From 074301848fb9da1767a5dd93c1fee0239278f5ff Mon Sep 17 00:00:00 2001
From: ben-jaynes <1btjaynes@gmail.com>
Date: Mon, 5 Oct 2026 08:52:57 -0700
Subject: [PATCH] vault backup: 2026-10-05 08:52:57

---
 .../AU 26/MATH 208 (Matrix Algebra)/Chapter 1.2 Notes.md | 9 +++++++--
 1 file changed, 7 insertions(+), 2 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 66e212e..b11c3c0 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	
@@ -31,8 +31,13 @@
 	- There can also be non-trivial solutions 
 
 
+To find when the system $$\begin{gathered} 3x_{1} + 6x_{2} = 2 \\ hx_{1} + 12x_{2} = k \end{gathered}$$ has no solutions
 $$\begin{bmatrix}
 3 & 6 & 2 \\ h & 12 & k
 \end{bmatrix} \sim \begin{bmatrix}
-
-\end{bmatrix}$$
\ No newline at end of file
+h & 2h & \frac{2h}{3} \\ h & 12 & k
+\end{bmatrix} \sim \begin{bmatrix}
+h & 2h & \frac{2h}{3} \\ 0 & 12 - 2h & k - \frac{2h}{3}
+\end{bmatrix} $$
+There are no solutions the system of equations when there is no pivot in a row and the value on the right is not equal to zero (if the value on the right is also equal to zero than that equation is essentially $0=0$ which is always true).
+This means that there are no solutions when $$12-2h = 0 \to h=6$$ and $$k-\frac{2h}{3} \ne 0$$
\ No newline at end of file