Exercise 1.2.7. Start with equation (4) from p. 8 of the text:
where a solution can be found for b = (2, 5, 7) and cannot be found for b = (2, 5, 6). Find two more values of b for which a solution can be found, and two more for which no solution is possible.
Answer: As noted in the text, the three columns (1, 2, 3), (1, 0, 1), and (1, 3, 4) lie in a plane, since three times (1, 2, 3)
plus (1, 0, 1) equals twice (1, 3, 4). In other words, one of the three columns can be expressed as a linear combination of the other two. The value b = (2, 5, 7) is also in the same plane, being expressible as (1, 2, 3) plus (1, 3, 4).
So to find additional b for which a solution is possible, we can simply take any b in the same plane as the other vectors. One possibility is twice (2, 5, 7) or (4, 10, 14). Another is (1, 0, 1) plus (1, 3, 4) or (2, 3, 5).
We can then take those solutions and find values of b for which the equations cannot be solved, by looking for values of b not in the same plane as the previous vectors. One possibility is (4, 10, 15), another is (2, 3, 6).
UPDATE: Corrected error in first paragraph of answer; thanks to freaky4you for the fix!
NOTE: This continues a series of posts containing worked out exercises from the (out of print) book Linear Algebra and Its Applications, Third Edition by Gilbert Strang.
If you find these posts useful I encourage you to also check out the more current Linear Algebra and Its Applications, Fourth Edition, Dr Strang’s introductory textbook Introduction to Linear Algebra, Fourth Edition and the accompanying free online course, and Dr Strang’s other books.