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If \[A = \left[ {\begin{array}{*{20}{c}}
  1&0&1 \\
  0&2&0 \\
  1&{( - 1)}&4
\end{array}} \right],A = B + C,B = {B^T},C = - {C^T}\] then \[{\text{C = }}\]
A. \[\left[ {\begin{array}{*{20}{c}}
  0&{0.5}&0 \\
  { - 0.5}&0&0 \\
  0&0&0
\end{array}} \right]\]
B. \[\left[ {\begin{array}{*{20}{c}}
  0&0&0 \\
  0&0&{0.5} \\
  0&{ - 0.5}&0
\end{array}} \right]\]
C. \[\left[ {\begin{array}{*{20}{c}}
  0&{ - 0.5}&{0.5} \\
  {0.5}&0&{0.5} \\
  { - 05}&0&0
\end{array}} \right]\]
D. \[\left[ {\begin{array}{*{20}{c}}
  0&{0.5}&0 \\
  { - 0.5}&0&{0.5} \\
  0&{ - 0.5}&0
\end{array}} \right]\]

Answer Verified Verified
Hint: Using the concept of symmetric and asymmetric matrix, we can write matrix A as \[A = \dfrac{1}{2}(A + A') + \dfrac{1}{2}(A - A')\], and hence the second part in the question is of asymmetric form and so in order to find matrix C we just have to calculate matrix of \[\dfrac{1}{2}(A - A')\].

Complete step by step answer:

As the given matrix is \[A = \left[ {\begin{array}{*{20}{c}}
  1&0&1 \\
  0&2&0 \\
  1&{( - 1)}&4
\end{array}} \right]\]
And the above matrix is given as
\[{\text{A = B + C}}\], in which B is a symmetric matrix while C is an asymmetric matrix.
So, indirectly we have to represent A as a sum of symmetric and asymmetric matrix as \[A = \dfrac{1}{2}(A + A') + \dfrac{1}{2}(A - A')\]
In the question we have to calculate C, which is of asymmetric form similar to \[\dfrac{1}{2}(A - A')\].
So, \[\dfrac{1}{2}(A - A') = C\]
As, \[A' = \left[ {\begin{array}{*{20}{c}}
  1&0&1 \\
  0&2&{ - 1} \\
  1&0&4
\end{array}} \right]\]
Now on calculating,
\[C = \dfrac{1}{2}(A - A')\]
On substituting the value of A and A’, we get,
\[ \Rightarrow {\text{C = }}\dfrac{{\text{1}}}{{\text{2}}}{\text{(}}\left[ {\begin{array}{*{20}{c}}
  {\text{1}}&{\text{0}}&{\text{1}} \\
  {\text{0}}&{\text{2}}&{\text{0}} \\
  {\text{1}}&{{\text{( - 1)}}}&{\text{4}}
\end{array}} \right]{\text{ - }}\left[ {\begin{array}{*{20}{c}}
  {\text{1}}&{\text{0}}&{\text{1}} \\
  {\text{0}}&{\text{2}}&{{\text{ - 1}}} \\
  {\text{1}}&{\text{0}}&{\text{4}}
\end{array}} \right])\]
On simplifying we get,
\[ \Rightarrow C = \dfrac{1}{2}(\left[ {\begin{array}{*{20}{c}}
  0&0&0 \\
  0&0&1 \\
  0&{( - 1)}&0
\end{array}} \right])\]
\[ \Rightarrow C = \left[ {\begin{array}{*{20}{c}}
  0&0&0 \\
  0&0&{0.5} \\
  0&{ - 0.5}&0
\end{array}} \right]\]
Hence, option (B) is the correct answer.

Note: In linear algebra, a symmetric matrix is a square matrix that is equal to its transpose.In mathematics, particularly in linear algebra, a skew-symmetric matrix is a square matrix whose transpose equals its negative. The zero matrix has that property, so it is a skew-symmetric matrix. Skew-symmetric matrices also form a vector space, and the zero matrix is the zero vector. In fact, the zero matrix is only a matrix which is both symmetric and skew-symmetric.
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