Solving Linear Equations

Lecture 6

Author
Affiliation

Minjae Park

Auburn University
MATH 2660 - Spring 2026

Published

January 21, 2026

Recap

Linear transformations

  • A linear transformation \(A:\mathbb{R}^k \to \mathbb{R}^n\) maps a vector \(\vec{x}\in\mathbb{R}^k\) to \(A\vec{x}\in\mathbb{R}^n\) via multiplication by an \(n\times k\) matrix \(A\).
  • The transformation is completely determined by its action on the coordinate vectors \(\vec{e}_i\in\mathbb{R}^k\), sending each to a vector \(\vec{v}_i\in\mathbb{R}^n\).
  • Each vector \(\vec{v}_i=A\vec{e}_i\) appears as the \(i\)-th column of \(A\). Equivalently, knowing \(A\vec{e}_i\) for all \(i\) determines the matrix.
  • In low dimensions—especially in the plane—these effects can often be visualized geometrically.

Example: \(A(x,y)=(3x+y,\,y)\)

  • What are \(A\vec{e}_1\) and \(A\vec{e}_2\)? They are exactly the column vectors of \(A\).

Solving linear equations

Motivation

  • A function \(f(x)=y\) describes the relationship between an input (\(x\)) and an output (\(y\)).
  • In practice, we often want to determine the input given an observed or measured output.
  • Example: “I spent $50 this month at coffee shops. How much coffee did I drink and how many bagels did I have? How should I adjust my habits to spend only $30 a month?”

Linear equations

  • Linear maps (transformations) are fundamental because many functions can be well approximated by linear ones, and linear problems are much easier to solve.
  • Fact: if a function is differentiable, then it can be approximated by a linear function; this also holds in higher dimensions (see multivariable calculus).
  • Therefore, solving equations of the form \[A\vec{x}=\vec{b}\] for a given \(\vec{b}\) is very important.
  • Such an equation is called a linear equation.

Notational convention

  • Let \(\vec{x}=\langle x_1,\dots,x_m \rangle\) be the vector of \(m\) unknown variables.
  • Let \[A=\begin{bmatrix} a_{11}&\cdots&a_{1m}\\ \vdots&\ddots&\vdots\\ a_{n1}&\cdots&a_{nm} \end{bmatrix}\] be an \(n\times m\) coefficient matrix.
  • Let \(\vec{b}=\langle b_1,\dots,b_n \rangle\) be a given vector.
  • Our goal is to solve \(A\vec{x}=\vec{b}\) for the unknowns \(x_1,\dots,x_m\).

System of linear equations

  • Writing the equation coordinate by coordinate gives the system of linear equations \[\left\{\begin{aligned} a_{11}x_1+\cdots+a_{1m}x_m &= b_1\\ &\vdots\\ a_{n1}x_1+\cdots+a_{nm}x_m &= b_n. \end{aligned}\right.\]
  • You are probably already familiar with solving equations of this form!

Gaussian elimination

  • Gaussian elimination is a systematic algorithm for solving systems of linear equations.
  • The idea is to eliminate variables step by step using existing equations.
  • In the end, the system may have a unique solution, infinitely many solutions, or no solution.

Elementary row operations

  • Interchange two equations.
    Notation: \(R_i\leftrightarrow R_j\)
  • Multiply an equation by a nonzero constant.
    Notation: \(R_i\leftarrow cR_i\), \(c\neq0\)
  • Add a multiple of one equation to another.
    Notation: \(R_i\leftarrow R_i+cR_j\)

Steps for Gaussian elimination

  • Ensure the first equation has a nonzero coefficient of \(x_1\) (swap equations if necessary).
  • Scale this equation to obtain a convenient leading coefficient.
  • Subtract suitable multiples of this equation from the others to eliminate \(x_1\).
  • Repeat for the remaining variables \(x_2,\dots,x_m\).

Augmented matrix

  • To keep track of these steps efficiently, we record only the coefficients and constants.
  • The \(n\times(m+1)\) matrix \[[A\mid\vec{b}] = \left[\begin{array}{ccc|c} a_{11}&\cdots&a_{1m}&b_1\\ \vdots&\ddots&\vdots&\vdots\\ a_{n1}&\cdots&a_{nm}&b_n \end{array}\right]\] is called the augmented matrix.

