Lecture 24
Auburn University
MATH 2660 - Spring 2026
March 16, 2026
$$ % Colors
% Coordinate vectors and matrices
% Common sets
% Abstract vector symbols
% Norms / absolute value
% Optional: dot product spacing (looks nicer in slides)
% Operators $$
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:
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 \[ \left\{\begin{pmatrix}1\\-1\\1\end{pmatrix}\right\}. \]
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