Post Break Review

Lecture 24

Author
Affiliation

Minjae Park

Auburn University
MATH 2660 - Spring 2026

Published

March 16, 2026

Welcome back 🏖️

Let’s catch up

  • Share with your classmates what you did over spring break.
  • Did you discover any hidden gems around Auburn?

Recap

Overview

Determinants

  • For an \(n \times n\) matrix \(A\):
    • \(\det(A)\) can be computed by cofactor expansion along any row or column. In practice, choose one with many zeros to simplify the computation.
    • \(|\det(A)|\) equals the hypervolume of the parallelepiped formed by the \(n\) column vectors.
    • The sign of \(\det(A)\) records the orientation of the column vectors.
    • If \(\det(A)=0\), the columns are linearly dependent.
    • If \(\det(A)\neq 0\), the columns are linearly independent and \(A\) is invertible.

Homogeneous linear equations

  • Let \(A = [\vec{u}_1 \cdots \vec{u}_n]\) be an \(n \times n\) matrix with columns \(\vec{u}_i \in \mathbb{R}^n\).
  • The system \(A\vec{x} = \vec{0}\) has a unique solution \(\vec{x}=\vec{0}\) if:
    • the vectors \(\vec{u}_i\) are linearly independent
    • \(\mathrm{RREF}(A) = I_n\)
    • \(\det(A) \ne 0\)
  • The system \(A\vec{x} = \vec{0}\) has infinitely many solutions if:
    • the vectors \(\vec{u}_i\) are linearly dependent
    • \(\mathrm{RREF}(A) \ne I_n\)
    • \(\det(A) = 0\)

Example

  • Let \[ A=\begin{pmatrix} 1 & 2 & 1\\ 2 & 4 & 2\\ 1 & 1 & 0 \end{pmatrix}. \] Determine how many solutions there are for \(A\vec{x}=\vec{0}\) by
    • computing \(\det(A)\)
    • using the Gauss-Jordan elimination.
  • Using elementary matrices used in the previous part, recover \(\det(A)\).
  • Find a basis of the nullspace \(N(A)\) by solving the equation.

Solution

Let \(\vec{x}=\begin{pmatrix}x_1\\x_2\\x_3\end{pmatrix}\).

  • Using the determinant: Expand along the first row: \[ \det(A)= 1\begin{vmatrix}4&2\\1&0\end{vmatrix} -2\begin{vmatrix}2&2\\1&0\end{vmatrix} +1\begin{vmatrix}2&4\\1&1\end{vmatrix} =1(0-2)-2(0-2)+(2-4)=0. \] Since \(\det(A)=0\), \(A\) is singular, so the homogeneous system \(A\vec{x}=\vec{0}\) has infinitely many solutions.

  • Using Gauss-Jordan: \[ \begin{pmatrix} 1&2&1\\ 2&4&2\\ 1&1&0 \end{pmatrix} \xrightarrow{R_2\to R_2-2R_1} \begin{pmatrix} 1&2&1\\ 0&0&0\\ 1&1&0 \end{pmatrix} \xrightarrow{R_3\to R_3-R_1} \begin{pmatrix} 1&2&1\\ 0&0&0\\ 0&-1&-1 \end{pmatrix} \] \[ \xrightarrow{R_3\to -R_3} \begin{pmatrix} 1&2&1\\ 0&0&0\\ 0&1&1 \end{pmatrix} \xrightarrow{R_1\to R_1-2R_3} \begin{pmatrix} 1&0&-1\\ 0&0&0\\ 0&1&1 \end{pmatrix} \xrightarrow{R_2\leftrightarrow R_3} \begin{pmatrix} 1&0&-1\\ 0&1&1\\ 0&0&0 \end{pmatrix}. \] There is one free variable, so again \(A\vec{x}=\vec{0}\) has infinitely many solutions.

  • Recovering \(\det(A)\) from elementary matrices: The row operations above were:

    • \(R_2\to R_2-2R_1\) (determinant factor \(1\))
    • \(R_3\to R_3-R_1\) (determinant factor \(1\))
    • \(R_3\to -R_3\) (determinant factor \(-1\))
    • \(R_1\to R_1-2R_3\) (determinant factor \(1\))
    • \(R_2\leftrightarrow R_3\) (determinant factor \(-1\))

    So if \(E_5E_4E_3E_2E_1A=R\), then \[ \det(R)=\det(E_5)\det(E_4)\det(E_3)\det(E_2)\det(E_1)\det(A). \] Here \[ \det(E_5)\det(E_4)\det(E_3)\det(E_2)\det(E_1)=(-1)(1)(-1)(1)(1)=1. \] Hence \(\det(R)=\det(A)\). But \(R\) has a zero row, so \(\det(R)=0\). Therefore \(\det(A)=0\).

  • Finding a basis of \(\mathop{\mathrm{null}}(A)\): From the RREF, \[ \begin{aligned} x_1-x_3&=0,\\ x_2+x_3&=0. \end{aligned} \] Let \(x_3=t\). Then \[ x_1=t,\qquad x_2=-t,\qquad x_3=t. \] Therefore \[ \vec{x}=t\begin{pmatrix}1\\-1\\1\end{pmatrix}. \] So \[ N(A)=\mathop{\mathrm{span}}(\left\{\begin{pmatrix}1\\-1\\1\end{pmatrix}\right\}), \] and a basis is \[ \left\{\begin{pmatrix}1\\-1\\1\end{pmatrix}\right\}. \]

This week’s plan

  • Next class we will discuss some applications of determinants (Cramer’s rule, cross product, etc). Please review the cofactor formula for determinants (especially for \(3\times 3\) matrices), their geometric meaning, and how to compute determinants using elementary matrices from Gauss–Jordan elimination.
  • This week’s homework will cover determinants and their applications. It will be due by Friday and may overlap with Quiz 2 if you want to start early.
  • On Friday we will begin a new topic: eigenvalues and eigenvectors. Please review the relationship between solutions of linear systems and determinants.