If $A = \begin{bmatrix} -2 \\ -1 \\ -4 \end{bmatrix}$, $B = [-1 \quad 2 \quad 3]$, then the value of $A'B'$ is

Given: $A = \begin{bmatrix} -2 \\ -1 \\ -4 \end{bmatrix}$ and $B = [-1 \quad 2 \quad 3]$

The prime symbol (') denotes the transpose operation. For a column vector, the transpose produces a row vector, and for a row vector, the transpose produces a column vector.

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$A = \begin{bmatrix} -2 \\ -1 \\ -4 \end{bmatrix}$ is a column vector with 3 rows and 1 column.

$A' = \begin{bmatrix} -2 & -1 & -4 \end{bmatrix}$ is a row vector with 1 row and 3 columns.

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$B = \begin{bmatrix} -1 & 2 & 3 \end{bmatrix}$ is a row vector with 1 row and 3 columns.

$B' = \begin{bmatrix} -1 \\ 2 \\ 3 \end{bmatrix}$ is a column vector with 3 rows and 1 column.

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$A'B' = \begin{bmatrix} -2 & -1 & -4 \end{bmatrix} \times \begin{bmatrix} -1 \\ 2 \\ 3 \end{bmatrix}$

$A'B' = (-2)(-1) + (-1)(2) + (-4)(3)$

$A'B' = 2 - 2 - 12$

$A'B' = -12$

Therefore, the value of $A'B'$ is $-12$.