Systems of Linear Equations

See how linear algebra naturally arises from solving multiple equations

You have been solving systems of equations for years, probably without realizing you were doing linear algebra all along. Every time you figured out the prices of two items from two receipts, or determined where two lines cross, you were engaging with the core problem that gave birth to the entire field. Linear algebra did not emerge from abstract theorizing; it grew directly from the practical need to solve many equations at once, efficiently and systematically.

Here is the beautiful thing: the methods you already know (substitution, elimination) are exactly the methods that linear algebra organizes and extends. The difference is that linear algebra provides a language and framework that scales effortlessly from two equations to two thousand. Instead of manipulating individual equations, you learn to work with matrices and vectors, structures that encode entire systems in compact form. This shift in perspective transforms a tedious algebraic grind into elegant matrix operations.

In this lesson, we revisit systems of equations through the lens of linear algebra. You will see how matrices naturally emerge as the right way to think about these problems, and why the equation $A\vec{x} = \vec{b}$ is one of the most important expressions in all of mathematics.

Core Concepts

Review: What Is a Linear Equation?

Before we tackle systems, let us be precise about what “linear” means. A linear equation in variables $x_1, x_2, \ldots, x_n$ is an equation of the form:

$$a_1 x_1 + a_2 x_2 + \cdots + a_n x_n = b$$

where $a_1, a_2, \ldots, a_n$ and $b$ are constants (real numbers). The key features of linearity are:

  • Each variable appears to the first power only (no $x^2$, no $\sqrt{x}$, no $x_1 \cdot x_2$)
  • Variables are only multiplied by constants and added together
  • There are no products of different variables

For example, $3x + 2y - z = 7$ is linear, but $xy = 4$ and $x^2 + y = 1$ are not.

Why does this matter? Because linear equations have beautiful, predictable behavior. Their solutions form straight lines (in 2D), planes (in 3D), and hyperplanes (in higher dimensions). This geometric regularity is what makes linear algebra so powerful.

Systems with Two Variables: Intersecting Lines

Consider two linear equations in two variables:

$$x + y = 5$$ $$2x - y = 1$$

Geometrically, each equation represents a line in the $xy$-plane. A solution to the system is a point $(x, y)$ that lies on both lines simultaneously. In other words, we are looking for the intersection point.

This geometric picture explains the three possible outcomes:

  1. The lines intersect at exactly one point: There is exactly one solution. This is the typical case when the lines have different slopes.

  2. The lines are parallel (never intersect): There is no solution. The equations are asking for something impossible, like finding a point that is simultaneously on two different parallel streets.

  3. The lines are the same (coincide completely): There are infinitely many solutions. Every point on the line satisfies both equations, because the equations are really saying the same thing in different words.

These three cases correspond to the fundamental trichotomy that appears throughout linear algebra: one solution, no solution, or infinitely many solutions. There is never any other option for systems of linear equations.

Systems with Three Variables: Intersecting Planes

With three variables, each linear equation represents a plane in three-dimensional space:

$$x + y + z = 6$$ $$2x - y + z = 3$$ $$x + 2y - z = 2$$

A solution must satisfy all three equations, meaning it must lie on all three planes. Geometrically, we are looking for where three planes intersect.

The possibilities become richer:

  • One solution: All three planes meet at a single point. Think of three walls meeting at a corner.
  • No solution: The planes have no common point. This can happen if two planes are parallel, or if three planes form a “triangular prism” shape where no single point lies on all three.
  • Infinitely many solutions: The planes share a common line (they all pass through the same line) or they are all the same plane.

This geometric intuition helps explain why some systems have solutions and others do not. But as we add more variables and equations, visualization becomes impossible. This is where the algebraic machinery of matrices becomes essential.

Consistent vs. Inconsistent Systems

We have terminology to describe whether a system can be solved:

  • A system is consistent if it has at least one solution (either exactly one or infinitely many).
  • A system is inconsistent if it has no solution at all.

The key insight is that consistency depends on whether the equations are asking for something geometrically possible. Parallel lines cannot intersect; inconsistent systems are asking the mathematical equivalent of finding their intersection.

When solving a system algebraically, you can detect inconsistency when the equations reduce to something impossible, like $0 = 5$. This is the algebraic signature that the geometry does not work out.

