In Fundamental of Linear Algebra by James B. Carrell, there is a typical Gaussian reduction (his way of saying elimination) with homogeneous system of equation. The concept I understand, but one example to explain it got me confused.
Example 2.6. Solve the following homogeneous linear system: $$\begin{equation} \left\{ \begin{aligned} 0x + x_2 + 2x_3 + 0x_4 + 3x_5 - x_6 = 0\\ 0x_1 + 0x_2 + 0x_3 +x_4 + 2x_5 + x_6 = 0 \end{aligned} \right. \end{equation}$$
So yes, the augmented coefficient matrix is already in reduced row echelon form: $$ \begin{bmatrix} 0 & 1 & 2 & 0 & 3 & -1 & 0\\ 0 & 0 & 0 & 1 & 2 & 0 & 0 \end{bmatrix} $$ But the author use a different coefficient matrix, which is $$ \begin{bmatrix} 0 & 1 & 2 & 0 & 3 & -1 \\ 0 & 0 & 0 & 1 & 2 & 0 \end{bmatrix} $$ without the last column of the other side of the equality. Next, he expresses the corner variable in terms of the free variables, i.e. the variables without coefficient $1$. Thus: $$\begin{equation} \left\{ \begin{aligned} x_2 = -2x_3 - 3x_5 + x_6\\ x_4 = -2x_5 \end{aligned} \right. \end{equation}$$
Let $\textbf{x}= (x_1,x_2,x_3,x_4,x_5,x_6)^{T}$ be the solution vector of the system with $\textbf{x}\in \mathbb{R}^6$. Expressing the corners variables in terms of the free variables results in: $$\textbf{x}= (x_1,-2x_3 - 3x_5 + x_6,x_3,-2x_5,x_5,x_6)^{T}$$
which only depends on the free variable. Up to this point, everything seems alright. But the next thing is kind of confusing.
The general solution vector now depends only on the free variables, and there is a solution for any choice of these variables. Using a little algebra, we can compute the vector coefficients of each one of the free variables in x. These vectors are called the fundamental solutions. In this example, the general solution vector $\textbf{x}$ has the form $$\textbf{x} = x_1 \mathbf{f_1} + x_3 \mathbf{f_2} + x_4 \mathbf{f_3} + x_5 \mathbf{f_4},$$
Now, at this point, I have some question:
- So, we are having a linear combination? Because I see vectors $\mathbf{f_1}$, $\mathbf{f_2}$, $\mathbf{f_3}$ and $\mathbf{f_4}$, with somewhat scalar (variable) multiplier. I thought Gaussian reduction gives us different form of solution?
- Of the above, we expressed corner variable in terms of the free variables, which is $x_2$ and $x_4$. However, the solution only has $x_1$, $x_3$, somehow $x_4$ and $x_5$. What happened to the others?
- What is the little algebra he is referring to? I do not understand how he did this part of the calculation.
Continuing, he present the exact coefficient vectors (somehow, it changed to vector coefficients), which is
$$ \mathbf{f}_1=\left(\begin{array}{l} 1 \\ 0 \\ 0 \\ 0 \\ 0 \\ 0 \end{array}\right), \mathbf{f}_2=\left(\begin{array}{c} 0 \\ 0 \\ -2 \\ 1 \\ 0 \\ 0 \end{array}\right), \mathbf{f}_3=\left(\begin{array}{c} 0 \\ -3 \\ 0 \\ -2 \\ 1 \\ 0 \end{array}\right) \text {, and } \mathbf{f}_4=\left(\begin{array}{c} 0 \\ -1 \\ 0 \\ 0 \\ 0 \\ 1 \end{array}\right) . $$
I tried to decompose the solution, which is just adding them all up: $$ \left( \begin{array} xx_1 \\ -3x_4 - x_5 \\ -2x_3 \\ x_3 - 2x_4 \\ x_4 \\ x_5 \end{array} \right) $$ which does not resemble that much the above steps given. I literally stuck here, and don't know how to explain, or try to reverse engineer it.
Any help is incredibly welcome, but I also want to have explanation and hints for me to follow through, trying to do it myself. Additionally, I want to know if the text is good enough, because personally, I may have to switch to another textbook.