From ordinary equations to the grammar of linear algebra
Guiding question.
How can many small equations become one geometric object, one computational problem, and one algebraic structure?
Linear algebra begins with a simple human problem: we know several pieces of information, and we want to find the unknowns.
At first this looks like algebra from high school:
\[ \begin{cases} x+y=1,\\ x-y=3. \end{cases} \]
But in this book we will learn to see this system in several deeper ways.
It is a collection of equations.
It is an intersection of geometric objects.
It is a matrix equation.
It is a question about a function.
It is a problem over a field.
It is a computational task.
It is the first example of how linear algebra turns information into structure.
This chapter builds the language we need for the rest of the book. We begin with linear systems, then introduce sets, functions, algebraic objects, fields, and Gauss–Jordan elimination. These are not separate topics. They are the grammar of linear algebra.
This version keeps the story-driven style, but the definitions and theorems are written more precisely. The goal is to help you move between three levels:
Students may use Python, MATLAB, or AI tools to check row reductions and explore examples. The purpose is not to replace mathematical thinking. The purpose is to ask precise questions, verify answers, find mistakes, and explain why a computation is correct.
We first fix a field \(\mathbb F\). In this course the main examples are
\[ \mathbb R \quad \text{and} \quad \mathbb C. \]
Most geometric examples in the beginning use \(\mathbb R\), while later chapters on eigenvalues and Schur decomposition naturally use \(\mathbb C\).
Let \(\mathbb F\) be a field. A linear equation in the unknowns \(x_1,x_2,\ldots,x_n\) over \(\mathbb F\) is an equation of the form
\[ a_1x_1+a_2x_2+\cdots+a_nx_n=b, \]
where \(a_1,a_2,\ldots,a_n,b\in \mathbb F\) are fixed scalars and \(x_1,x_2,\ldots,x_n\) are unknown scalars.
Equivalently, a linear equation is an equation in which each unknown appears only to the first power, and no products such as \(x_ix_j\) occur.
A linear system of \(m\) equations in \(n\) unknowns over \(\mathbb F\) is a collection of equations
\[ \begin{aligned} a_{11}x_1+a_{12}x_2+\cdots+a_{1n}x_n &= b_1,\\ a_{21}x_1+a_{22}x_2+\cdots+a_{2n}x_n &= b_2,\\ &\vdots\\ a_{m1}x_1+a_{m2}x_2+\cdots+a_{mn}x_n &= b_m, \end{aligned} \]
where \(a_{ij},b_i\in \mathbb F\).
A solution of the system is a vector
\[ \mathbf{x}=\begin{bmatrix}x_1\\x_2\\ \vdots\\x_n\end{bmatrix}\in \mathbb F^n \]
such that all \(m\) equations are satisfied.
The solution set is the subset of \(\mathbb F^n\) consisting of all solutions.
This system can be written as one matrix equation:
\[ A\mathbf{x}=\mathbf{b}, \]
where
\[ A= \begin{bmatrix} a_{11} & a_{12} & \cdots & a_{1n}\\ a_{21} & a_{22} & \cdots & a_{2n}\\ \vdots & \vdots & \ddots & \vdots\\ a_{m1} & a_{m2} & \cdots & a_{mn} \end{bmatrix}\in \mathbb F^{m\times n}, \qquad \mathbf{x}= \begin{bmatrix} x_1\\x_2\\ \vdots\\x_n \end{bmatrix}\in \mathbb F^n, \qquad \mathbf{b}= \begin{bmatrix} b_1\\b_2\\ \vdots\\b_m \end{bmatrix}\in \mathbb F^m. \]
The matrix \(A\) is called the coefficient matrix. The matrix
\[ [A\mid \mathbf b] \]
is called the augmented matrix. It remembers both the coefficients and the target vector. It is the object we row-reduce.
