Let A=left[begin{array}{rrr}-1 & -2 & -3 3 & | vedclass

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(C) Given,$A=\left[\begin{array}{rrr}-1 & -2 & -3 \ 3 & 4 & 5 \ 4 & 5 & 6\end{array}ight]$.Calculating the determinant of $A$:$|A| = -1(24-25) + 2

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$b < a < c$

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(C) Given,$A=\left[\begin{array}{rrr}-1 & -2 & -3 \ 3 & 4 & 5 \ 4 & 5 & 6\end{array}ight]$.Calculating the determinant of $A$:$|A| = -1(24-25) + 2(18-20) - 3(15-16) = 1 - 4 + 3 = 0$.Since $|A| = 0$,the rank of $A$ is less than $3$. We check a $2 	imes 2$ minor: $\left|\begin{array}{rr}4 & 5 \ 5 & 6\end{array}ight| = 24 - 25 = -1 
eq 0$.Therefore,the rank of $A$ is $a = 2$.Given,$B=\left[\begin{array}{rr}1 & -2 \ -1 & 2\end{array}ight]$.Calculating the determinant of $B$:$|B| = (1)(2) - (-2)(-1) = 2 - 2 = 0$.Since $|B| = 0$ and there exists at least one non-zero element,the rank of $B$ is $b = 1$.Given,$C=\left[\begin{array}{rrr}2 & 0 & 0 \ 0 & 2 & 0 \ 0 & 0 & 2\end{array}ight]$.Calculating the determinant of $C$:$|C| = 2(4-0) = 8 
eq 0$.Since $C$ is a $3 	imes 3$ matrix with a non-zero determinant,the rank of $C$ is $c = 3$.Comparing the values: $b = 1, a = 2, c = 3$.Thus,$b < a < c$.

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