Question

# 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. $\left[ {\begin{array}{*{20}{c}} 0&{0.5}&0 \\ { - 0.5}&0&{0.5} \\ 0&{ - 0.5}&0 \end{array}} \right]$

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 $\dfrac{1}{2}(A - A')$.

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.