If left[begin{array}{ccc}0 & 2 & a b & 0 & 4 | vedclass

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(C) Let $A = \left[\begin{array}{ccc}0 & 2 & a \ b & 0 & 4 \ -3 & c & 0\end{array}ight]$ be a skew-symmetric matrix.By definition,$A = -A^T$.$\lef

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$\left[\begin{array}{cc}2 & -8 \ -8 & 2\end{array}ight]$

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(C) Let $A = \left[\begin{array}{ccc}0 & 2 & a \ b & 0 & 4 \ -3 & c & 0\end{array}ight]$ be a skew-symmetric matrix.By definition,$A = -A^T$.$\left[\begin{array}{ccc}0 & 2 & a \ b & 0 & 4 \ -3 & c & 0\end{array}ight] = -\left[\begin{array}{ccc}0 & b & -3 \ 2 & 0 & c \ a & 4 & 0\end{array}ight] = \left[\begin{array}{ccc}0 & -b & 3 \ -2 & 0 & -c \ -a & -4 & 0\end{array}ight]$.Comparing the corresponding elements,we get:$-b = 2 \Rightarrow b = -2$$a = 3$$c = -4$Now,substitute these values into the expression:$\left[\begin{array}{cc}a & b \ b & a\end{array}ight]\left[\begin{array}{cc}b & c \ c & b\end{array}ight] = \left[\begin{array}{cc}3 & -2 \ -2 & 3\end{array}ight]\left[\begin{array}{cc}-2 & -4 \ -4 & -2\end{array}ight]$$= \left[\begin{array}{cc}(3)(-2) + (-2)(-4) & (3)(-4) + (-2)(-2) \ (-2)(-2) + (3)(-4) & (-2)(-4) + (3)(-2)\end{array}ight]$$= \left[\begin{array}{cc}-6 + 8 & -12 + 4 \ 4 - 12 & 8 - 6\end{array}ight] = \left[\begin{array}{cc}2 & -8 \ -8 & 2\end{array}ight]$.

If $\left[\begin{array}{ccc}0 & 2 & a \\ b & 0 & 4 \\ -3 & c & 0\end{array}\right]$ is a skew-symmetric matrix,then $\left[\begin{array}{cc}a & b \\ b & a\end{array}\right]\left[\begin{array}{cc}b & c \\ c & b\end{array}\right]=$

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