§1 Linear Equations and Linear Systems

1. The Big Idea

Linear algebra begins with a simple question:

Which values make several linear equations true at the same time?

A linear equation contains only first-power variables. No $x^2$, no $xy$, no $\frac{1}{x}$. Geometrically, it is always a flat object: a point, line, plane, or a higher-dimensional version of those.

The type of answer depends on how many variables are involved:

Everything that follows is about organizing equations so the solution becomes easier to see.

---

2. Linear Equations in One Variable

A one-variable linear equation has the form

$$ax=b$$

where $a$ and $b$ are constants. The number $a$ decides what kind of equation we have:

Example: One Solution

$$3x=6$$

Divide by $3$:

$$x=2$$

The coefficient of $x$ is nonzero, so the equation forces $x$ to be one specific value.

Example: No Solution

$$0x=5$$

The left side is always $0$, no matter what $x$ is. So this really says

$$0=5$$

That is impossible, so there is no solution.

Example: Infinitely Many Solutions

$$0x=0$$

This says

$$0=0$$

which is always true. The equation gives no restriction on $x$, so every real number is a solution.

---

3. Linear Equations in Several Variables

When there is more than one variable, a single equation usually cannot fix every variable. Instead, it describes a whole family of possible solutions.

Two Variables: A Line

A linear equation in two variables has the form

$$ax+by=c$$

where $a,b,c$ are constants and $a,b$ are not both zero.

For example:

$$2x+y=7$$

A solution is an ordered pair $(x,y)$ that makes the equation true. If $x=2$, then

$$2(2)+y=7$$

so

$$4+y=7$$

and therefore

$$y=3$$

So $(2,3)$ is one solution.

But it is not the only solution. If we pick a different $x$, we get a different $y$. As $x$ varies, the solutions sweep out all the points on the line.

Parameters: Using a Variable as a Slider

A parameter is a free variable that lets us describe infinitely many solutions without listing them one at a time.

For

$$2x+y=7$$

choose $x$ freely. Let

$$x=t$$

Then

$$2t+y=7$$

so

$$y=7-2t$$

The full solution set is

$$(x,y)=(t,7-2t),\qquad t\in\mathbb R$$

Think of $t$ as a slider. Every time you choose a real number for $t$, you get one point on the line.

Three Variables: A Plane

A linear equation in three variables has the form

$$ax+by+cz=d$$

where $a,b,c,d$ are constants and $a,b,c$ are not all zero.

For example:

$$x+2y+3z=5$$

One equation with three variables usually has two free variables. Let

$$y=s,\qquad z=t$$

Then

$$x+2s+3t=5$$

so

$$x=5-2s-3t$$

The solution set is

$$(x,y,z)=(5-2s-3t,s,t),\qquad s,t\in\mathbb R$$

This describes a plane in three-dimensional space.

---

4. Systems of Linear Equations

A system of linear equations is a collection of equations that must all be true simultaneously.

Example:

$$\begin{cases} x+y=5\\ 2x-y=1 \end{cases}$$

A solution must satisfy every equation, not just one of them.

Systems have only three possible outcomes:

Outcome 1: Exactly One Solution

$$\begin{cases} x+y=5\\ x-y=1 \end{cases}$$

Adding the equations gives $2x=6$, so $x=3$. Then $y=2$.

The solution is

$$(x,y)=(3,2)$$

Geometrically, two lines meet at one point.

Outcome 2: No Solution

$$\begin{cases} x+y=3\\ x+y=7 \end{cases}$$

The same expression cannot equal both $3$ and $7$. These equations contradict each other.

In row-reduction, this kind of contradiction appears as something like

$$0=5$$

or, in augmented-matrix form,

$$[0\ 0\ 0\mid 5]$$

That row says "zero equals five," so the system is inconsistent.

Outcome 3: Infinitely Many Solutions

$$\begin{cases} x+y=3\\ 2x+2y=6 \end{cases}$$

The second equation is just twice the first. It does not add new information.

So instead of one intersection point, the equations describe the same line. Every point on that line is a solution.

---

5. Augmented Matrices

An augmented matrix is a compact way to write a system. We keep the coefficients on the left and the constants on the right.

For example,

$$\begin{cases} 2x+y=7\\ -x+3y=1 \end{cases}$$

becomes

$$\left[ \begin{array}{cc|c} 2&1&7\\ -1&3&1 \end{array} \right]$$

Read it like this:

• each row is one equation
• each coefficient column belongs to one variable
• the vertical bar separates coefficients from constants

For the matrix above, the variable order is $x,y$. That means the first column is the coefficient of $x$, and the second column is the coefficient of $y$.

Variable order matters. If the variables are $x,y,z$, keep that order consistently when building and reading the matrix.

---

6. Elementary Row Operations

Row-reducing a matrix means repeatedly applying three legal moves called elementary row operations.

These operations change how the system looks, but they do not change the solution set.

That is the whole point: we keep rewriting the system into an easier version until the answer is visible.

