a. Yes b. the third column of A is a linear combination of the columns of A
Work Step by Step
a. Yes, because the columns of $A$ are linearly independent. b. All linear combinations of A's columns can be represented as $x_1a_1 + x_2a_2 + x_3a_3$. $a_3 = 0a_1 + 0a_2 + 1a_3$, thus it is a linear combination of the columns of A, and thus it is in $W$.