Which of the following matrices has rank 3? | vedclass

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(C) To find the rank of a $3 	imes 3$ matrix,we calculate its determinant. If the determinant is non-zero,the rank is $3$.Option $(A)$: Let $A = \lef

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$\left[\begin{array}{ccc}0 & 1 & 2 \ -1 & 0 & 5 \ -2 & 7 & 0\end{array}ight]$

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(C) To find the rank of a $3 	imes 3$ matrix,we calculate its determinant. If the determinant is non-zero,the rank is $3$.Option $(A)$: Let $A = \left[\begin{array}{ccc}10 & 11 & 12 \ 11 & 12 & 13 \ 12 & 13 & 14\end{array}ight]$. The rows are in arithmetic progression. $R_2 - R_1 = [1, 1, 1]$ and $R_3 - R_2 = [1, 1, 1]$. Since $R_3 - R_2 = R_2 - R_1$,the rows are linearly dependent. Thus,$\det(A) = 0$ and $	ext{rank}(A) Option $(B)$: Let $B = \left[\begin{array}{ccc}0 & -51 & 101 \ 51 & 0 & -581 \ -101 & 581 & 0\end{array}ight]$. This is a skew-symmetric matrix of odd order $(3 	imes 3)$. The determinant of a skew-symmetric matrix of odd order is always $0$. Thus,$	ext{rank}(B) Option $(C)$: Let $C = \left[\begin{array}{ccc}0 & 1 & 2 \ -1 & 0 & 5 \ -2 & 7 & 0\end{array}ight]$. Calculate the determinant: $\det(C) = 0(0 - 35) - 1(0 - (-10)) + 2(-7 - 0) = 0 - 10 - 14 = -24$. Since $\det(C) 
eq 0$,the rank of matrix $C$ is $3$.Option $(D)$: Let $D = \left[\begin{array}{lll}1 & 2 & 3 \ 2 & 4 & 6 \ 3 & 6 & 9\end{array}ight]$. Here,$R_2 = 2R_1$ and $R_3 = 3R_1$. The rows are linearly dependent. Thus,$\det(D) = 0$ and $	ext{rank}(D) = 1$.Therefore,the correct option is $(C)$.

Which of the following matrices has rank $3$?

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