Think of \(A\) as a machine. The input is \(\mathbf{x}\). The output is \(A\mathbf{x}\). Solving \(A\mathbf{x}=\mathbf b\) asks:
Which inputs produce the desired output \(\mathbf b\)?
A linear system has only three possible types of solution sets:
There is no fourth possibility.
Let \(A\in \mathbb F^{m\times n}\) and \(\mathbf b\in \mathbb F^m\). The solution set of the linear system
\[ A\mathbf{x}=\mathbf b \]
is exactly one of the following:
More precisely, if the system is consistent and \(\mathbf{x}_p\) is one particular solution, then the solution set is
\[ \mathbf{x}_p+\ker(A)=\{\mathbf{x}_p+\mathbf z: \mathbf z\in \ker(A)\}. \]
Thus it has exactly one solution if \(\ker(A)=\{\mathbf 0\}\) and infinitely many solutions if \(\ker(A)\neq \{\mathbf 0\}\).
If one solution \(\mathbf{x}_p\) exists, then every other solution differs from \(\mathbf{x}_p\) by a solution of the homogeneous system \(A\mathbf z=\mathbf 0\). So the solution set is a translated copy of \(\ker(A)\). A vector space is either the zero space or contains infinitely many vectors over \(\mathbb R\) or \(\mathbb C\).
Consider the following systems over \(\mathbb R\).
No solution:
\[ \begin{cases} x+y=1,\\ x+y=2. \end{cases} \]
The two equations ask for the same left-hand side to equal two different numbers. Geometrically, these are two parallel lines.
One solution:
\[ \begin{cases} x+y=1,\\ x-y=3. \end{cases} \]
Adding the two equations gives \(2x=4\), so \(x=2\). Then \(y=-1\). The solution is
\[ (x,y)=(2,-1). \]
Geometrically, two nonparallel lines meet at one point.
Infinitely many solutions:
\[ \begin{cases} x+y=1,\\ 2x+2y=2. \end{cases} \]
The second equation is just twice the first. The solution set is the whole line
\[ (x,y)=(t,1-t),\qquad t\in\mathbb R. \]
In two variables, each nonzero linear equation is usually a line. In three variables, each nonzero linear equation is usually a plane. In higher dimensions, each nonzero linear equation defines a hyperplane.
Solving a linear system means finding an intersection.
That is the first geometric idea of linear algebra.
The system
\[ \begin{cases} x+y=1,\\ x-y=3 \end{cases} \]
can be written as
\[ \begin{bmatrix} 1 & 1\\ 1 & -1 \end{bmatrix} \begin{bmatrix} x\\y \end{bmatrix} = \begin{bmatrix} 1\\3 \end{bmatrix}. \]
The augmented matrix is
\[ \left[ \begin{array}{cc|c} 1 & 1 & 1\\ 1 & -1 & 3 \end{array} \right]. \]
The matrix form is not just shorthand. It changes the problem. Once a system is written as a matrix equation, we can study it using row operations, functions, vector spaces, rank, geometry, numerical algorithms, and computation.
Let \(A=[a_{ij}]\in \mathbb F^{m\times n}\) and \(\mathbf{x}=(x_1,\ldots,x_n)^T\in \mathbb F^n\). The product \(A\mathbf{x}\in \mathbb F^m\) is defined by
\[ (A\mathbf{x})_i=\sum_{j=1}^n a_{ij}x_j, \qquad i=1,\ldots,m. \]
Equivalently, if \(\mathbf a_1,\ldots,\mathbf a_n\) are the columns of \(A\), then
\[ A\mathbf{x}=x_1\mathbf a_1+x_2\mathbf a_2+\cdots+x_n\mathbf a_n. \]
The equation \(A\mathbf{x}=\mathbf b\) asks whether \(\mathbf b\) can be written as a linear combination of the columns of \(A\).
A matrix is not merely an array of numbers. It is a compressed description of a linear process.
Before we go deeper, we need the language of sets.
