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(B) Given the matrix $A = \begin{bmatrix} 2 & 4 & 5 \ 4 & 8 & 10 \ -6 & -12 & -15 \end{bmatrix}$.First,we observe the rows of the matrix.Row $2$ $(R

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(B) Given the matrix $A = \begin{bmatrix} 2 & 4 & 5 \ 4 & 8 & 10 \ -6 & -12 & -15 \end{bmatrix}$.First,we observe the rows of the matrix.Row $2$ $(R_2)$ is $2 	imes R_1$ $(4, 8, 10 = 2 	imes (2, 4, 5))$.Row $3$ $(R_3)$ is $-3 	imes R_1$ $(-6, -12, -15 = -3 	imes (2, 4, 5))$.Since $R_2$ and $R_3$ are scalar multiples of $R_1$,the rank of the matrix cannot be $3$ because the determinant $|A| = 0$.Next,we check for any non-zero $2 	imes 2$ minor.All $2 	imes 2$ minors are formed by selecting two rows and two columns. For example,the minor $\begin{vmatrix} 2 & 4 \ 4 & 8 \end{vmatrix} = 16 - 16 = 0$.Similarly,all other $2 	imes 2$ minors such as $\begin{vmatrix} 4 & 5 \ 8 & 10 \end{vmatrix} = 40 - 40 = 0$ or $\begin{vmatrix} 4 & 8 \ -6 & -12 \end{vmatrix} = -48 - (-48) = 0$ are also zero.Since all $2 	imes 2$ minors are zero and there exists at least one non-zero element (e.g.,$2 
eq 0$),the rank of the matrix is $1$.

If $A = \begin{bmatrix} 2 & 4 & 5 \\ 4 & 8 & 10 \\ -6 & -12 & -15 \end{bmatrix}$,then the rank of $A$ is equal to

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