Types of Solutions: A Classification

Let us be systematic about the possible outcomes for a system of linear equations:

Case Number of Solutions Geometric Picture (2D) Terminology
Lines intersect Exactly one Two lines crossing Consistent, independent
Lines parallel None Two parallel lines Inconsistent
Lines coincide Infinitely many Same line Consistent, dependent

For systems to have exactly one solution, the equations must provide “enough independent information” to pin down each variable. When equations are redundant (saying the same thing differently), you get infinitely many solutions. When equations contradict each other, you get no solution.

Matrix Representation: The Birth of Linear Algebra

Here is where the perspective shift happens. Consider this system:

$$2x + 3y = 7$$ $$x - y = 2$$

Instead of thinking about two separate equations, we can encode the entire system in a single matrix equation:

$$\begin{pmatrix} 2 & 3 \ 1 & -1 \end{pmatrix} \begin{pmatrix} x \ y \end{pmatrix} = \begin{pmatrix} 7 \ 2 \end{pmatrix}$$

Let us verify this works. When you multiply the matrix by the vector, you get:

$$\begin{pmatrix} 2 \cdot x + 3 \cdot y \ 1 \cdot x + (-1) \cdot y \end{pmatrix} = \begin{pmatrix} 2x + 3y \ x - y \end{pmatrix}$$

Setting this equal to the right-hand side gives us exactly our original equations. This is the form $A\vec{x} = \vec{b}$, where:

  • $A$ is the coefficient matrix containing the coefficients of the variables
  • $\vec{x}$ is the variable vector (the unknowns we are solving for)
  • $\vec{b}$ is the constant vector (the right-hand sides of the equations)

This notation is not just compact; it reveals structure. The matrix $A$ captures all the relationships between variables, separate from what specific values we are trying to achieve (encoded in $\vec{b}$). This separation turns out to be profoundly useful.

The Augmented Matrix

For computational purposes, we often combine $A$ and $\vec{b}$ into a single augmented matrix, separated by a vertical line:

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

The augmented matrix contains all the information in the system. The vertical line is just a visual reminder of where the equals sign was. When we perform row operations to solve the system, we work with this augmented matrix, performing the same operations on both the coefficients and the constants.

Why Matrices Make This Easier

You might wonder: why bother with matrices when substitution and elimination already work? The answer comes down to organization and scalability.

Organization: Matrices strip away the variable names and focus on the numbers that matter: the coefficients and constants. This makes it easier to see patterns and apply systematic procedures.

Systematic procedures: The process of Gaussian elimination (which you will study next) becomes a mechanical procedure on the augmented matrix. You do not have to make clever choices about which equation to substitute where; you follow a clear algorithm that always works.

Scalability: For two equations, substitution is fine. For three, it is manageable. For ten equations in ten unknowns? A hundred equations? Matrix methods scale linearly with problem size, while naive substitution becomes exponentially more complex.

Deeper insight: The matrix $A$ in $A\vec{x} = \vec{b}$ tells you everything about the structure of the system, independent of any particular $\vec{b}$. Properties of $A$ (like its determinant, rank, or invertibility) determine whether the system has solutions and how many. This structural insight is impossible to see when looking at individual equations.

Computer implementation: Matrices map directly to arrays in programming. Algorithms for solving $A\vec{x} = \vec{b}$ are among the most optimized in all of numerical computing. Every time you use a spreadsheet solver or engineering software, matrices are doing the heavy lifting.

From Equations to Geometry and Back

Linear algebra gives you multiple ways to view the same system:

Row picture (equations): Each equation defines a line/plane/hyperplane. The solution is where they all intersect.

Column picture (vectors): Write the system as $x_1 \vec{a}_1 + x_2 \vec{a}_2 + \cdots = \vec{b}$, where $\vec{a}_i$ are the columns of $A$. The solution tells you how to combine the column vectors to produce $\vec{b}$.

Matrix picture: The matrix $A$ represents a transformation. Solving $A\vec{x} = \vec{b}$ asks: what input $\vec{x}$ gives output $\vec{b}$?

These perspectives are all equivalent mathematically, but each illuminates different aspects of the problem. Developing fluency in moving between them is a core skill in linear algebra.

