Question

Let $A = |\begin{matrix} 3 & 2 & 5 \\ 4 & 1 & 3 \\ 0 & 6 & 7 \end{matrix}| .$ Express A as sum of two matrices such that one is symmetric and the other is skew symmetric.

Solution

$A = |\begin{matrix} 3 & 2 & 5 \\ 4 & 1 & 3 \\ 0 & 6 & 7 \end{matrix}| A' = |\begin{matrix} 3 & 4 & 0 \\ 2 & 1 & 6 \\ 5 & 3 & 7 \end{matrix}|$

Now, A can be written as:

$A = \frac{1}{2} A + A' + \frac{1}{2} A = A' A + A' = |\begin{matrix} 3 & 2 & 5 \\ 4 & 1 & 3 \\ 0 & 6 & 7 \end{matrix}| + |\begin{matrix} 3 & 4 & 0 \\ 2 & 1 & 6 \\ 5 & 3 & 7 \end{matrix}| = |\begin{matrix} 3 + 3 & 2 + 4 & 5 + 0 \\ 4 + 2 & 1 + 1 & 3 + 6 \\ 0 + 5 & 6 + 3 & 7 + 7 \end{matrix}| = |\begin{matrix} 6 & 6 & 5 \\ 6 & 2 & 9 \\ 5 & 9 & 14 \end{matrix}| \frac{1}{2} A + A' = \frac{1}{2} |\begin{matrix} 6 & 6 & 5 \\ 6 & 2 & 9 \\ 5 & 9 & 14 \end{matrix}| = |\begin{matrix} 3 & 3 & \frac{5}{2} \\ 3 & 1 & \frac{9}{2} \\ \frac{5}{2} & \frac{9}{2} & 7 \end{matrix}| = P, say Now, P' = |\begin{matrix} 3 & 3 & \frac{5}{2} \\ 3 & 1 & \frac{9}{2} \\ \frac{5}{2} & \frac{9}{2} & 7 \end{matrix}| Thus, P = \frac{1}{2} A + A' is a symmetric matrix .$

$Now, A - A' = |\begin{matrix} 3 - 3 & 2 - 4 & 5 - 0 \\ 4 - 2 & 1 - 1 & 3 - 6 \\ 0 - 5 & 6 - 3 & 7 - 7 \end{matrix}| = |\begin{matrix} 0 & - 2 & 5 \\ 2 & 0 & - 3 \\ - 5 & 3 & 0 \end{matrix}| \frac{1}{2} A - A' = |\begin{matrix} 0 & - 1 & \frac{5}{2} \\ 1 & 0 & \frac{- 3}{2} \\ \frac{- 5}{2} & \frac{3}{2} & 0 \end{matrix}| = Q, say Now, Q' = |\begin{matrix} 0 & 1 & \frac{- 5}{2} \\ - 1 & 0 & \frac{3}{2} \\ \frac{5}{2} & \frac{- 3}{2} & 0 \end{matrix}| = - |\begin{matrix} 0 & - 1 & \frac{- 5}{2} \\ 1 & 0 & \frac{- 3}{2} \\ \frac{- 5}{2} & \frac{3}{2} & 0 \end{matrix}| = - Q Thus, Q = \frac{1}{2} A - A' is a skew symmetric matrix . ∴ A = |\begin{matrix} 3 & 2 & 5 \\ 4 & 1 & 3 \\ 0 & 6 & 7 \end{matrix}| = |\begin{matrix} 3 & 3 & \frac{5}{2} \\ 3 & 1 & \frac{9}{2} \\ \frac{5}{2} & \frac{- 3}{2} & 7 \end{matrix}| + |\begin{matrix} 0 & - 1 & \frac{- 5}{2} \\ 1 & 0 & \frac{- 3}{2} \\ \frac{- 5}{2} & \frac{3}{2} & 0 \end{matrix}|$

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