A set is a well-defined collection of distinct objects, called its elements.
If \(x\) is an element of a set \(S\), we write \(x\in S\). If \(x\) is not an element of \(S\), we write \(x\notin S\).
A set \(A\) is a subset of a set \(S\), written \(A\subseteq S\), if every element of \(A\) is also an element of \(S\).
Examples:
\[ \{1,2,3\},\qquad \mathbb R,\qquad \mathbb R^2,\qquad \{x\in\mathbb R:x^2<4\}. \]
If \(A\) and \(B\) are subsets of a universe \(S\), then:
\[ A\cup B=\{x\in S:x\in A\text{ or }x\in B\}, \]
\[ A\cap B=\{x\in S:x\in A\text{ and }x\in B\}, \]
\[ A^c=\{x\in S:x\notin A\}. \]
For sets \(X\) and \(Y\), the Cartesian product is
\[ X\times Y=\{(x,y):x\in X,\ y\in Y\}. \]
For example, \(\mathbb R\times\mathbb R=\mathbb R^2\).
Later, vector spaces, matrix spaces, solution sets, subspaces, eigenspaces, function spaces, and Grassmannians will all be sets with additional structure.
Let \(A\) and \(B\) be sets. A function from \(A\) to \(B\) is a rule \(f\) that assigns to each element \(a\in A\) exactly one element \(f(a)\in B\).
We write
\[ f:A\to B. \]
The set \(A\) is the domain. The set \(B\) is the codomain. The image or range of \(f\) is
\[ \operatorname{im}(f)=\{f(a):a\in A\}\subseteq B. \]
In linear algebra, the most important functions are linear maps:
\[ T:V\to W. \]
A matrix is one way of representing a linear map.
Let \(f:A\to B\) be a function.
\(f\) is injective or one-to-one if \[ f(a_1)=f(a_2) \implies a_1=a_2 \] for all \(a_1,a_2\in A\).
\(f\) is surjective or onto if \[ \operatorname{im}(f)=B. \] Equivalently, for every \(b\in B\), there exists \(a\in A\) such that \(f(a)=b\).
\(f\) is bijective if it is both injective and surjective.
For a linear map \(T:V\to W\):
For a square matrix \(A\), invertibility means the map \(\mathbf{x}\mapsto A\mathbf{x}\) is bijective.
Let
\[ f:A\to B,\qquad g:B\to C. \]
The composition of \(f\) followed by \(g\) is the function
\[ g\circ f:A\to C \]
defined by
\[ (g\circ f)(a)=g(f(a)). \]
In matrix language, composition of linear maps becomes matrix multiplication.
If \(T(\mathbf{x})=A\mathbf{x}\) and \(S(\mathbf{y})=B\mathbf{y}\), then
\[ (S\circ T)(\mathbf{x})=B(A\mathbf{x})=(BA)\mathbf{x}. \]
This is why matrix multiplication appears in the order \(BA\): first apply \(A\), then apply \(B\).
A function \(f:A\to B\) is invertible if there exists a function \(f^{-1}:B\to A\) such that
\[ f^{-1}(f(a))=a \quad \text{for all } a\in A, \]
and
\[ f(f^{-1}(b))=b \quad \text{for all } b\in B. \]
A function is invertible if and only if it is bijective.
For matrices, this becomes
\[ A^{-1}A=I, \qquad AA^{-1}=I. \]
Linear algebra is not only about numbers. It is about operations that obey rules.
A binary operation on a set \(S\) is a function
\[ *:S\times S\to S. \]
For \(a,b\in S\), the output is usually written \(a*b\).
Examples:
A set with an operation may have algebraic structure.
A monoid is a set \(M\) together with a binary operation \(*\) such that:
Square matrices under multiplication form a monoid:
\[ A(BC)=(AB)C, \qquad AI=IA=A. \]
Not every matrix has a multiplicative inverse, so this is not a group.
