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(B) Given the matrix $A = \begin{bmatrix} 1 & 2 & 3 & 0 \ 2 & 4 & 3 & 2 \ 3 & 2 & 1 & 3 \ 6 & 8 & 7 & \alpha \end{bmatrix}$.Applying row operations

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$5$

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(B) Given the matrix $A = \begin{bmatrix} 1 & 2 & 3 & 0 \ 2 & 4 & 3 & 2 \ 3 & 2 & 1 & 3 \ 6 & 8 & 7 & \alpha \end{bmatrix}$.Applying row operations $R_2 ightarrow R_2 - 2R_1$,$R_3 ightarrow R_3 - 3R_1$,and $R_4 ightarrow R_4 - 2R_3$:$A \sim \begin{bmatrix} 1 & 2 & 3 & 0 \ 0 & 0 & -3 & 2 \ 0 & -4 & -8 & 3 \ 0 & 4 & 5 & \alpha - 6 \end{bmatrix}$.Applying $R_4 ightarrow R_4 + R_3$:$A \sim \begin{bmatrix} 1 & 2 & 3 & 0 \ 0 & 0 & -3 & 2 \ 0 & -4 & -8 & 3 \ 0 & 0 & -3 & \alpha - 3 \end{bmatrix}$.Applying $R_4 ightarrow R_4 - R_2$:$A \sim \begin{bmatrix} 1 & 2 & 3 & 0 \ 0 & 0 & -3 & 2 \ 0 & -4 & -8 & 3 \ 0 & 0 & 0 & \alpha - 5 \end{bmatrix}$.Since the rank of matrix $A$ is $3$,the last row must be a zero row.Therefore,$\alpha - 5 = 0$,which implies $\alpha = 5$.

If the matrix $A = \begin{bmatrix} 1 & 2 & 3 & 0 \\ 2 & 4 & 3 & 2 \\ 3 & 2 & 1 & 3 \\ 6 & 8 & 7 & \alpha \end{bmatrix}$ is of rank $3$,then $\alpha$ equals to

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