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Question: If \[A = \left[ {\begin{array}{*{20}{c}} 1&0&1 \\\ 0&2&0 \\\ 1&{( - 1)}&4 \end{arra...

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 C = {\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]

Explanation

Solution

Using the concept of symmetric and asymmetric matrix, we can write matrix A as A=12(A+A)+12(AA)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 12(AA)\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
A = B + C{\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=12(A+A)+12(AA)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 12(AA)\dfrac{1}{2}(A - A').
So, 12(AA)=C\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=12(AA)C = \dfrac{1}{2}(A - A')
On substituting the value of A and A’, we get,

{\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.