Notation and Terminology

Term Meaning Example
System of equations Multiple equations, same variables $x + y = 5$, $2x - y = 1$
Linear equation Equation where variables appear to first power only $3x - 2y + z = 7$
Consistent Has at least one solution System with intersecting lines
Inconsistent Has no solution Parallel lines
Independent Equations provide distinct constraints $x + y = 5$ and $x - y = 1$
Dependent One equation is a multiple of another $x + y = 5$ and $2x + 2y = 10$
Coefficient matrix Matrix $A$ of variable coefficients $\begin{pmatrix} 1 & 1 \ 2 & -1 \end{pmatrix}$
Variable vector Column vector $\vec{x}$ of unknowns $\begin{pmatrix} x \ y \end{pmatrix}$
Constant vector Column vector $\vec{b}$ of right-hand sides $\begin{pmatrix} 5 \ 1 \end{pmatrix}$
Augmented matrix Matrix with both coefficients and constants $\left[\begin{array}{cc|c} 1 & 1 & 5 \ 2 & -1 & 1 \end{array}\right]$

Examples

Example 1: Solving by Substitution

Solve the system $x + y = 5$ and $x - y = 1$.

Solution:

From the first equation, solve for $x$: $$x = 5 - y$$

Substitute into the second equation: $$(5 - y) - y = 1$$ $$5 - 2y = 1$$ $$-2y = -4$$ $$y = 2$$

Back-substitute to find $x$: $$x = 5 - y = 5 - 2 = 3$$

Verification:

  • First equation: $3 + 2 = 5$ (True)
  • Second equation: $3 - 2 = 1$ (True)

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

This system has exactly one solution because the two lines (with slopes $-1$ and $1$) intersect at a single point.

Example 2: Writing a System as an Augmented Matrix

Write the system $2x + 3y = 7$ and $x - y = 2$ as an augmented matrix.

Solution:

First, identify the coefficient matrix $A$ and constant vector $\vec{b}$:

$$A = \begin{pmatrix} 2 & 3 \ 1 & -1 \end{pmatrix}, \quad \vec{b} = \begin{pmatrix} 7 \ 2 \end{pmatrix}$$

The augmented matrix combines these with a vertical line:

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

The first column contains the coefficients of $x$, the second column contains the coefficients of $y$, and the column after the vertical line contains the constants.

Note: The matrix form of this system is:

$$\begin{pmatrix} 2 & 3 \ 1 & -1 \end{pmatrix} \begin{pmatrix} x \ y \end{pmatrix} = \begin{pmatrix} 7 \ 2 \end{pmatrix}$$

Example 3: Determining the Number of Solutions

Determine if the system $x + 2y = 4$ and $2x + 4y = 8$ has one solution, no solution, or infinitely many solutions.

Solution:

Let us examine the relationship between the two equations.

Multiply the first equation by 2: $$2(x + 2y) = 2(4)$$ $$2x + 4y = 8$$

This is exactly the second equation! The two equations are not independent; the second is just the first multiplied by 2. They represent the same line.

Geometric interpretation: Both equations describe the same line in the $xy$-plane. Every point on that line satisfies both equations.

Algebraic verification: If we try elimination by subtracting twice the first equation from the second: $$2x + 4y - 2(x + 2y) = 8 - 2(4)$$ $$2x + 4y - 2x - 4y = 8 - 8$$ $$0 = 0$$

We get a true statement that tells us nothing new about $x$ or $y$. This is the signature of infinitely many solutions.

Solution: The system has infinitely many solutions. We can express them as: $(x, y) = (4 - 2t, t)$ for any real number $t$. For example, $(4, 0)$, $(2, 1)$, and $(0, 2)$ are all solutions.

Example 4: From Matrix Equation to System of Equations

Write $\begin{pmatrix} 2 & 1 \ 3 & -2 \end{pmatrix}\begin{pmatrix} x \ y \end{pmatrix} = \begin{pmatrix} 5 \ 4 \end{pmatrix}$ as a system of equations.

Solution:

When we multiply the matrix by the vector, we use the rule: each entry of the result is the dot product of a row with the column vector.

