If the matrix left[begin{array}{rr}2 & 3 5 & | vedclass

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(D) Given,$C = \left[\begin{array}{rr}2 & 3 \ 5 & -1\end{array}ight] = A+B$. We know that any square matrix $C$ can be uniquely expressed as the su

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$\left[\begin{array}{rr}0 & -1 \ 1 & 0\end{array}ight]$

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(D) Given,$C = \left[\begin{array}{rr}2 & 3 \ 5 & -1\end{array}ight] = A+B$. We know that any square matrix $C$ can be uniquely expressed as the sum of a symmetric matrix $A$ and a skew-symmetric matrix $B$,where $A = \frac{1}{2}(C+C^T)$ and $B = \frac{1}{2}(C-C^T)$. First,find the transpose of $C$: $C^T = \left[\begin{array}{rr}2 & 5 \ 3 & -1\end{array}ight]$. Now,calculate $B = \frac{1}{2}(C-C^T)$: $B = \frac{1}{2} \left( \left[\begin{array}{rr}2 & 3 \ 5 & -1\end{array}ight] - \left[\begin{array}{rr}2 & 5 \ 3 & -1\end{array}ight] ight)$ $B = \frac{1}{2} \left[\begin{array}{rr}2-2 & 3-5 \ 5-3 & -1-(-1)\end{array}ight]$ $B = \frac{1}{2} \left[\begin{array}{rr}0 & -2 \ 2 & 0\end{array}ight]$ $B = \left[\begin{array}{rr}0 & -1 \ 1 & 0\end{array}ight]$.

If the matrix $\left[\begin{array}{rr}2 & 3 \\ 5 & -1\end{array}\right]=A+B$,where $A$ is symmetric and $B$ is skew-symmetric,then $B$ is equal to

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