\displaystyle \textbf{Question 1:} 
\displaystyle \text{If } A=\begin{bmatrix}2 & 3\\4 & 5\end{bmatrix}, \text{ prove that } A-A^T \text{ is a skew-symmetric matrix.}

\displaystyle \text{Answer:}

\displaystyle \text{Given: } A=\begin{bmatrix}2&3\\4&5\end{bmatrix} A^{T}=\begin{bmatrix}2&4\\3&5\end{bmatrix} \\[10pt] \text{Now,} (A-A^{T})= \begin{bmatrix}2&3\\4&5\end{bmatrix} - \begin{bmatrix}2&4\\3&5\end{bmatrix} = \begin{bmatrix} 2-2 & 3-4\\ 4-3 & 5-5 \end{bmatrix} = \begin{bmatrix} 0 & -1\\ 1 & 0 \end{bmatrix} \quad (1) \\ [10pt] (A-A^{T})^{T} = \begin{bmatrix} 0 & -1\\ 1 & 0 \end{bmatrix}^{T} = \begin{bmatrix} 0 & 1\\ -1 & 0 \end{bmatrix} = - \begin{bmatrix} 0 & -1\\ 1 & 0 \end{bmatrix} \\ [10pt] \Rightarrow (A-A^{T})^{T}=-(A-A^{T}) \\[10pt] \quad [\text{Using (1)}] \therefore\ (A-A^{T}) \text{ is a skew-symmetric matrix.}

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\displaystyle \textbf{Question 2:}
\displaystyle \text{If } A=\begin{bmatrix}3 & -4\\1 & -1\end{bmatrix}, \text{ show that } A-A^T \text{ is a skew-symmetric matrix.}

\displaystyle \text{Answer:}

\displaystyle \text{Given } A=\begin{bmatrix}3 & -4\\ 1 & -1\end{bmatrix} \quad A^{T}=\begin{bmatrix}3 & 1\\ -4 & -1\end{bmatrix} \\ [10pt] \text{Now,} \quad A-A^{T} =\begin{bmatrix}3 & -4\\ 1 & -1\end{bmatrix} -\begin{bmatrix}3 & 1\\ -4 & -1\end{bmatrix} =\begin{bmatrix}3-3 & -4-1\\ 1+4 & -1+1\end{bmatrix} =\begin{bmatrix}0 & -5\\ 5 & 0\end{bmatrix} \quad (1) \\ [10pt] \text{Then,} \quad (A-A^{T})^{T} =\begin{bmatrix}0 & -5\\ 5 & 0\end{bmatrix}^{T} =\begin{bmatrix}0 & 5\\ -5 & 0\end{bmatrix} =-\begin{bmatrix}0 & -5\\ 5 & 0\end{bmatrix} \\ [10pt] \Rightarrow (A-A^{T})^{T}=-(A-A^{T}) \\ [10pt] \text{Hence, } (A-A^{T}) \text{ is a skew-symmetric matrix.}

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\displaystyle \textbf{Question 3:} 
\displaystyle \text{If the matrix } A=\begin{bmatrix} 5 & 2 & x\\ y & z & -3\\ 4 & t & -7 \end{bmatrix} \text{ is a symmetric matrix, find } x,y,z \text{ and } t.

\displaystyle \text{Answer:}

\displaystyle \text{Given : } A = \begin{bmatrix} 5 & 2 & x \\ y & z & -3 \\ 4 & t & -7 \end{bmatrix} \Rightarrow A^T = \begin{bmatrix} 5 & y & 4 \\ 2 & z & t \\ x & -3 & -7 \end{bmatrix} \\ [10pt] \text{Since } A \text{ is a symmetric matrix, } A^T = A. \\ [10pt] \Rightarrow \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} \\ [10pt] \text{Comparing corresponding elements:} x = 4,\quad y = 2,\quad t = -3,\quad z = z \\ [10pt] \text{Hence, } x = 4,\ y = 2,\ t = -3 \text{ and } z \text{ can have any value.}

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\displaystyle \textbf{Question 4:} 
\displaystyle \text{Let } A=\begin{bmatrix} 3 & 2 & 7\\ 1 & 4 & 3\\ -2 & 5 & 8 \end{bmatrix}. \text{ Find matrices } X \text{ and } Y \text{ such that } X+Y=A, \text{ where } X \text{ is symmetric and } \\ Y \text{ is skew-symmetric.}

