\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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