Question
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 =
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]
Solution
Using the concept of symmetric and asymmetric matrix, we can write matrix A as A=21(A+A′)+21(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 21(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, 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=21(A+A′)+21(A−A′)
In the question we have to calculate C, which is of asymmetric form similar to 21(A−A′).
So, 21(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=21(A−A′)
On substituting the value of A and A’, we get,