If the ranks of the matrices A = begin{bmatri | vedclass

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(B) First,we find the rank of matrix $A = \begin{bmatrix} 1 & 0 & 1 \ 2 & 1 & 2 \ 1 & 0 & -1 \end{bmatrix}$.Calculating the determinant of $A$:$|A|

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$1$

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(B) First,we find the rank of matrix $A = \begin{bmatrix} 1 & 0 & 1 \ 2 & 1 & 2 \ 1 & 0 & -1 \end{bmatrix}$.Calculating the determinant of $A$:$|A| = 1(1(-1) - 2(0)) - 0(2(-1) - 2(1)) + 1(2(0) - 1(1)) = 1(-1) - 0 + 1(-1) = -1 - 1 = -2$.Since $|A| 
eq 0$,the rank of $A$ $(r_1)$ is $3$.Next,we find the rank of matrix $B = \begin{bmatrix} 1 & 2 & 3 & 4 \ 2 & 4 & 6 & -8 \end{bmatrix}$.This is a $2 	imes 4$ matrix. The maximum possible rank is $2$.We check for a non-zero $2 	imes 2$ minor. Consider the minor formed by the last two columns:$\begin{vmatrix} 3 & 4 \ 6 & -8 \end{vmatrix} = (3)(-8) - (4)(6) = -24 - 24 = -48 
eq 0$.Since there exists a non-zero $2 	imes 2$ minor,the rank of $B$ $(r_2)$ is $2$.Finally,$r_1 - r_2 = 3 - 2 = 1$.

If the ranks of the matrices $A = \begin{bmatrix} 1 & 0 & 1 \\ 2 & 1 & 2 \\ 1 & 0 & -1 \end{bmatrix}$ and $B = \begin{bmatrix} 1 & 2 & 3 & 4 \\ 2 & 4 & 6 & -8 \end{bmatrix}$ are $r_1$ and $r_2$ respectively,then $r_1 - r_2 =$

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