First row times the vector: $$2 \cdot x + 1 \cdot y = 2x + y$$

Setting this equal to the first entry of the result vector: $$2x + y = 5$$

Second row times the vector: $$3 \cdot x + (-2) \cdot y = 3x - 2y$$

Setting this equal to the second entry of the result vector: $$3x - 2y = 4$$

The system of equations is: $$2x + y = 5$$ $$3x - 2y = 4$$

Verification: You can solve this system (by substitution or elimination) to get $x = 2$, $y = 1$. Check: $\begin{pmatrix} 2 & 1 \ 3 & -2 \end{pmatrix}\begin{pmatrix} 2 \ 1 \end{pmatrix} = \begin{pmatrix} 4 + 1 \ 6 - 2 \end{pmatrix} = \begin{pmatrix} 5 \ 4 \end{pmatrix}$ (Correct)

Example 5: Finding Values for Infinitely Many Solutions

For what value(s) of $k$ does the system $x + 2y = 3$ and $2x + ky = 6$ have infinitely many solutions?

Solution:

For a system of two linear equations to have infinitely many solutions, the equations must represent the same line. This happens when one equation is a scalar multiple of the other.

If the second equation is a multiple of the first, then multiplying $x + 2y = 3$ by some constant $c$ should give $2x + ky = 6$.

Multiplying by $c$: $$c(x + 2y) = c \cdot 3$$ $$cx + 2cy = 3c$$

Comparing with $2x + ky = 6$:

  • Coefficients of $x$: $c = 2$
  • Coefficients of $y$: $2c = k$
  • Constants: $3c = 6$

From $c = 2$:

  • Check constants: $3(2) = 6$ (Consistent)
  • Find $k$: $k = 2c = 2(2) = 4$

Verification: With $k = 4$, the system becomes: $$x + 2y = 3$$ $$2x + 4y = 6$$

Dividing the second equation by 2 gives $x + 2y = 3$, which is identical to the first equation. They represent the same line.

Alternative approach using the augmented matrix:

$$\left[\begin{array}{cc|c} 1 & 2 & 3 \ 2 & k & 6 \end{array}\right]$$

For infinitely many solutions, the second row must be a multiple of the first. If we subtract 2 times the first row from the second:

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

For infinitely many solutions (not just the trivial equation $0 = 0$), we need $k - 4 = 0$, so $k = 4$.

Answer: $k = 4$

Key Properties and Rules

Recognizing Solution Types

When solving a system algebraically:

What You Get Meaning
Unique values for all variables Exactly one solution (consistent, independent)
A contradiction like $0 = 5$ No solution (inconsistent)
A tautology like $0 = 0$ with free variables Infinitely many solutions (consistent, dependent)

Converting Between Forms

From system to matrix form $A\vec{x} = \vec{b}$:

  1. List variables in a consistent order (e.g., $x$, $y$, $z$)
  2. Extract coefficients into matrix $A$ (row by row, matching equation order)
  3. Form the variable vector $\vec{x}$
  4. Extract constants into $\vec{b}$

From matrix form to system:

  1. Multiply each row of $A$ by $\vec{x}$ using the dot product
  2. Set each result equal to the corresponding entry of $\vec{b}$

Conditions for Solution Types (Two Equations, Two Variables)

For the system $a_1 x + b_1 y = c_1$ and $a_2 x + b_2 y = c_2$:

  • One solution: The lines have different slopes, meaning $\frac{a_1}{a_2} \neq \frac{b_1}{b_2}$ (or equivalently, $a_1 b_2 \neq a_2 b_1$)
  • No solution: The lines are parallel but distinct, meaning $\frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2}$
  • Infinitely many solutions: The lines are identical, meaning $\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}$

The Determinant Test (Preview)

For a $2 \times 2$ coefficient matrix $A = \begin{pmatrix} a & b \ c & d \end{pmatrix}$, the determinant is:

$$\det(A) = ad - bc$$

  • If $\det(A) \neq 0$: The system $A\vec{x} = \vec{b}$ has exactly one solution for any $\vec{b}$
  • If $\det(A) = 0$: The system has either no solution or infinitely many, depending on $\vec{b}$

This is a powerful preview of how matrix properties determine system behavior.

