Linear Algebra Problem

[Adamantine Chains]

Problem 2, Problem Set 1.2, From Strang's Linear Algebra and Its
Applications:

Solve to find a combination of the columns that equals b:

u - v - w = b1
v + w = b2
w = b3.

[Apologies for the poorly reproduced spacing; that should look like a
Triangular system]

(1) I do not know what form the solution should take. Thus far in the
chapter his examples involve unknown x (in Ax = b), not unknown b. This
looks as if it has an infinite number of solutions.

(2) There is no answer in the appendix. I cannot reverse engineer the
solution and thereby gain an understanding of the concepts.

(3) I found an answer online, but it does not provide enough explanation
for understanding. Here is that answer:

The answer will depend on what b is. So, your answer should
somehow involve b_1, b_2 and b_3.

To solve, row reduce the matrix

[1 -1 -1 | b_1]

[0 1 1 | b_2]

[0 0 1 | b_3]

to get

[1 0 0 | b_1 + b_2]

[0 1 0 | b_2 - b_3 ]

[0 0 1 | b_3 ]

The values in the right column give you the proper coefficients to write
(b_1, b_2, b_3) as a linear combination of the columns. In other words,
(b_1 + b_2)(1, 0, 0) + (b_2 - b_3)(-1, 1, 0) + (b_3)(-1, 1, 1) = (b_1,
b_2, b_3)

I don't understand how this solves anything. The row reduction is clear
(that did help to a degree), but I don't see why (b1 + b2), etc., are
coefficients (x), and I don't see why this should be viewed as a solution.

I'm missing some theory here. Any help would be appreciated.

 

[stapel]

To solve, row reduce the matrix:

. . .\(\displaystyle \left[ \begin{array}{rrr|r}1&-1&-1&b_1\\0&1&1&b_2\\0&0&1&b_3 \end{array} \right]\)

...to get:

. . .\(\displaystyle \left[ \begin{array}{rrr|r}1&0&0&b_1\, +\, b_2\\0&1&0&b_2\, -\, b_3\\0&0&1&b_3 \end{array} \right]\)

The values in the right column give you the proper coefficients to write (b₁, b₂, b₃) as a linear combination of the columns. In other words, (b₁ + b₂)(1, 0, 0) + (b₂ - b₃)(-1, 1, 0) + (b₃)(-1, 1, 1) = (b₁, b₂, b₃)

I don't understand how this solves anything. The row reduction is clear (that did help to a degree), but I don't see why (b₁ + b₂), etc., are coefficients (x), and I don't see why this should be viewed as a solution.

How did they arrive at the row-reduced solution? What steps did they take? What did these steps do to the fourth column?

Do the steps yourself, and it would become clear how they arrived at this final form.

 

[Adamantine Chains]

!

The row reduction left the values of u, v, and w: (1, 0, 0) = (b1 + b2) =
u; (0, 1, 0) = (b2 - b3) = v; (0, 0, 1) = b3 = w.

That is beautiful in its elegance, I must say.

Thanks for the heuristic reply!