Homogeneous and Nonhomogeneous Systems
A homogeneous system of linear equations is one in which all of the constant terms are zero. A homogeneous system always has at least one solution, namely the zero vector. When a row operation is applied to a homogeneous system, the new system is still homogeneous. It is important to note that when we represent a homogeneous system as a matrix, we often leave off the final column of constant terms, since applying row operations would not modify that column. So, we use a regular matrix instead of an augmented matrix. Of course, when looking for a solution, it's important to take the constant zero terms into account.
A nonhomogeneous system has an associated homogeneous system, which you get by replacing the constant term in each equation with zero. Section 1.I.3 in the textbook is about understanding the structure of solution sets of homogeneous and non-homogeneous systems. The main theorems that are proved in this section are:
Theorem: The solution set of a homogeneous linear system with $n$ variables is of the form $\{a_1\vec v_1 + a_2\vec v_2 + \cdots + a_k\vec v_k \,|\, a_1,a_2,\dots,a_k\in\R \}$, where $k$ is the number of free variables in an echelon form of the system and $\vec v_1,\vec v_2,\dots,\vec v_k$ are [constant] vectors in $\R^n.$
Theorem: Consider a system of linear equations in $n$ variables, and suppose that $\vec p$ is a solution of the system. Then the solution set of the system is of the form $\{\vec p + a_1\vec v_1 + a_2\vec v_2 + \cdots + a_k\vec v_k \,|\, a_1,a_2,\dots,a_k\in\R \}$, where $\{a_1\vec v_1 + a_2\vec v_2 + \cdots + a_k\vec v_k \,|\, a_1,a_2,\dots,a_k\in\R \}$ is the solution set of the associated homogeneous system. (So, $k$ is still the number of free variables in an echelon form of the system.)
The vector $\vec p$ in the second theorem is said to be a particular solution of the system. (Remember that for a nonhomogeneous system, it is possible that no particular solution exists, and the solution set is empty.) A homogeneous system always has $\vec 0$ as a particular solution, and the second theorem applies to homogeneous systems by taking $\vec p=\vec 0$. Note that for a given system, the vectors $\vec p$ and $\vec v_i$ are not unique. There can be many different sequences of row operatation that could be used to put the system into echelon form. The $\vec p$ and $\vec v_i$ that you get can depend on the specific sequence of row operations that you use. However, you can only get different ways of writing the same solution set. (A possibly surprising fact, which has not yet been proved, is that no matter what sequence of row operations you use to put the system into echelon form, you always get the same number of free variables. This means that the number $k$ in the system is uniquely determined by the system.)
These theorems do not really change the way that you go about solving a linear system, but it does help us understand the structure of the solution set of the system, and in particular the geometry of the solution set. The solution set of a homogeneous system with $n$ variables is a linear space in $\R^n$ that contains the origin. Adding the vector $\vec p$ to all the points in that linear space gives a "parallel" linear space that contains $\vec p$. See the second picture in the previous sectio].
The other significant topic in Section 1.I.3 of the textbook is singular versus nonsingular matrices. This is an issue only for square matrices. (A square matrix is one in which the number of rows is equal to the number of columns.) A square matrix is the associated matrix of some homogeneous system. Since the matrix is square, the homogeneous system has the same number of equations as there are variables. The homogeneous system will either have $\vec 0$ as its only solution, or it will have an infinite number of solutions. The matrix is said to be nonsingular if the system has a unique solution. It is said to be singular if the system has an infinite number of solutions. (The terms "singular" and "nonsingular" only apply to square matrices.) Note that, by the above theorems, a square matrix is singular if and only it has at least one free variable when it is put into echelon form, which in turn is true if and only if an echelon form of the matrix has at least one row containing only zeros.
I would like to add some geometric perspective to all this, using ideas from the previous section. You should try to develop your higher-dimensional geometric intuition! Suppose we have a homogeneous system of $n$ equations in $n$ variables. The solution set for each individual equation in the system is a linear space in $\R^n$ of dimension of $n-1$ (unless the equation is the trivial equation, $0=0$.) And since $\vec 0$ is a solution, that linear space passes through the origin. The solution set for the entire system is the intersection of the solution sets for all of the individual equations; that is, it is the intersection of $n$ linear spaces of dimension $n-1$.
