\displaystyle \textbf{Question 1: }\text{If }A=\begin{bmatrix}2&3\\4&5\end{bmatrix},
\displaystyle \text{prove that }A-A^T\text{ is a skew-symmetric matrix.}
\displaystyle \text{Answer:}
\displaystyle A=\begin{bmatrix}2&3\\4&5\end{bmatrix},\qquad A^T=\begin{bmatrix}2&4\\3&5\end{bmatrix}
\displaystyle \text{Let }Q=A-A^T.
\displaystyle Q=\begin{bmatrix}2&3\\4&5\end{bmatrix}-\begin{bmatrix}2&4\\3&5\end{bmatrix}=\begin{bmatrix}0&-1\\1&0\end{bmatrix}
\displaystyle Q^T=(A-A^T)^T
\displaystyle =A^T-(A^T)^T\qquad\left[\because\ (A-B)^T=A^T-B^T\right]
\displaystyle =A^T-A\qquad\left[\because\ (A^T)^T=A\right]
\displaystyle =-(A-A^T)=-Q
\displaystyle \therefore\ Q^T=-Q.
\displaystyle \therefore\ Q=A-A^T\text{ is a skew-symmetric matrix.}
\displaystyle \\

\displaystyle \textbf{Question 2: }\text{If }A=\begin{bmatrix}3&-4\\1&-1\end{bmatrix},
\displaystyle \text{show that }A-A^T\text{ is a skew-symmetric matrix.}
\displaystyle \text{Answer:}
\displaystyle A=\begin{bmatrix}3&-4\\1&-1\end{bmatrix},\qquad A^T=\begin{bmatrix}3&1\\-4&-1\end{bmatrix}
\displaystyle \text{Let }Q=A-A^T.
\displaystyle Q=\begin{bmatrix}3&-4\\1&-1\end{bmatrix}-\begin{bmatrix}3&1\\-4&-1\end{bmatrix}=\begin{bmatrix}0&-5\\5&0\end{bmatrix}
\displaystyle Q^T=(A-A^T)^T
\displaystyle =A^T-(A^T)^T\qquad\left[\because\ (A-B)^T=A^T-B^T\right]
\displaystyle =A^T-A\qquad\left[\because\ (A^T)^T=A\right]
\displaystyle =-(A-A^T)=-Q
\displaystyle \therefore\ Q^T=-Q.
\displaystyle \therefore\ Q=A-A^T\text{ is a skew-symmetric matrix.}
\displaystyle \\

\displaystyle \textbf{Question 3: }\text{If the matrix }A=\begin{bmatrix}5&2&x\\y&z&-3\\4&t&-7\end{bmatrix}
\displaystyle \text{is symmetric, find }x,y,z\text{ and }t.
\displaystyle \text{Answer:}
\displaystyle A=\begin{bmatrix}5&2&x\\y&z&-3\\4&t&-7\end{bmatrix}
\displaystyle \therefore\ A^T=\begin{bmatrix}5&y&4\\2&z&t\\x&-3&-7\end{bmatrix}
\displaystyle \text{Since }A\text{ is symmetric, }A^T=A.
\displaystyle \therefore\ \begin{bmatrix}5&y&4\\2&z&t\\x&-3&-7\end{bmatrix}=\begin{bmatrix}5&2&x\\y&z&-3\\4&t&-7\end{bmatrix}
\displaystyle \text{Comparing the corresponding off-diagonal elements,}
\displaystyle y=2,\qquad x=4,\qquad t=-3
\displaystyle \text{The diagonal element }z\text{ is unrestricted and may have any value.}
\displaystyle \therefore\ x=4,\quad y=2,\quad t=-3,\quad z\in\mathbb{R}.
\displaystyle \\

