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Special Determinants

If matrix A= ...

Question

If matrix $A = ⎡ ⎢ ⎣ \begin{matrix} 0 & 1 & - 1 4 & - 3 & 4 3 & - 3 & 4 \end{matrix} ⎤ ⎥ ⎦ = B + C$ , where $B$ is symmetric matrix and $C$ is skew-symmetric matrix.
Then the matrices $B$ and $C$ are:

A

B = ⎡ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣ \begin{matrix} 0 & \frac{5}{2} & 1 \frac{5}{2} & - 3 & \frac{1}{2} 1 & \frac{1}{2} & 4 \end{matrix} ⎤ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦

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B

B = ⎡ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣ \begin{matrix} 0 & - \frac{5}{2} & - 1 \frac{5}{2} & 3 & \frac{1}{2} 1 & - \frac{1}{2} & 4 \end{matrix} ⎤ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦

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C

C = ⎡ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣ \begin{matrix} 0 & - \frac{3}{2} & - 2 \frac{3}{2} & 0 & \frac{7}{2} 2 & - \frac{7}{2} & 0 \end{matrix} ⎤ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦

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D

C = ⎡ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣ \begin{matrix} 0 & - \frac{3}{2} & 2 - \frac{3}{2} & 0 & - \frac{7}{2} 2 & - \frac{7}{2} & 0 \end{matrix} ⎤ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦

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Solution

The correct option is C $C = ⎡ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣ \begin{matrix} 0 & - \frac{3}{2} & - 2 \frac{3}{2} & 0 & \frac{7}{2} 2 & - \frac{7}{2} & 0 \end{matrix} ⎤ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦$
Here matrix $A$ is expressed as sum of symmetric and skew-symmetric matrix.

Then $B = \frac{1}{2} (A + A^{T})$ which is the symmetric part, and $C = \frac{1}{2} (A - A^{T})$ which is the skew-symmetric part.

Now $A = ⎡ ⎢ ⎣ \begin{matrix} 0 & 1 & - 1 4 & - 3 & 4 3 & - 3 & 4 \end{matrix} ⎤ ⎥ ⎦ \Rightarrow A^{T} = ⎡ ⎢ ⎣ \begin{matrix} 0 & 4 & 3 1 & - 3 & - 3 - 1 & 4 & 4 \end{matrix} ⎤ ⎥ ⎦$

$\Rightarrow B = \frac{1}{2} ⎛ ⎜ ⎝ ⎡ ⎢ ⎣ \begin{matrix} 0 & 1 & - 1 4 & - 3 & 4 3 & - 3 & 4 \end{matrix} ⎤ ⎥ ⎦ + ⎡ ⎢ ⎣ \begin{matrix} 0 & 4 & 3 1 & - 3 & - 3 - 1 & 4 & 4 \end{matrix} ⎤ ⎥ ⎦ ⎞ ⎟ ⎠ = \frac{1}{2} ⎡ ⎢ ⎣ \begin{matrix} 0 & 5 & 2 5 & - 6 & 1 2 & 1 & 8 \end{matrix} ⎤ ⎥ ⎦$
$\Rightarrow B = ⎡ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣ \begin{matrix} 0 & \frac{5}{2} & 1 \frac{5}{2} & - 3 & \frac{1}{2} 1 & \frac{1}{2} & 4 \end{matrix} ⎤ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦$

and $C = \frac{1}{2} ⎛ ⎜ ⎝ ⎡ ⎢ ⎣ \begin{matrix} 0 & 1 & - 1 4 & - 3 & 4 3 & - 3 & 4 \end{matrix} ⎤ ⎥ ⎦ - ⎡ ⎢ ⎣ \begin{matrix} 0 & 4 & 3 1 & - 3 & - 3 - 1 & 4 & 4 \end{matrix} ⎤ ⎥ ⎦ ⎞ ⎟ ⎠ = \frac{1}{2} ⎡ ⎢ ⎣ \begin{matrix} 0 & - 3 & - 4 3 & 0 & 7 4 & - 7 & 0 \end{matrix} ⎤ ⎥ ⎦$
$\Rightarrow C = ⎡ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣ \begin{matrix} 0 & - \frac{3}{2} & - 2 \frac{3}{2} & 0 & \frac{7}{2} 2 & - \frac{7}{2} & 0 \end{matrix} ⎤ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦$

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