Real-World Applications

Balancing Chemical Equations

When you balance a chemical equation like:

$$a \text{H}_2 + b \text{O}_2 \rightarrow c \text{H}_2\text{O}$$

you are solving a system of linear equations. Each element must be conserved:

  • Hydrogen: $2a = 2c$
  • Oxygen: $2b = c$

This gives you the system that determines the coefficients. For complex reactions with many compounds, matrix methods become essential.

Network Flow Problems

Consider a network of roads or pipes where flow must be conserved at each junction (what flows in must flow out). At each node, you get a linear equation. A city’s water distribution system or internet routing network involves hundreds or thousands of such equations, all linear, all solvable using the matrix methods you are learning.

Economics: Supply and Demand Equilibrium

In economics, supply and demand are often modeled as linear functions:

  • Supply: $Q_s = a + bP$ (quantity supplied depends on price)
  • Demand: $Q_d = c - dP$ (quantity demanded depends on price)

At equilibrium, $Q_s = Q_d$. This is a system of linear equations whose solution gives the equilibrium price and quantity. When multiple goods interact (the price of one affecting demand for another), you get larger systems that require matrix methods.

Circuit Analysis (Kirchhoff’s Laws)

In electrical engineering, Kirchhoff’s laws state:

  • The sum of currents entering a junction equals the sum leaving (conservation of charge)
  • The sum of voltage drops around any closed loop equals zero (conservation of energy)

Each law gives linear equations in the unknown currents. Analyzing even a moderately complex circuit produces a system of many equations, routinely solved using matrix techniques.

Computer Graphics and Transformations

Every time you see a 3D model rotate, scale, or move on a computer screen, matrices are at work. The position of each vertex is a vector, and transformations are matrix multiplications. Solving $A\vec{x} = \vec{b}$ answers questions like: “What original position would end up at this screen location after the transformation?”

Self-Test Problems

Problem 1: Write the system $3x - 2y = 5$ and $x + 4y = -1$ as an augmented matrix.

Show Answer

The coefficient matrix has the coefficients of $x$ in the first column and $y$ in the second column. The constant vector contains the right-hand sides.

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

Problem 2: Write $\begin{pmatrix} 1 & 3 \ 4 & -1 \end{pmatrix}\begin{pmatrix} x \ y \end{pmatrix} = \begin{pmatrix} 2 \ 7 \end{pmatrix}$ as a system of equations, then solve it.

Show Answer

System of equations: $$x + 3y = 2$$ $$4x - y = 7$$

Solving by substitution: From the first equation, $x = 2 - 3y$. Substitute into the second: $$4(2 - 3y) - y = 7$$ $$8 - 12y - y = 7$$ $$8 - 13y = 7$$ $$-13y = -1$$ $$y = \frac{1}{13}$$

Back-substitute: $x = 2 - 3 \cdot \frac{1}{13} = 2 - \frac{3}{13} = \frac{26 - 3}{13} = \frac{23}{13}$

Solution: $(x, y) = \left(\frac{23}{13}, \frac{1}{13}\right)$

Verification: $\frac{23}{13} + 3 \cdot \frac{1}{13} = \frac{26}{13} = 2$ and $4 \cdot \frac{23}{13} - \frac{1}{13} = \frac{92 - 1}{13} = \frac{91}{13} = 7$ (Both correct)

Problem 3: Determine whether the system $2x - 3y = 4$ and $-4x + 6y = 5$ is consistent or inconsistent.

Show Answer

Let us check if one equation is a multiple of the other.

Multiply the first equation by $-2$: $$-2(2x - 3y) = -2(4)$$ $$-4x + 6y = -8$$

Compare with the second equation: $-4x + 6y = 5$

The left sides are identical, but $-8 \neq 5$. This means the equations are contradictory.

The system is inconsistent (no solution).

Geometrically, these are parallel lines (same slope, different y-intercepts) that never intersect.

Problem 4: For the system $x + y + z = 6$, $x - y + z = 2$, and $2x + z = 8$, write the coefficient matrix $A$ and the constant vector $\vec{b}$.

