Find the complete solution to Ax = b, b = (-1,0,1)

[s3a]

Homework Statement

Find the complete solution to Ax = b for b = (-1,0,1).

Homework Equations

Reduced-row echelon form procedure.
Matrix multiplication procedure.

The Attempt at a Solution

I have no idea what is being done in the solution attached (other than the reduced-row echelon and matrix multiplication procedures).

If someone could help me understand what the attached solution is saying, I would very much appreciate it!

 

[HallsofIvy]

The row reduction gives
$$\begin{bmatrix}1 & 2 & 3 & 4 & b_1 \\ 0 & 1 & 2 & 3 & \frac{3}{2}b_1-\frac{1}{2}b_2 \\ 0 & 0 & 0 & 0 & -\frac{1}{4}b_1+ \frac{1}{2}b_2- \frac{1}{4}b_3 \end{bmatrix}$$.

That last line is equivalent to
$$0b_1+ 0b_2+ 0b_3= -\frac{1}{4}b_1+ \frac{1}{2}b_2- \frac{1}{4}b_3$$
The left side is clearly 0 for any $b_1$, $b_2$, $b_3$ so, in order for this equation to be true, the right side must also be 0: $-\frac{1}{4}b_1+ \frac{1}{2}b_2- \frac{1}{4}b_3= 0$. Since we could choose any values for two of $b_1$, $b_2$, $b_3$ and solve that equation for the other, the solution set had dimension 2.

 

[Mark44]

The row reduction gives
$$\begin{bmatrix}1 & 2 & 3 & 4 & b_1 \\ 0 & 1 & 2 & 3 & \frac{3}{2}b_1-\frac{1}{2}b_2 \\ 0 & 0 & 0 & 0 & -\frac{1}{4}b_1+ \frac{1}{2}b_2- \frac{1}{4}b_3 \end{bmatrix}$$.

That last line is equivalent to
$$0b_1+ 0b_2+ 0b_3= -\frac{1}{4}b_1+ \frac{1}{2}b_2- \frac{1}{4}b_3$$

The left side shouldn't be in terms of b₁, etc.

The last line in the augmented matrix above represents $0x_1 + 0x_2 + 0x_3 + 0x_4 = -\frac{1}{4}b_1+ \frac{1}{2}b_2- \frac{1}{4}b_3$
As stated below, the left side is identically zero for any choices of $x_1, x_2, x_3, x_4$, which means that the right side must be zero. Since we have one equation ($-\frac{1}{4}b_1+ \frac{1}{2}b_2- \frac{1}{4}b_3 = 0$) in three variables, two of the variables are arbitrary, which gives us a solution space of dimension 2.

The left side is clearly 0 for any $b_1$, $b_2$, $b_3$ so, in order for this equation to be true, the right side must also be 0: $-\frac{1}{4}b_1+ \frac{1}{2}b_2- \frac{1}{4}b_3= 0$. Since we could choose any values for two of $b_1$, $b_2$, $b_3$ and solve that equation for the other, the solution set had dimension 2.