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(C) The principal diagonal elements of matrix $A$ are $a^2, b^2, c^2$ and the principal diagonal elements of matrix $B$ are $2a, 2b, 2c-3$.According t

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$-4$

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(C) The principal diagonal elements of matrix $A$ are $a^2, b^2, c^2$ and the principal diagonal elements of matrix $B$ are $2a, 2b, 2c-3$.According to the given condition,the sum of the principal diagonal elements of $A$ is equal to the sum of the principal diagonal elements of $B$:$a^2+b^2+c^2 = 2a+2b+2c-3$Rearranging the terms,we get:$(a^2-2a+1) + (b^2-2b+1) + (c^2-2c+1) = 0$$(a-1)^2 + (b-1)^2 + (c-1)^2 = 0$Since the sum of squares of real numbers is zero only if each term is zero,we have:$a-1=0, b-1=0, c-1=0$Therefore,$a=1, b=1, c=1$.The product of the principal diagonal elements of $B$ is $(2a)(2b)(2c-3)$.Substituting the values $a=1, b=1, c=1$:$= (2 	imes 1)(2 	imes 1)(2 	imes 1 - 3)$$= 2 	imes 2 	imes (-1) = -4$.

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