From a5ec51b20391ede76d230de632ab3334a993ab96 Mon Sep 17 00:00:00 2001
From: ben-jaynes <1btjaynes@gmail.com>
Date: Mon, 5 Oct 2026 09:39:55 -0700
Subject: [PATCH] vault backup: 2026-10-05 09:39:55

---
 College/AU 26/MATH 208 (Matrix Algebra)/Chapter 1.2 Notes.md | 5 +----
 1 file changed, 1 insertion(+), 4 deletions(-)

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 a34552a..1150d21 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	
@@ -40,7 +40,4 @@ h & 2h & \frac{2h}{3} \\ h & 12 & k
 h & 2h & \frac{2h}{3} \\ 0 & 12 - 2h & k - \frac{2h}{3}
 \end{array}\right] $$
 There are no solutions the system of equations when there is no pivot in a row and the value on the right is not equal to zero (if the value on the right is also equal to zero than that equation is essentially $0=0$ which is always true).
-This means that there are no solutions when $$12-2h = 0 \to h=6$$ and $$ \begin{gather} k-\frac{2h}{3} \ne 0 \\  k-4 \ne 0 \\[1.5ex]  k \ne 4 \end{gather}$$
-
-
-
+This means that there are no solutions when $$\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}$$