A group is a set \(G\) together with a binary operation \(*\) such that:
If also \(a*b=b*a\) for all \(a,b\in G\), then \(G\) is an abelian group.
The set of all invertible \(n\times n\) matrices over a field \(\mathbb F\) is called the general linear group:
\[ GL(n,\mathbb F)=\{A\in \mathbb F^{n\times n}:A\text{ is invertible}\}. \]
This group will appear naturally whenever we study change of basis.
A ring is a set \(R\) with two operations, addition and multiplication, such that:
Some authors require a multiplicative identity in the definition of a ring. In this book, when a multiplicative identity is needed, we will say so explicitly.
A field is a set \(\mathbb F\) with two operations, addition and multiplication, such that:
In a field, addition, subtraction, multiplication, and division by nonzero elements are all possible.
Important examples:
\[ \mathbb Q, \qquad \mathbb R, \qquad \mathbb C. \]
The rational numbers, real numbers, and complex numbers are fields.
The integers \(\mathbb Z\) are not a field because, for example, \(2\) has no multiplicative inverse inside \(\mathbb Z\).
Many theorems in linear algebra depend on the field. A matrix may have real eigenvalues over \(\mathbb R\), or only complex eigenvalues over \(\mathbb C\). Later, this difference becomes important in spectral theory and Schur decomposition.
The main computational method for solving linear systems is row reduction.
The three elementary row operations on a matrix are:
These operations are performed on the augmented matrix of a linear system.
Elementary row operations on the augmented matrix of a linear system do not change the solution set of the system.
Each row represents one equation. Swapping equations does not change the set of common solutions. Multiplying an equation by a nonzero scalar gives an equivalent equation. Replacing one equation by itself plus a multiple of another equation also gives an equivalent system because the operation can be reversed.
The goal is to transform the augmented matrix into a simpler matrix.
Let \(M\) be a matrix. The leading entry of a nonzero row is the first nonzero entry in that row, reading from left to right. A pivot position is the position of a leading entry in a row-echelon form of \(M\).
A matrix is in row-echelon form if:
A matrix is in reduced row-echelon form if it is in row-echelon form and, in addition:
The reduced row-echelon form of a matrix \(M\) is denoted by \(\operatorname{rref}(M)\).
Every matrix over a field is row equivalent to exactly one reduced row-echelon matrix.
The leading variables are called pivot variables. The remaining variables are free variables.
Solve
\[ \begin{cases} x+y+z=2,\\ 2x+3y+z=5,\\ x+2y+3z=6. \end{cases} \]
The augmented matrix is
\[ \left[ \begin{array}{ccc|c} 1&1&1&2\\ 2&3&1&5\\ 1&2&3&6 \end{array} \right]. \]
Perform row operations:
\[ \left[ \begin{array}{ccc|c} 1&1&1&2\\ 2&3&1&5\\ 1&2&3&6 \end{array} \right] \xrightarrow{R_2\leftarrow R_2-2R_1,\ R_3\leftarrow R_3-R_1} \left[ \begin{array}{ccc|c} 1&1&1&2\\ 0&1&-1&1\\ 0&1&2&4 \end{array} \right]. \]
Then
\[ \left[ \begin{array}{ccc|c} 1&1&1&2\\ 0&1&-1&1\\ 0&1&2&4 \end{array} \right] \xrightarrow{R_3\leftarrow R_3-R_2} \left[ \begin{array}{ccc|c} 1&1&1&2\\ 0&1&-1&1\\ 0&0&3&3 \end{array} \right] \xrightarrow{R_3\leftarrow \frac13R_3} \left[ \begin{array}{ccc|c} 1&1&1&2\\ 0&1&-1&1\\ 0&0&1&1 \end{array} \right]. \]
Back substitution gives
\[ z=1, \qquad y-z=1, \qquad x+y+z=2. \]
Thus
\[ z=1, \qquad y=2, \qquad x=-1. \]
So the solution is
\[ \mathbf{x}= \begin{bmatrix} -1\\2\\1 \end{bmatrix}. \]
The rank of a matrix \(A\), denoted \(\operatorname{rank}(A)\), is the number of pivot columns in a row-echelon form of \(A\).