\displaystyle \textbf{Question 4: }\text{Let }A=\begin{bmatrix}3&2&7\\1&4&3\\-2&5&8\end{bmatrix}.
\displaystyle \text{Find matrices }X\text{ and }Y\text{ such that }X+Y=A,\text{ where }X
\displaystyle \text{is symmetric and }Y\text{ is skew-symmetric.}
\displaystyle \text{Answer:}
\displaystyle A=\begin{bmatrix}3&2&7\\1&4&3\\-2&5&8\end{bmatrix}
\displaystyle \therefore\ A^T=\begin{bmatrix}3&1&-2\\2&4&5\\7&3&8\end{bmatrix}
\displaystyle \text{Let }X=\frac{1}{2}(A+A^T).
\displaystyle X=\frac{1}{2}\left(\begin{bmatrix}3&2&7\\1&4&3\\-2&5&8\end{bmatrix}+\begin{bmatrix}3&1&-2\\2&4&5\\7&3&8\end{bmatrix}\right)
\displaystyle =\frac{1}{2}\begin{bmatrix}6&3&5\\3&8&8\\5&8&16\end{bmatrix}
\displaystyle =\begin{bmatrix}3&\frac{3}{2}&\frac{5}{2}\\\frac{3}{2}&4&4\\\frac{5}{2}&4&8\end{bmatrix}
\displaystyle X^T=\left[\frac{1}{2}(A+A^T)\right]^T
\displaystyle =\frac{1}{2}\left[A^T+(A^T)^T\right]
\displaystyle =\frac{1}{2}(A^T+A)=\frac{1}{2}(A+A^T)=X
\displaystyle \therefore\ X\text{ is symmetric.}
\displaystyle \text{Let }Y=\frac{1}{2}(A-A^T).
\displaystyle Y=\frac{1}{2}\left(\begin{bmatrix}3&2&7\\1&4&3\\-2&5&8\end{bmatrix}-\begin{bmatrix}3&1&-2\\2&4&5\\7&3&8\end{bmatrix}\right)
\displaystyle =\frac{1}{2}\begin{bmatrix}0&1&9\\-1&0&-2\\-9&2&0\end{bmatrix}
\displaystyle =\begin{bmatrix}0&\frac{1}{2}&\frac{9}{2}\\-\frac{1}{2}&0&-1\\-\frac{9}{2}&1&0\end{bmatrix}
\displaystyle Y^T=\left[\frac{1}{2}(A-A^T)\right]^T
\displaystyle =\frac{1}{2}\left[A^T-(A^T)^T\right]
\displaystyle =\frac{1}{2}(A^T-A)=-\frac{1}{2}(A-A^T)=-Y
\displaystyle \therefore\ Y\text{ is skew-symmetric.}
\displaystyle X+Y=\frac{1}{2}(A+A^T)+\frac{1}{2}(A-A^T)
\displaystyle =\frac{1}{2}(2A)=A
\displaystyle \therefore\ A=X+Y,
\displaystyle X=\begin{bmatrix}3&\frac{3}{2}&\frac{5}{2}\\\frac{3}{2}&4&4\\\frac{5}{2}&4&8\end{bmatrix},\qquad Y=\begin{bmatrix}0&\frac{1}{2}&\frac{9}{2}\\-\frac{1}{2}&0&-1\\-\frac{9}{2}&1&0\end{bmatrix}.
\displaystyle \\

\displaystyle \textbf{Question 5: }\text{Express the matrix }A=\begin{bmatrix}4&2&-1\\3&5&7\\1&-2&1\end{bmatrix}
\displaystyle \text{as the sum of a symmetric and a skew-symmetric matrix.}
\displaystyle \text{Answer:}
\displaystyle A=\begin{bmatrix}4&2&-1\\3&5&7\\1&-2&1\end{bmatrix}
\displaystyle \therefore\ A^T=\begin{bmatrix}4&3&1\\2&5&-2\\-1&7&1\end{bmatrix}
\displaystyle \text{Let }X=\frac{1}{2}(A+A^T).
\displaystyle X=\frac{1}{2}\left(\begin{bmatrix}4&2&-1\\3&5&7\\1&-2&1\end{bmatrix}+\begin{bmatrix}4&3&1\\2&5&-2\\-1&7&1\end{bmatrix}\right)
\displaystyle =\frac{1}{2}\begin{bmatrix}8&5&0\\5&10&5\\0&5&2\end{bmatrix}
\displaystyle =\begin{bmatrix}4&\frac{5}{2}&0\\\frac{5}{2}&5&\frac{5}{2}\\0&\frac{5}{2}&1\end{bmatrix}
\displaystyle X^T=\left[\frac{1}{2}(A+A^T)\right]^T
\displaystyle =\frac{1}{2}\left[A^T+(A^T)^T\right]
\displaystyle =\frac{1}{2}(A^T+A)=\frac{1}{2}(A+A^T)=X
\displaystyle \therefore\ X\text{ is symmetric.}
\displaystyle \text{Let }Y=\frac{1}{2}(A-A^T).
\displaystyle Y=\frac{1}{2}\left(\begin{bmatrix}4&2&-1\\3&5&7\\1&-2&1\end{bmatrix}-\begin{bmatrix}4&3&1\\2&5&-2\\-1&7&1\end{bmatrix}\right)
\displaystyle =\frac{1}{2}\begin{bmatrix}0&-1&-2\\1&0&9\\2&-9&0\end{bmatrix}
\displaystyle =\begin{bmatrix}0&-\frac{1}{2}&-1\\\frac{1}{2}&0&\frac{9}{2}\\1&-\frac{9}{2}&0\end{bmatrix}
\displaystyle Y^T=\left[\frac{1}{2}(A-A^T)\right]^T
\displaystyle =\frac{1}{2}\left[A^T-(A^T)^T\right]
\displaystyle =\frac{1}{2}(A^T-A)=-\frac{1}{2}(A-A^T)=-Y
\displaystyle \therefore\ Y\text{ is skew-symmetric.}
\displaystyle X+Y=\frac{1}{2}(A+A^T)+\frac{1}{2}(A-A^T)=A
\displaystyle \therefore\ A=\begin{bmatrix}4&\frac{5}{2}&0\\\frac{5}{2}&5&\frac{5}{2}\\0&\frac{5}{2}&1\end{bmatrix}+\begin{bmatrix}0&-\frac{1}{2}&-1\\\frac{1}{2}&0&\frac{9}{2}\\1&-\frac{9}{2}&0\end{bmatrix}.
\displaystyle \\