For example, if $n=2$, we are looking at the intersection of two lines through the origin; the possibilities are that the lines intersect only at the origin [nonsingular matrix] or that the lines are actually identical [singular matrix]. Of course, if you pick two lines at random, it will be very unlikely that they are identical. That means that if you pick a $2\times2$ matrix at random, it is very unlikely that it will be singular.
For $n=3$, we are intersecting three planes that contain the origin. The intersection of two planes through the origin is a line, unless the planes happen to be identical. When you add the third plane to the intersection, you are most likely intersecting that plane with a line and the result will be a single point (namely the origin), except in the unlikely case that the line happens to lie entirely in the plane.
For dimension $n$, the situation is similar. The solution set to the first equation is a linear space in $\R^n$ of dimension $n-1$. Taking the intersection of that space with the solution set of the second equation is likely to give a linear space of dimension $n-2$. As the solution set for each equation is added to the intersection, the dimension of the intersection is likely to go down by one. When you get to the intersection of the solution sets for all $n$ equations, you are likely to have a space of dimension zero—a single point, namely the origin. Again, if you pick an $n\times n$ matrix at random, it is very unlikely to be singular. (The word "singular" means "notably unusual.")
When you look a nonhomogeneous linear system of $n$ equations in $n$ variables, you are intersecting solution sets that do not necessarily contain the origin. The mostly likely possibility for the intersection is still a single point, and an infinite intersection is still possible. But you also have the possiblility of an empty intersection—no solution—as would happen for example if you intersect two parallel lines.
We can also think about what happens when we apply row reduction to put an $n\times n$ matrix into echelon form. Consider the matrix as the matrix for a homogeneous linear system, written without the constant zero terms from the right sides of the equations. $$\begin{pmatrix} c_{11} & c_{12} & \cdots & c_{1n}\\ c_{21} & c_{22} & \cdots & c_{2n}\\ \vdots&\vdots&\ddots&\vdots\\ c_{n1} & c_{n2} & \cdots & c_{nn} \end{pmatrix}$$ Remember that the matrix is square, with the same number of rows as of columns. The echelon form matrix that results from row reduction will have the form $$ \begin{pmatrix} d_{11} & d_{12} & \cdots & d_{1n}\\ 0 & d_{22} & \cdots & d_{2n}\\ \vdots &\vdots &\ddots &\vdots\\ 0 & \cdots & 0 & d_{nn} \end{pmatrix}$$ where all the entries below the diagonal are zero (and some of the $d_{ij}$ could also be zero). But there might be all-zero rows at the bottom. That is, $d_{nn}$ might be zero. If $d_{nn}\ne0$, then there are no free variables, and the homogeneous system has $\vec 0$ as its only solution. If $d_{nn}=0$, there is at least one non-zero row, and at most $n-1$ of the $n$ rows are non-zero. So, there are at most $n-1$ leading variables, which implies that there is at least one free variable; the system has an infinite number of solutions.
Now, suppose that we have a nonhomogeneous system with the same matrix of coefficients. The augmented matrix for the nonhomogeneous system has the form $$\left(\begin{array}{cccc|c} c_{11} & c_{12} & \cdots & c_{1n} & a_1\\ c_{21} & c_{22} & \cdots & c_{2n} & a_2\\ \vdots&\vdots&\ddots&\vdots&\vdots\\ c_{n1} & c_{n2} & \cdots & c_{nn} & a_n \end{array}\right)$$ If we apply Gauss's method using the same row operations that we used for the homogeneous system, we get a matrix of the form $$\left(\begin{array}{cccc|c} d_{11} & d_{12} & \cdots & d_{1n} & b_1\\ 0 & d_{22} & \cdots & d_{2n} & b_2\\ \vdots &\vdots &\ddots &\vdots & \vdots\\ 0 & \cdots & 0 & d_{nn} & b_n \end{array}\right)$$ Now, in the case $d_{nn}\ne0$, there are no free variables and we can solve the system to get a unique solution. (Thus, a linear system whose matrix of coefficients is a square, nonsingular matrix will always have a unique solution.) In the case $d_{nn}=0$, we have to consider the constant terms. When $d_{nn}=0$ and $b_n\ne 0$, we have an equation of the form $0=k$ where $k\ne 0$, and there is no solution. When $d_{nn}=0$ and $b_n=0$, we have an all-zero row. However, there might still be no solution because one of the previous equations might still be of the form $0=k$, where $k\ne0$. However, we can say that if there are no such rows, then we have a solvable system with at least one free variable, and the number of solutions is infinite.