The symmetric part of the matrix A = begin{bm | vedclass

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(B) The symmetric part of a square matrix $A$ is given by the formula $\frac{1}{2}(A + A^T)$.Given $A = \begin{bmatrix} 1 & 2 & 4 \ 6 & 8 & 2 \ 2 &

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$ \begin{bmatrix} 1 & 4 & 3 \ 4 & 8 & 0 \ 3 & 0 & 7 \end{bmatrix} $

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(B) The symmetric part of a square matrix $A$ is given by the formula $\frac{1}{2}(A + A^T)$.Given $A = \begin{bmatrix} 1 & 2 & 4 \ 6 & 8 & 2 \ 2 & -2 & 7 \end{bmatrix}$,its transpose $A^T$ is $\begin{bmatrix} 1 & 6 & 2 \ 2 & 8 & -2 \ 4 & 2 & 7 \end{bmatrix}$.Now,$A + A^T = \begin{bmatrix} 1+1 & 2+6 & 4+2 \ 6+2 & 8+8 & 2-2 \ 2+4 & -2+2 & 7+7 \end{bmatrix} = \begin{bmatrix} 2 & 8 & 6 \ 8 & 16 & 0 \ 6 & 0 & 14 \end{bmatrix}$.Finally,$\frac{1}{2}(A + A^T) = \frac{1}{2} \begin{bmatrix} 2 & 8 & 6 \ 8 & 16 & 0 \ 6 & 0 & 14 \end{bmatrix} = \begin{bmatrix} 1 & 4 & 3 \ 4 & 8 & 0 \ 3 & 0 & 7 \end{bmatrix}$.

The symmetric part of the matrix $A = \begin{bmatrix} 1 & 2 & 4 \\ 6 & 8 & 2 \\ 2 & -2 & 7 \end{bmatrix}$ is

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