}{c}}
  1&6&3 \\
  2&8&5 \\
  4&1&7
\end{array}} \right]$
We know that matrix addition is done by adding the corresponding terms.
 $ \Rightarrow A + A' = \left[ {\begin{array}{*{20}{c}}
  2&8&7 \\
  8&{16}&6 \\
  7&6&{14}
\end{array}} \right]$
On dividing with 2, we get,
 $ \Rightarrow \dfrac{1}{2}\left( {A + A'} \right) = \dfrac{1}{2}\left[ {\begin{array}{*{20}{c}}
  2&8&7 \\
  8&{16}&6 \\
  7&6&{14}
\end{array}} \right]$
 $ \Rightarrow \dfrac{1}{2}\left( {A + A'} \right) = \left[ {\begin{array}{*{20}{c}}
  1&4&{\dfrac{3}{2}} \\
  4&8&3 \\
  {\dfrac{3}{2}}&3&7
\end{array}} \right]$
Now we can find $A - A'$ ,
 $ \Rightarrow A - A' = \left[ {\begin{array}{*{20}{c}}
  1&2&4 \\
  6&8&1 \\
  3&5&7
\end{array}} \right] - \left[ {\begin{array}{*{20}{c}}
  1&6&3 \\
  2&8&5 \\
  4&1&7
\end{array}} \right]$
We know that matrix subtraction is done by subtracting the corresponding terms.
 $ \Rightarrow A - A' = \left[ {\begin{array}{*{20}{c}}
  0&{ - 4}&1 \\
  4&0&{ - 4} \\
  1&4&0
\end{array}} \right]$
On dividing with 2, we get,
 $ \Rightarrow \dfrac{1}{2}\left( {A - A'} \right) = \dfrac{1}{2}\left[ {\begin{array}{*{20}{c}}
  0&{ - 4}&1 \\
  4&0&{ - 4} \\
  1&4&0
\end{array}} \right]$
 $ \Rightarrow \dfrac{1}{2}\left( {A - A'} \right) = \left[ {\begin{array}{*{20}{c}}
  0&{ - 2}&{\dfrac{1}{2}} \\
  2&0&{ - 2} \\
  {\dfrac{1}{2}}&2&0
\end{array}} \right]$
Now we can write A as
 $ \Rightarrow A = \dfrac{1}{2}\left( {A + A'} \right) + \dfrac{1}{2}\left( {A - A'} \right)$
 $ \Rightarrow A = \left[ {\begin{array}{*{20}{c}}
  1&4&{\dfrac{3}{2}} \\
  4&8&3 \\
  {\dfrac{3}{2}}&3&7
\end{array}} \right] + \left[ {\begin{array}{*{20}{c}}
  0&{ - 2}&{\dfrac{1}{2}} \\
  2&0&{ - 2} \\
  {\dfrac{1}{2}}&2&0
\end{array}} \right]$
Thus the symmetric and skew symmetric parts of A are $\left[ {\begin{array}{*{20}{c}}
  1&4&{\dfrac{3}{2}} \\
  4&8&3 \\
  {\dfrac{3}{2}}&3&7
\end{array}} \right]$ and $\left[ {\begin{array}{*{20}{c}}
  0&{ - 2}&{\dfrac{1}{2}} \\
  2&0&{ - 2} \\
  {\dfrac{1}{2}}&2&0
\end{array}} \right]$ respectively.

Note: A symmetric matrix is the matrix having its transpose equal to the matrix itself. It will be symmetric along the diagonal entries. A skew-symmetric matrix is the matrix that has its transpose equal to the negative of the matrix. The diagonal entries of a skew-symmetric matrix will be always equal to zero. All the square matrices can be expressed as the sum of a symmetric and a skew symmetric matrix. All these are applicable only for square matrices. Matrix addition is defined only for matrices of the same order. Sum of 2 matrices is given by the sum of the corresponding elements in the matrix. The product of a scalar with a matrix is given by the product of the scale with all the elements of the matrix.