Solve. 2a + 3b = 5 6= a -5 a = 4 b = -1 1 a = 6 b=1 a=6 6 = -1 a=4 6 = 1
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would be a basis for
if
(1) the vectors in are independent, and
(2) the vectors span .
- Linear independence requires that
is the only solution to
These vectors are not linearly independent because if ,
, and
, we have
so is not a basis for
.