Given,

$$A = \left[ {\matrix{ 1 & { - 2} & \alpha \cr \alpha & 2 & { - 1} \cr } } \right]$$

and $$B = \left[ {\matrix{ 2 & \alpha \cr { - 1} & 2 \cr 4 & { - 5} \cr } } \right]$$

$$AB = \left[ {\matrix{ 1 & { - 2} & \alpha \cr \alpha & 2 & { - 1} \cr } } \right]\left[ {\matrix{ 2 & \alpha \cr { - 1} & 2 \cr 4 & { - 5} \cr } } \right]$$

$$ = \left[ {\matrix{ {4 + 4\alpha } & { - 4\alpha - 4} \cr {2\alpha - 6} & {{\alpha ^2} + 9} \cr } } \right]$$

Given,

$$|AB| = 0$$

$$\therefore$$ $$\left| {\matrix{ {4 + 4\alpha } & { - 4\alpha - 4} \cr {2\alpha - 6} & {{\alpha ^2} + 9} \cr } } \right| = 0$$

$$ \Rightarrow (4\alpha + 4)\left| {\matrix{ 1 & { - 1} \cr {2\alpha - 6} & {{\alpha ^2} + 9} \cr } } \right| = 0$$

$$ \Rightarrow (4\alpha + 4)({\alpha ^2} + 9 + 2\alpha - 6) = 0$$

$$ \Rightarrow (4\alpha + 4)({\alpha ^2} + 2\alpha + 3) = 0$$

$$\therefore$$ $$\alpha - = - 1$$

or $${\alpha ^2} + 2\alpha + 3 = 0$$

$${\alpha _1} + {\alpha _2} = - 2$$

$$\therefore$$ Sum of all values of $$\alpha = - 1 - 2 = - 3$$

$$\therefore$$ Absolute value of $$\alpha = | - 3| = 3$$