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(B) Given $A = \begin{vmatrix} -1 & 2 & 4 \ 3 & 1 & 0 \ -2 & 4 & 2 \end{vmatrix}$.We observe the matrix $B = \begin{vmatrix} -2 & 4 & 2 \ 6 & 2 & 0

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$B = -4A$

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(B) Given $A = \begin{vmatrix} -1 & 2 & 4 \ 3 & 1 & 0 \ -2 & 4 & 2 \end{vmatrix}$.We observe the matrix $B = \begin{vmatrix} -2 & 4 & 2 \ 6 & 2 & 0 \ -2 & 4 & 8 \end{vmatrix}$.Step $1$: Multiply $R_2$ of $A$ by $2$: $\begin{vmatrix} -1 & 2 & 4 \ 6 & 2 & 0 \ -2 & 4 & 2 \end{vmatrix} = 2A$.Step $2$: Multiply $R_3$ of the resulting matrix by $4$: $\begin{vmatrix} -1 & 2 & 4 \ 6 & 2 & 0 \ -8 & 16 & 8 \end{vmatrix} = 2 	imes 4 A = 8A$.Wait,let us re-evaluate the transformation: $B$ is obtained from $A$ by swapping $R_1$ and $R_3$ (which introduces a factor of $-1$),multiplying $R_2$ by $2$,and multiplying $R_3$ by $4$.Actually,comparing $A$ and $B$: $R_1$ of $B$ is $2 	imes R_3$ of $A$,$R_2$ of $B$ is $2 	imes R_2$ of $A$,and $R_3$ of $B$ is $4 	imes R_1$ of $A$.Thus,$B = (2 	imes 2 	imes 4) 	imes (-1) A = -16A$ is incorrect. Let us check the determinant values directly.$det(A) = -1(2-0) - 2(6-0) + 4(12 - (-2)) = -2 - 12 + 56 = 42$.$det(B) = -2(16-0) - 4(48-0) + 2(24 - (-4)) = -32 - 192 + 56 = -168$.Since $-168 = -4 	imes 42$,we have $B = -4A$.

If $A = \begin{vmatrix} -1 & 2 & 4 \\ 3 & 1 & 0 \\ -2 & 4 & 2 \end{vmatrix}$ and $B = \begin{vmatrix} -2 & 4 & 2 \\ 6 & 2 & 0 \\ -2 & 4 & 8 \end{vmatrix}$,then $B$ is given by

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