Example

  • Solve the system of linear equations \[\left\{\begin{aligned} x+y+z &= 6\\ 2x-y+z &= 3\\ -x+2y+z &= 6 \end{aligned}\right.\]

Gaussian elimination

  • Eliminate \(x\) from the second and third rows: \[\left[\begin{array}{ccc|c} 1&1&1&6\\ 2&-1&1&3\\ -1&2&1&6 \end{array}\right]\\ \xrightarrow{\substack{R_2\leftarrow R_2-2R_1\\[2pt] R_3\leftarrow R_3+R_1}} \left[\begin{array}{ccc|c} 1&1&1&6\\ 0&-3&-1&-9\\ 0&3&2&12 \end{array}\right].\]

Continue elimination

  • Eliminate \(y\) from the third row: \[\xrightarrow{R_3\leftarrow R_3+R_2} \left[\begin{array}{ccc|c} 1&1&1&6\\ 0&-3&-1&-9\\ 0&0&1&3 \end{array}\right].\]

Back substitution

  • Solve from the bottom up: \[\begin{align*} &z=3,\\ -3y-z=-9\ \Rightarrow\ &y=2,\\ x+y+z=6\ \Rightarrow\ &x=1.\end{align*}\]
  • This is called the back substitution.

Gauss–Jordan elimination

  • By incorporating back substitution into elimination, we can simplify the augmented matrix further.
  • Eliminate variables above each leading entry (the first nonzero entry of each row).
  • Scale each row so the leading coefficient is \(1\).
  • This procedure is called Gauss–Jordan elimination.

Reduced row-echelon form

  • The final matrix obtained after Gauss–Jordan elimination is in reduced row-echelon form (RREF).
  • Each leading entry is \(1\) and is the only nonzero entry in its column.
  • Any rows consisting entirely of zeros for coefficients (if present) appear at the bottom.

RREF (Case 1: unique solution)

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

  • Every row has a leading entry (also called a pivot).
  • Each variable is uniquely determined by the augmented values.
  • The number of equations matches the number of variables, so the system is neither overdetermined nor underdetermined.

RREF (Case 2: infinitely many solutions)

\[ \left[\begin{array}{cccc|c} 1 & 0 & 3 & 0 & 2\\ 0 & 0 & 0 & 1 & -1\\ 0 & 0 & 0 & 0 & 0 \end{array}\right] \quad\text{or}\quad \left[\begin{array}{cccc|c} 1 & 0 & -1 & 2 & 4\\ 0 & 1 & 3 & -1 & -2 \end{array}\right] \]

  • The number of leading entries (nonzero rows) is less than the number of variables.
  • Any variable without a leading entry is a free variable and may take arbitrary values.
  • All other variables are uniquely determined in terms of the free variables.
  • Free variables are also called parameters.

RREF (Case 3: possibly no solution)

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

  • Some bottom rows have no leading entry in their coefficient part.
  • If \(c\neq0\), such a row represents the equation \(0=c\), so the system is inconsistent and has no solution.
  • If \(c=0\), the row represents the identity \(0=0\) and can be ignored.
  • After removing all zero rows, the system falls into either the unique-solution case or the infinitely-many-solutions case.

Summary

  • A linear equation has the form \(A\vec{x}=\vec{b}\), where \(A\) is an \(n\times m\) matrix, \(\vec{x}\in\mathbb{R}^m\) is the vector of unknowns, and \(\vec{b}\in\mathbb{R}^n\) is the given vector.
  • The terms linear equation and system of linear equations are often used interchangeably.
  • The corresponding augmented matrix is written as \([A\mid\vec{b}]\).
  • To solve a system of linear equations, we apply Gauss–Jordan elimination to the augmented matrix using elementary row operations to obtain the reduced row-echelon form (RREF).
  • Depending on the final RREF, the system may have a unique solution, infinitely many solutions, or no solution.

Comments

  • Working through examples on your own is one of the most effective ways to understand these concepts. For practice, try HW 2 or study the worked examples in Larson’s book. (You may also ask GPT to generate examples step by step and focus on understanding each step.)
  • Depending on your background in algebra, this material may take time to fully sink in, so do not hesitate to try multiple examples.
  • Accurately computing the RREF by hand is not the main goal—computers can always perform these calculations for you.
  • What matters is understanding that any linear system can be reduced to RREF, and that the RREF determines whether solutions exist. If a computer reports that there is no solution, you should understand why: the system was reduced to an inconsistent row.