\displaystyle \textbf{Question 6: }\text{Define a symmetric matrix. Prove that for }
\displaystyle A=\begin{bmatrix}2&4\\5&6\end{bmatrix},\ A+A^T\text{ is a symmetric matrix,}
\displaystyle \text{where }A^T\text{ is the transpose of }A.
\displaystyle \text{Answer:}
\displaystyle \text{A square matrix }A\text{ is called symmetric if }A^T=A.
\displaystyle A=\begin{bmatrix}2&4\\5&6\end{bmatrix},\qquad A^T=\begin{bmatrix}2&5\\4&6\end{bmatrix}
\displaystyle A+A^T=\begin{bmatrix}2&4\\5&6\end{bmatrix}+\begin{bmatrix}2&5\\4&6\end{bmatrix}
\displaystyle =\begin{bmatrix}4&9\\9&12\end{bmatrix}
\displaystyle \text{Let }P=A+A^T.
\displaystyle P^T=(A+A^T)^T
\displaystyle =A^T+(A^T)^T\qquad\left[\because\ (A+B)^T=A^T+B^T\right]
\displaystyle =A^T+A\qquad\left[\because\ (A^T)^T=A\right]
\displaystyle =A+A^T=P
\displaystyle \therefore\ P^T=P.
\displaystyle \therefore\ A+A^T\text{ is a symmetric matrix.}
\displaystyle \\

\displaystyle \textbf{Question 7: }\text{Express the matrix }A=\begin{bmatrix}3&-4\\1&-1\end{bmatrix}
\displaystyle \text{as the sum of a symmetric and a skew-symmetric matrix.}
\displaystyle \text{Answer:}
\displaystyle A=\begin{bmatrix}3&-4\\1&-1\end{bmatrix}
\displaystyle \therefore\ A^T=\begin{bmatrix}3&1\\-4&-1\end{bmatrix}
\displaystyle \text{Let }X=\frac{1}{2}(A+A^T).
\displaystyle X=\frac{1}{2}\left(\begin{bmatrix}3&-4\\1&-1\end{bmatrix}+\begin{bmatrix}3&1\\-4&-1\end{bmatrix}\right)
\displaystyle =\frac{1}{2}\begin{bmatrix}6&-3\\-3&-2\end{bmatrix}
\displaystyle =\begin{bmatrix}3&-\frac{3}{2}\\-\frac{3}{2}&-1\end{bmatrix}
\displaystyle X^T=\begin{bmatrix}3&-\frac{3}{2}\\-\frac{3}{2}&-1\end{bmatrix}=X
\displaystyle \therefore\ X\text{ is symmetric.}
\displaystyle \text{Let }Y=\frac{1}{2}(A-A^T).
\displaystyle Y=\frac{1}{2}\left(\begin{bmatrix}3&-4\\1&-1\end{bmatrix}-\begin{bmatrix}3&1\\-4&-1\end{bmatrix}\right)
\displaystyle =\frac{1}{2}\begin{bmatrix}0&-5\\5&0\end{bmatrix}
\displaystyle =\begin{bmatrix}0&-\frac{5}{2}\\\frac{5}{2}&0\end{bmatrix}
\displaystyle Y^T=\begin{bmatrix}0&\frac{5}{2}\\-\frac{5}{2}&0\end{bmatrix}=-Y
\displaystyle \therefore\ Y\text{ is skew-symmetric.}
\displaystyle X+Y=\begin{bmatrix}3&-\frac{3}{2}\\-\frac{3}{2}&-1\end{bmatrix}+\begin{bmatrix}0&-\frac{5}{2}\\\frac{5}{2}&0\end{bmatrix}
\displaystyle =\begin{bmatrix}3&-4\\1&-1\end{bmatrix}=A
\displaystyle \therefore\ A=\begin{bmatrix}3&-\frac{3}{2}\\-\frac{3}{2}&-1\end{bmatrix}+\begin{bmatrix}0&-\frac{5}{2}\\\frac{5}{2}&0\end{bmatrix}.
\displaystyle \\