---

7. REF, RREF, and Pivots

Row-reduction aims for a clean staircase pattern.

Row Echelon Form (REF)

A matrix is in row echelon form if:

1. All zero rows sit at the bottom.
2. Each leading nonzero entry is to the right of the one above it.
3. Entries below each leading entry are zero.

This creates a staircase shape.

Reduced Row Echelon Form (RREF)

A matrix is in reduced row echelon form if:

1. It is already in REF.
2. Each leading entry is $1$.
3. Each leading $1$ has zeros above and below it in its column.

RREF is the cleanest form because it usually lets you read the solution directly.

Pivots and Free Variables

A pivot is a leading entry in a row. In RREF, pivots are leading $1$s; variables with pivot columns are fixed, and the others are free.

• A variable with a pivot in its column is a pivot variable.
• A variable without a pivot in its column is a free variable.
• Free variables become parameters.

This pivot/free-variable idea is one of the most important habits in linear algebra.

---

8. Gaussian vs. Gauss-Jordan Elimination

The names sound similar, but the difference is simple:

Gaussian Elimination

Gaussian elimination reduces a matrix to REF. Once you have a staircase, you start from the bottom row and work upward.

This is called back substitution.

Gauss-Jordan Elimination

Gauss-Jordan elimination keeps going until the matrix is in RREF. The goal is to make each pivot column as clean as possible.

For example,

$$\left[ \begin{array}{cc|c} 1&0&3\\ 0&1&2 \end{array} \right]$$

already says

$$x=3,\qquad y=2$$

No extra work is needed.

---

9. Reading Solutions from RREF

Once a matrix is reduced, do not just stare at the numbers. Translate each row back into an equation.

Case A: One Solution

$$\left[ \begin{array}{cc|c} 1&0&3\\ 0&1&2 \end{array} \right]$$

This means

$$x=3,\qquad y=2$$

Both variables have pivots, so the system has exactly one solution:

$$(x,y)=(3,2)$$

Case B: No Solution

$$\left[ \begin{array}{ccc|c} 1&0&2&4\\ 0&1&-1&3\\ 0&0&0&5 \end{array} \right]$$

The last row means

$$0x+0y+0z=5$$

or simply

$$0=5$$

That is impossible, so the system has no solution.

Case C: Infinitely Many Solutions

$$\left[ \begin{array}{ccc|c} 1&0&2&5\\ 0&1&-1&1 \end{array} \right]$$

With variables ordered as $x,y,z$, this means

$$x+2z=5$$

and

$$y-z=1$$

The $z$-column has no pivot, so $z$ is free. Let

$$z=t$$

Then

$$x=5-2t$$

and

$$y=1+t$$

So the solution set is

$$(x,y,z)=(5-2t,1+t,t),\qquad t\in\mathbb R$$

This describes infinitely many solutions because every real value of $t$ gives a different solution.

---

10. Homogeneous Systems

A homogeneous system has zeros on the right-hand side.

Example:

$$\begin{cases} x+2y-z=0\\ 3x-y+4z=0 \end{cases}$$

In matrix form, homogeneous systems look like

$$A\vec{x}=\vec{0}$$

Every homogeneous system has at least one solution:

$$\vec{x}=\vec{0}$$

This is called the trivial solution.

The more interesting question is:

Are there any nontrivial solutions?

A nontrivial solution is a solution where at least one variable is not zero.

More Unknowns Than Equations

If a homogeneous system has more unknowns than equations, then at least one variable is free, so it has infinitely many solutions.

Why? There are not enough equations to force every variable to be a pivot variable.

So for a homogeneous system:

$$\text{more variables than equations} \quad\Longrightarrow\quad \text{infinitely many solutions}$$

That also means there must be nontrivial solutions.

---

11. Problem-Solving Routine

When solving a system by row-reduction, use this checklist:

1. Write the system clearly. Make sure each equation is in the same variable order.
2. Build the augmented matrix. Coefficients go left of the bar, constants go right.
3. Row-reduce. Use elementary row operations to reach REF or RREF.
4. Look for contradiction rows. A row like $[0\ 0\ 0\mid 5]$ means no solution.
5. Identify pivots. Pivot columns correspond to pivot variables.
6. Name the free variables. Usually use parameters like $s$ and $t$.
7. Write the solution clearly. Use a point for one solution, or a parameterized form for infinitely many solutions.

---

12. Key Takeaways

• A solution to a system must satisfy every equation at the same time.
• Row operations preserve the solution set, even though they change the matrix.
• REF gives a staircase; RREF gives the cleanest possible version.
• Pivot variables are fixed by the equations.
• Free variables become parameters.
• A contradiction row means the system has no solution.
• Homogeneous systems always have the trivial solution, but may also have nontrivial solutions.

---

Mini-Self-Check

1. What does the row $[0\ 0\ 0\mid 4]$ mean?

2. If a system with variables $x,y,z$ has pivots in the $x$- and $y$-columns only, which variable is free?

3. Why does a homogeneous system always have at least one solution?