\displaystyle \text{Answer:}

\displaystyle \text{Given: } A= \begin{bmatrix} 3 & 2 & 7\\ 1 & 4 & 3\\ -2 & 5 & 8 \end{bmatrix} \Rightarrow A^{T}= \begin{bmatrix} 3 & 1 & -2\\ 2 & 4 & 5\\ 7 & 3 & 8 \end{bmatrix} \\ [10pt] \text{Let } X=\frac{1}{2}(A+A^{T}) =\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) = \begin{bmatrix} 3 & \tfrac{3}{2} & \tfrac{5}{2}\\ \tfrac{3}{2} & 4 & 4\\ \tfrac{5}{2} & 4 & 8 \end{bmatrix} \\ [10pt] \text{Let } Y=\frac{1}{2}(A-A^{T}) =\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) = \begin{bmatrix} 0 & \tfrac{1}{2} & \tfrac{9}{2}\\ -\tfrac{1}{2} & 0 & -1\\ -\tfrac{9}{2} & 1 & 0 \end{bmatrix} \\ [10pt] X^{T}=X,\quad Y^{T}=-Y \\ [10pt] \text{Hence, } X \text{ is symmetric and } Y \text{ is skew-symmetric.} \\ [10pt] \text{Now, } X+Y= \begin{bmatrix} 3 & 2 & 7\\ 1 & 4 & 3\\ -2 & 5 & 8 \end{bmatrix} =A \\ [10pt] \therefore\; A=X+Y, \quad X= \begin{bmatrix} 3 & \tfrac{3}{2} & \tfrac{5}{2}\\ \tfrac{3}{2} & 4 & 4\\ \tfrac{5}{2} & 4 & 8 \end{bmatrix}, \quad Y= \begin{bmatrix} 0 & \tfrac{1}{2} & \tfrac{9}{2}\\ -\tfrac{1}{2} & 0 & -1\\ -\tfrac{9}{2} & 1 & 0 \end{bmatrix}.

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\displaystyle \textbf{Question 5:} 
\displaystyle \text{Express the matrix } A=\begin{bmatrix} 4 & 2 & -1\\ 3 & 5 & 7\\ 1 & -2 & 1 \end{bmatrix} \text{ as the sum of a symmetric and a skew-symmetric matrix.}

\displaystyle \text{Answer:}

\displaystyle \text{Given: } A=\begin{bmatrix} 4 & 2 & -1\\ 3 & 5 & 7\\ 1 & -2 & 1 \end{bmatrix}

\displaystyle 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}) =\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) =\begin{bmatrix} 4 & \tfrac{5}{2} & 0\\ \tfrac{5}{2} & 5 & \tfrac{5}{2}\\ 0 & \tfrac{5}{2} & 1 \end{bmatrix}

\displaystyle X^{T}= \begin{bmatrix} 4 & \tfrac{5}{2} & 0\\ \tfrac{5}{2} & 5 & \tfrac{5}{2}\\ 0 & \tfrac{5}{2} & 1 \end{bmatrix} =X

\displaystyle \text{Let } Y=\frac{1}{2}(A-A^{T}) =\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) =\begin{bmatrix} 0 & -\tfrac{1}{2} & -1\\ \tfrac{1}{2} & 0 & \tfrac{9}{2}\\ 1 & -\tfrac{9}{2} & 0 \end{bmatrix}

\displaystyle Y^{T}= \begin{bmatrix} 0 & \tfrac{1}{2} & 1\\ -\tfrac{1}{2} & 0 & -\tfrac{9}{2}\\ -1 & \tfrac{9}{2} & 0 \end{bmatrix} =-Y

\displaystyle \text{Thus, } X \text{ is a symmetric matrix and } Y \text{ is a skew-symmetric matrix.}

\displaystyle \text{Now, } X+Y= \begin{bmatrix} 4 & \tfrac{5}{2} & 0\\ \tfrac{5}{2} & 5 & \tfrac{5}{2}\\ 0 & \tfrac{5}{2} & 1 \end{bmatrix} + \begin{bmatrix} 0 & -\tfrac{1}{2} & -1\\ \tfrac{1}{2} & 0 & \tfrac{9}{2}\\ 1 & -\tfrac{9}{2} & 0 \end{bmatrix} = \begin{bmatrix} 4 & 2 & -1\\ 3 & 5 & 7\\ 1 & -2 & 1 \end{bmatrix} =A

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\displaystyle \textbf{Question 6:} 
\displaystyle \text{Define a symmetric matrix. Prove that for } A=\begin{bmatrix}2 & 4\\5 & 6\end{bmatrix}, \ A+A^T \text{ is a symmetric matrix where } \\ A^T \text{ is the transpose of } A.