Show Answer

Align the equations with variables in order $(x, y, z)$:

  • $1x + 1y + 1z = 6$
  • $1x + (-1)y + 1z = 2$
  • $2x + 0y + 1z = 8$

Coefficient matrix: $$A = \begin{pmatrix} 1 & 1 & 1 \ 1 & -1 & 1 \ 2 & 0 & 1 \end{pmatrix}$$

Constant vector: $$\vec{b} = \begin{pmatrix} 6 \ 2 \ 8 \end{pmatrix}$$

The matrix equation is $A\vec{x} = \vec{b}$ where $\vec{x} = \begin{pmatrix} x \ y \ z \end{pmatrix}$.

Problem 5: For what value(s) of $k$ does the system $2x + 4y = 10$ and $x + ky = 5$ have infinitely many solutions?

Show Answer

For infinitely many solutions, the second equation must be a scalar multiple of the first (or vice versa).

Divide the first equation by 2: $$x + 2y = 5$$

Compare with the second equation: $x + ky = 5$

For these to be the same equation:

  • Coefficients of $x$: $1 = 1$ (Already matches)
  • Coefficients of $y$: $k = 2$
  • Constants: $5 = 5$ (Already matches)

Answer: $k = 2$

Verification: With $k = 2$, both equations become $x + 2y = 5$, representing the same line.

Problem 6: A movie theater sells adult tickets for $12 and child tickets for $8. One showing sold 150 tickets for a total of $1,560. Set up this problem as a matrix equation $A\vec{x} = \vec{b}$.

Show Answer

Let $a$ = number of adult tickets and $c$ = number of child tickets.

Equations:

  • Total tickets: $a + c = 150$
  • Total revenue: $12a + 8c = 1560$

Matrix form: $$\begin{pmatrix} 1 & 1 \ 12 & 8 \end{pmatrix}\begin{pmatrix} a \ c \end{pmatrix} = \begin{pmatrix} 150 \ 1560 \end{pmatrix}$$

Solving: From $a + c = 150$, we get $a = 150 - c$. Substituting: $$12(150 - c) + 8c = 1560$$ $$1800 - 12c + 8c = 1560$$ $$1800 - 4c = 1560$$ $$-4c = -240$$ $$c = 60$$

So $a = 150 - 60 = 90$.

Answer: 90 adult tickets and 60 child tickets.

Problem 7: Explain geometrically why a system of three linear equations in two variables ($x$ and $y$) usually has no solution.

Show Answer

Each linear equation in two variables represents a line in the $xy$-plane. For a system of three equations to have a solution, all three lines must pass through a common point.

While two lines generically intersect at exactly one point, there is no reason for a third line to pass through that same point. Three randomly chosen lines will almost never all meet at a single point.

Conclusion: A system of three linear equations in two variables is typically inconsistent (no solution), unless the lines happen to share a common intersection point or some of the equations are redundant (dependent).

This illustrates an important principle: having more equations than variables often leads to inconsistency, while having more variables than equations often leads to infinitely many solutions.

Summary

  • A linear equation is one where variables appear only to the first power, with no products of different variables. Examples: $3x + 2y = 7$ is linear; $xy = 4$ is not.

  • A system of linear equations consists of multiple equations in the same variables. We seek values that satisfy all equations simultaneously.

  • Geometric interpretation: In 2D, each equation is a line; solutions are intersection points. In 3D, each equation is a plane; solutions are where all planes meet.

  • Systems have exactly three possible outcomes: one unique solution, no solution (inconsistent), or infinitely many solutions (dependent).

  • A system is consistent if it has at least one solution, and inconsistent if it has none.

  • The matrix form $A\vec{x} = \vec{b}$ compactly represents a system: $A$ is the coefficient matrix, $\vec{x}$ is the variable vector, and $\vec{b}$ is the constant vector.

  • The augmented matrix $[A | \vec{b}]$ combines all information for solving the system.

  • Why matrices matter: They provide organization, enable systematic solution procedures, scale to large systems, reveal structural properties, and form the basis for computer algorithms.

  • The determinant of the coefficient matrix (for square systems) indicates whether a unique solution exists: nonzero determinant means exactly one solution.

  • Real-world applications include balancing chemical equations, network flow analysis, economic equilibrium, circuit analysis, and computer graphics.

  • This lesson establishes the foundation for Gaussian elimination, where you will learn the systematic matrix algorithm for solving any system of linear equations.