If matrix D_1 = operatorname{diag}(a, b, c),m | vedclass

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(A) Given that $A$ is a skew-symmetric matrix of order $3 	imes 3$,its diagonal elements are all zero,i.e.,$\operatorname{diag}(A) = (0, 0, 0)$.Let $

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$2a + 2b + 2c - 9$

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(A) Given that $A$ is a skew-symmetric matrix of order $3 	imes 3$,its diagonal elements are all zero,i.e.,$\operatorname{diag}(A) = (0, 0, 0)$.Let $D_1 = \operatorname{diag}(a, b, c)$ and $D_2 = \operatorname{diag}(3, 3, 3) = 3I$,where $I$ is the identity matrix.The trace of a matrix is the sum of its diagonal elements.For any skew-symmetric matrix $A$,the product of a diagonal matrix $D$ and $A$ (i.e.,$DA$) results in a matrix where the diagonal elements are $d_{ii} 	imes a_{ii}$. Since $a_{ii} = 0$,the diagonal elements of $DA$ are $0$.Thus,$\operatorname{diag}(D_1 D_2 A) = (0, 0, 0)$,$\operatorname{diag}(D_1 A) = (0, 0, 0)$,and $\operatorname{diag}(D_2 A) = (0, 0, 0)$.Now,consider the expression inside the trace: $M = D_1 D_2 A + D_1 D_2 + D_1 A + D_2 A$.The diagonal elements of $M$ are $\operatorname{diag}(M) = \operatorname{diag}(D_1 D_2) + \operatorname{diag}(D_1 A) + \operatorname{diag}(D_2 A) + \operatorname{diag}(D_1 D_2 A)$.Since $\operatorname{diag}(D_1 A) = \operatorname{diag}(D_2 A) = \operatorname{diag}(D_1 D_2 A) = (0, 0, 0)$,we have $\operatorname{diag}(M) = \operatorname{diag}(D_1 D_2) = (3a, 3b, 3c)$.Therefore,$\operatorname{Tr}(M) = 3a + 3b + 3c$.We need to calculate $\operatorname{Tr}(M) - \operatorname{Tr}(D_1 + D_2)$.$\operatorname{Tr}(D_1 + D_2) = (a+3) + (b+3) + (c+3) = a + b + c + 9$.Finally,$\operatorname{Tr}(M) - \operatorname{Tr}(D_1 + D_2) = (3a + 3b + 3c) - (a + b + c + 9) = 2a + 2b + 2c - 9$.

If matrix $D_1 = \operatorname{diag}(a, b, c)$,matrix $D_2 = \operatorname{diag}(3, 3, 3)$ and $A$ is a skew-symmetric matrix of $3^{rd}$ order,then $\operatorname{Tr}(D_1 D_2 A + D_1 D_2 + D_1 A + D_2 A) - \operatorname{Tr}(D_1 + D_2) =$

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