Equivalently, \(\operatorname{rank}(A)\) is the dimension of the column space of \(A\).
For a system
\[ A\mathbf{x}=\mathbf b, \]
we compare the rank of \(A\) and the rank of the augmented matrix \([A\mid \mathbf b]\).
Let \(A\in \mathbb F^{m\times n}\) and \(\mathbf b\in \mathbb F^m\). The system
\[ A\mathbf{x}=\mathbf b \]
is consistent if and only if
\[ \operatorname{rank}(A)=\operatorname{rank}([A\mid \mathbf b]). \]
The augmented matrix has one additional column, the vector \(\mathbf b\). If adding this column creates a new pivot, then \(\mathbf b\) is not generated by the columns of \(A\), and the system is inconsistent. If it does not create a new pivot, then \(\mathbf b\) lies in the column space of \(A\), and the system is consistent.
Assume \(A\mathbf{x}=\mathbf b\) is consistent, where \(A\in \mathbb F^{m\times n}\). Then
\[ \text{number of free variables}=n-\operatorname{rank}(A). \]
Therefore:
Let \(V\) and \(W\) be vector spaces over the same field \(\mathbb F\). A function \(T:V\to W\) is called linear if, for all \(\mathbf u,\mathbf v\in V\) and all scalars \(c\in \mathbb F\),
\[ T(\mathbf u+\mathbf v)=T(\mathbf u)+T(\mathbf v), \]
and
\[ T(c\mathbf u)=cT(\mathbf u). \]
Equivalently,
\[ T(c_1\mathbf v_1+c_2\mathbf v_2)=c_1T(\mathbf v_1)+c_2T(\mathbf v_2) \]
for all \(\mathbf v_1,\mathbf v_2\in V\) and \(c_1,c_2\in\mathbb F\).
Every matrix \(A\in\mathbb F^{m\times n}\) defines a linear map
\[ T_A:\mathbb F^n\to\mathbb F^m, \qquad T_A(\mathbf{x})=A\mathbf{x}. \]
For \(A\in \mathbb F^{m\times n}\), define
\[ \operatorname{Col}(A)=\operatorname{im}(T_A)=\{A\mathbf{x}:\mathbf{x}\in\mathbb F^n\}\subseteq \mathbb F^m, \]
and
\[ \ker(A)=\{\mathbf{x}\in\mathbb F^n:A\mathbf{x}=\mathbf 0\}\subseteq \mathbb F^n. \]
The space \(\operatorname{Col}(A)\) is called the column space of \(A\). The space \(\ker(A)\) is called the kernel or null space of \(A\).
Therefore
\[ A\mathbf{x}=\mathbf b \]
has a solution if and only if
\[ \mathbf b\in \operatorname{Col}(A). \]
This is one of the most important conceptual shifts in linear algebra.
We stop asking only:
Can I solve the equations?
We start asking:
Is the target vector inside the space generated by the columns?
Let \(A\in\mathbb F^{m\times n}\) and \(\mathbf b\in\mathbb F^m\). Suppose \(\mathbf{x}_p\) is one solution of
\[ A\mathbf{x}=\mathbf b. \]
Then the full solution set is
\[ \mathbf{x}_p+\ker(A)=\{\mathbf{x}_p+\mathbf z:\mathbf z\in\ker(A)\}. \]
In words, every solution is one particular solution plus a solution of the homogeneous system \(A\mathbf z=\mathbf 0\).