\displaystyle \text{Answer:}

\displaystyle \text{A square matrix } A \text{ is called a symmetric matrix, if } A^T = A. \\ [10pt] \text{Given: } A = \begin{bmatrix} 2 & 4\\ 5 & 6 \end{bmatrix} \qquad A^T = \begin{bmatrix} 2 & 5\\ 4 & 6 \end{bmatrix} \\ [10pt] \text{Now, } A + A^T = \begin{bmatrix} 2 & 4\\ 5 & 6 \end{bmatrix} + \begin{bmatrix} 2 & 5\\ 4 & 6 \end{bmatrix} = \begin{bmatrix} 4 & 9\\ 9 & 12 \end{bmatrix} \quad \cdots (1) \\ [10pt] (A + A^T)^T = \begin{bmatrix} 4 & 9\\ 9 & 12 \end{bmatrix}^T = \begin{bmatrix} 4 & 9\\ 9 & 12 \end{bmatrix} \\ [10pt] \therefore\ (A + A^T)^T = (A + A^T) \\ [10pt] \text{Thus, } (A + A^T) \text{ is a symmetric matrix.}

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\displaystyle \textbf{Question 7:} 
\displaystyle \text{Express the matrix } A=\begin{bmatrix} 3 & -4\\ 1 & -1 \end{bmatrix} \text{ as the sum of a symmetric and a skew-symmetric matrix.}

\displaystyle \text{Answer:}

\displaystyle \text{Given: } A=\begin{bmatrix}3 & -4\\ 1 & -1\end{bmatrix} \Rightarrow A^{T}=\begin{bmatrix}3 & 1\\ -4 & -1\end{bmatrix} \\ [10pt] \text{Let } X=\frac12(A+A^{T}) =\frac12\left( \begin{bmatrix}3 & -4\\ 1 & -1\end{bmatrix} + \begin{bmatrix}3 & 1\\ -4 & -1\end{bmatrix} \right) = \begin{bmatrix} 3 & -\frac32\\ -\frac32 & -1 \end{bmatrix} \\ [10pt] X^{T} = \begin{bmatrix} 3 & -\frac32\\ -\frac32 & -1 \end{bmatrix} =X \\ [10pt] \text{Let } Y=\frac12(A-A^{T}) =\frac12\left( \begin{bmatrix}3 & -4\\ 1 & -1\end{bmatrix} - \begin{bmatrix}3 & 1\\ -4 & -1\end{bmatrix} \right) = \begin{bmatrix} 0 & -\frac52\\ \frac52 & 0 \end{bmatrix} \\ [10pt] Y^{T} = \begin{bmatrix} 0 & \frac52\\ -\frac52 & 0 \end{bmatrix} = -\,Y \\ [10pt] \text{Therefore, } X \text{ is a symmetric matrix and } Y \text{ is a skew-symmetric matrix.} X+Y = \begin{bmatrix} 3 & -\frac32\\ -\frac32 & -1 \end{bmatrix} + \begin{bmatrix} 0 & -\frac52\\ \frac52 & 0 \end{bmatrix} = \begin{bmatrix} 3 & -4\\ 1 & -1 \end{bmatrix} =A

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\displaystyle \textbf{Question 8:} 
\displaystyle \text{Express the following matrix as the sum of a symmetric and a skew-symmetric} \\ \text{matrix and verify your result:} \begin{bmatrix} 3 & -2 & -4\\ 3 & -2 & -5\\ -1 & 1 & 2 \end{bmatrix}

\displaystyle \text{Answer:}

\displaystyle \textit{Here,}

\displaystyle A=\begin{bmatrix} 3&-2&-4\\ 3&-2&-5\\ -1&1&2 \end{bmatrix}

\displaystyle \Rightarrow A^{T}=\begin{bmatrix} 3&3&-1\\ -2&-2&1\\ -4&-5&2 \end{bmatrix}

\displaystyle \text{Let }X=\frac{1}{2}\left(A+A^{T}\right) =\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) = \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 \text{Let }Y=\frac{1}{2}\left(A-A^{T}\right) =\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) = \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 \Rightarrow X \text{ is a symmetric matrix and } Y \text{ is a skew-symmetric matrix.}

\displaystyle \textit{Now,}

\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} = \begin{bmatrix} 3&-2&-4\\ 3&-2&-5\\ -1&1&2 \end{bmatrix} =A

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