\displaystyle \textbf{Question 8: }\text{Express the following matrix as the sum of a symmetric}
\displaystyle \text{and a skew-symmetric matrix, and verify your result:}
\displaystyle \begin{bmatrix}3&-2&-4\\3&-2&-5\\-1&1&2\end{bmatrix}
\displaystyle \text{Answer:}
\displaystyle A=\begin{bmatrix}3&-2&-4\\3&-2&-5\\-1&1&2\end{bmatrix}
\displaystyle \therefore\ A^T=\begin{bmatrix}3&3&-1\\-2&-2&1\\-4&-5&2\end{bmatrix}
\displaystyle \text{Let }X=\frac{1}{2}(A+A^T).
\displaystyle X=\frac{1}{2}\left(\begin{bmatrix}3&-2&-4\\3&-2&-5\\-1&1&2\end{bmatrix}+\begin{bmatrix}3&3&-1\\-2&-2&1\\-4&-5&2\end{bmatrix}\right)
\displaystyle =\frac{1}{2}\begin{bmatrix}6&1&-5\\1&-4&-4\\-5&-4&4\end{bmatrix}
\displaystyle =\begin{bmatrix}3&\frac{1}{2}&-\frac{5}{2}\\\frac{1}{2}&-2&-2\\-\frac{5}{2}&-2&2\end{bmatrix}
\displaystyle X^T=\begin{bmatrix}3&\frac{1}{2}&-\frac{5}{2}\\\frac{1}{2}&-2&-2\\-\frac{5}{2}&-2&2\end{bmatrix}=X
\displaystyle \therefore\ X\text{ is symmetric.}
\displaystyle \text{Let }Y=\frac{1}{2}(A-A^T).
\displaystyle Y=\frac{1}{2}\left(\begin{bmatrix}3&-2&-4\\3&-2&-5\\-1&1&2\end{bmatrix}-\begin{bmatrix}3&3&-1\\-2&-2&1\\-4&-5&2\end{bmatrix}\right)
\displaystyle =\frac{1}{2}\begin{bmatrix}0&-5&-3\\5&0&-6\\3&6&0\end{bmatrix}
\displaystyle =\begin{bmatrix}0&-\frac{5}{2}&-\frac{3}{2}\\\frac{5}{2}&0&-3\\\frac{3}{2}&3&0\end{bmatrix}
\displaystyle Y^T=\begin{bmatrix}0&\frac{5}{2}&\frac{3}{2}\\-\frac{5}{2}&0&3\\-\frac{3}{2}&-3&0\end{bmatrix}=-Y
\displaystyle \therefore\ Y\text{ is skew-symmetric.}
\displaystyle X+Y=\begin{bmatrix}3&\frac{1}{2}&-\frac{5}{2}\\\frac{1}{2}&-2&-2\\-\frac{5}{2}&-2&2\end{bmatrix}+\begin{bmatrix}0&-\frac{5}{2}&-\frac{3}{2}\\\frac{5}{2}&0&-3\\\frac{3}{2}&3&0\end{bmatrix}
\displaystyle =\begin{bmatrix}3&-2&-4\\3&-2&-5\\-1&1&2\end{bmatrix}=A
\displaystyle \therefore\ A=\begin{bmatrix}3&\frac{1}{2}&-\frac{5}{2}\\\frac{1}{2}&-2&-2\\-\frac{5}{2}&-2&2\end{bmatrix}+\begin{bmatrix}0&-\frac{5}{2}&-\frac{3}{2}\\\frac{5}{2}&0&-3\\\frac{3}{2}&3&0\end{bmatrix}.
\displaystyle \\

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