Let \(\mathbf{x}\) be any solution. Then
\[ A\mathbf{x}=\mathbf b, \qquad A\mathbf{x}_p=\mathbf b. \]
Subtracting gives
\[ A(\mathbf{x}-\mathbf{x}_p)=\mathbf 0. \]
Thus \(\mathbf{x}-\mathbf{x}_p\in\ker(A)\), so \(\mathbf{x}=\mathbf{x}_p+\mathbf z\) for some \(\mathbf z\in\ker(A)\).
Conversely, if \(\mathbf z\in\ker(A)\), then
\[ A(\mathbf{x}_p+\mathbf z)=A\mathbf{x}_p+A\mathbf z=\mathbf b+\mathbf 0=\mathbf b. \]
So every vector of the form \(\mathbf{x}_p+\mathbf z\) is a solution.
This is an affine subspace: a translated version of the null space.
Solving a nonhomogeneous system means finding one particular solution and then adding all solutions of the homogeneous system.
We now use Python to solve the worked example.
import sympy as sp
A = sp.Matrix([
[1, 1, 1],
[2, 3, 1],
[1, 2, 3]
])
b = sp.Matrix([2, 5, 6])
Aug = A.row_join(b)
Aug_rref, pivots = Aug.rref()
Aug_rref, pivots(Matrix([
[1, 0, 0, -1],
[0, 1, 0, 2],
[0, 0, 1, 1]]),
(0, 1, 2))The reduced row-echelon form tells us the solution directly.
solution = A.LUsolve(b)
solution\(\displaystyle \left[\begin{matrix}-1\\2\\1\end{matrix}\right]\)
We can also check the answer.
A * solution\(\displaystyle \left[\begin{matrix}2\\5\\6\end{matrix}\right]\)
import numpy as np
A_np = np.array([
[1, 1, 1],
[2, 3, 1],
[1, 2, 3]
], dtype=float)
b_np = np.array([2, 5, 6], dtype=float)
x_np = np.linalg.solve(A_np, b_np)
x_nparray([-1., 2., 1.])sympy performs exact symbolic computation when possible. numpy performs numerical computation using floating-point arithmetic. In small examples they often agree, but for large or ill-conditioned systems, numerical issues become important.
A = sp.Matrix([
[1, 2, 1],
[2, 4, 2],
[1, 1, 0]
])
b = sp.Matrix([3, 6, 1])
Aug = A.row_join(b)
A.rank(), Aug.rank(), Aug.rref()(2,
2,
(Matrix([
[1, 0, -1, -1],
[0, 1, 1, 2],
[0, 0, 0, 0]]),
(0, 1)))The pair
\[ \bigl(\operatorname{rank}(A),\operatorname{rank}([A\mid \mathbf b])\bigr) \]
tells us whether the system is consistent.
Students now have access to AI tools. This changes how we study linear algebra.
AI can help generate examples, explain steps, and write code. But AI can also make mistakes, skip assumptions, or confuse similar concepts.
Here is a useful prompt:
Explain why elementary row operations do not change the solution set of a linear system. Give a numerical example and a geometric interpretation.
After reading the answer, check:
Do not ask AI only for the answer. Ask it for examples, counterexamples, visualizations, and alternative explanations. Then verify the mathematics yourself.
Choose one of the following systems:
\[ \begin{cases} x+y=1,\\ x-y=3, \end{cases} \qquad \begin{cases} x+y=1,\\ 2x+2y=2, \end{cases} \qquad \begin{cases} x+y=1,\\ x+y=2. \end{cases} \]
For your chosen system:
Here is starter code.
import numpy as np
import matplotlib.pyplot as plt
x = np.linspace(-4, 4, 400)
# Example: x + y = 1 and x - y = 3
y1 = 1 - x
y2 = x - 3
plt.figure()
plt.plot(x, y1, label=r"$x+y=1$")
plt.plot(x, y2, label=r"$x-y=3$")
plt.axhline(0, linewidth=0.5)
plt.axvline(0, linewidth=0.5)
plt.xlabel("x")
plt.ylabel("y")
plt.legend()
plt.title("A linear system as intersection of lines")
plt.grid(True)
plt.show()
This chapter introduced the first language of linear algebra.
