Let A=begin{bmatrix} -1 & -2 & -3 3 & 4 & 5 | vedclass

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(C) Given,$A = \begin{bmatrix} -1 & -2 & -3 \ 3 & 4 & 5 \ 4 & 5 & 6 \end{bmatrix}$.The determinant $|A| = -1(24-25) + 2(18-20) - 3(15-16) = 1 - 4 +

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$b < a < c$

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(C) Given,$A = \begin{bmatrix} -1 & -2 & -3 \ 3 & 4 & 5 \ 4 & 5 & 6 \end{bmatrix}$.The determinant $|A| = -1(24-25) + 2(18-20) - 3(15-16) = 1 - 4 + 3 = 0$.Since $|A| = 0$,the rank of $A$ is less than $3$.Consider the minor $\begin{vmatrix} 4 & 5 \ 5 & 6 \end{vmatrix} = 24 - 25 = -1 
eq 0$.Thus,the rank of $A$ is $a = 2$.Given,$B = \begin{bmatrix} 1 & -2 \ -1 & 2 \end{bmatrix}$.The determinant $|B| = (1)(2) - (-2)(-1) = 2 - 2 = 0$.Since $|B| = 0$ and there exists at least one non-zero element (e.g.,$1 
eq 0$),the rank of $B$ is $b = 1$.Given,$C = \begin{bmatrix} 2 & 0 & 0 \ 0 & 2 & 0 \ 0 & 0 & 2 \end{bmatrix}$.The determinant $|C| = 2(4 - 0) = 8 
eq 0$.Since $C$ is a $3 	imes 3$ matrix and $|C| 
eq 0$,the rank of $C$ is $c = 3$.Comparing the ranks: $b = 1, a = 2, c = 3$.Therefore,the correct order is $b < a < c$.

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