From 9d249419cf2c25a661d8647111fe836e838587ec Mon Sep 17 00:00:00 2001
From: ben-jaynes <1btjaynes@gmail.com>
Date: Wed, 7 Oct 2026 08:52:59 -0700
Subject: [PATCH] vault backup: 2026-10-07 08:52:59

---
 College/AU 26/MATH 208 (Matrix Algebra)/Chapter 1.2 Notes.md | 2 ++
 1 file changed, 2 insertions(+)

diff --git a/College/AU 26/MATH 208 (Matrix Algebra)/Chapter 1.2 Notes.md b/College/AU 26/MATH 208 (Matrix Algebra)/Chapter 1.2 Notes.md
index 1150d21..574e12d 100644
--- a/College/AU 26/MATH 208 (Matrix Algebra)/Chapter 1.2 Notes.md	
+++ b/College/AU 26/MATH 208 (Matrix Algebra)/Chapter 1.2 Notes.md	
@@ -41,3 +41,5 @@ h & 2h & \frac{2h}{3} \\ 0 & 12 - 2h & k - \frac{2h}{3}
 \end{array}\right] $$
 There are no solutions the system of equations when there is no pivot in a row and the value on the right is not equal to zero (if the value on the right is also equal to zero than that equation is essentially $0=0$ which is always true).
 This means that there are no solutions when $$\begin{gather} 12-2h = 0 \\[1.5ex] h=6 \end{gather}$$ and $$ \begin{gather} k-\frac{2h}{3} \ne 0 \\[1.5ex]  k-4 \ne 0 \\[1.5ex]  k \ne 4 \end{gather}$$
+
+