We learned that:
The next chapter begins the study of vectors and vector spaces. We will move from systems of equations to the spaces of all possible solutions.
Explain why a system of two linear equations in two variables cannot have exactly two solutions.
Give an example of a system with infinitely many solutions. Explain both algebraically and geometrically.
What is the difference between the coefficient matrix and the augmented matrix?
Explain why the equation \(A\mathbf{x}=\mathbf b\) has a solution if and only if \(\mathbf b\) is in the column space of \(A\).
Why is multiplying a row by zero not an elementary row operation?
\[ \begin{cases} 2x+3y=4,\\ x-y=1. \end{cases} \]
\[ \begin{cases} x+2y+z=1,\\ 2x+4y+2z=3. \end{cases} \]
\[ \left[ \begin{array}{ccc|c} 1&2&1&3\\ 2&4&2&6\\ 1&1&0&1 \end{array} \right]. \]
Determine the solution set.
\[ A= \begin{bmatrix} 1&2&3\\ 2&4&6\\ 1&1&1 \end{bmatrix}. \]
Prove that swapping two equations in a linear system does not change the solution set.
Prove that if \(\mathbf{x}_p\) is one solution of \(A\mathbf{x}=\mathbf b\), then every solution has the form
\[ \mathbf{x}_p+\mathbf{x}_h, \]
where \(A\mathbf{x}_h=\mathbf 0\).
Write a Python function that takes an augmented matrix and returns its reduced row-echelon form using sympy.
Generate three random \(3\times 3\) matrices. For each matrix, compute its rank and determinant. What do you observe when the determinant is zero?
Use Python to create a system with infinitely many solutions. Row-reduce the augmented matrix and identify the free variables.
Ask an AI tool:
“What is the relationship between rank and the number of solutions of a linear system?”
Then critique the answer. Did it mention the augmented matrix?
Ask an AI tool to produce a system with no solution, one solution, and infinitely many solutions. Verify each example by row reduction.
Ask an AI tool to explain the difference between symbolic and numerical computation. Then test its explanation by solving a system using both sympy and numpy.
Two distinct lines in the plane either do not meet, meet once, or are the same line. If they are the same line, they share infinitely many points. Therefore two linear equations in two variables cannot have exactly two solutions.
From
\[ \begin{cases} 2x+3y=4,\\ x-y=1, \end{cases} \]
the second equation gives \(x=y+1\). Substitute into the first:
\[ 2(y+1)+3y=4. \]
So
\[ 5y+2=4, \qquad 5y=2, \qquad y=\frac25. \]
Then
\[ x=1+\frac25=\frac75. \]
Thus
\[ (x,y)=\left(\frac75,\frac25\right). \]
Suppose \(\mathbf{x}\) is any solution of \(A\mathbf{x}=\mathbf b\) and \(\mathbf{x}_p\) is one particular solution. Then
\[ A\mathbf{x}=\mathbf b, \qquad A\mathbf{x}_p=\mathbf b. \]
Subtracting gives
\[ A(\mathbf{x}-\mathbf{x}_p)=\mathbf 0. \]
Thus \(\mathbf{x}-\mathbf{x}_p\in\ker(A)\), so
\[ \mathbf{x}=\mathbf{x}_p+\mathbf{x}_h \]
for some \(\mathbf{x}_h\in\ker(A)\).
Conversely, if \(\mathbf{x}_h\in\ker(A)\), then
\[ A(\mathbf{x}_p+\mathbf{x}_h)=A\mathbf{x}_p+A\mathbf{x}_h=\mathbf b+\mathbf 0=\mathbf b. \]
Therefore every vector of the form \(\mathbf{x}_p+\mathbf{x}_h\) is a solution.