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College/AU 26/MATH 208 (Matrix Algebra)/Chapter 1.2 Notes.md

@@ -8,9 +8,9 @@
 	- $\sim$ is used to indicate the transformation between equivalent linear systems
 - matrices can be used to simplify when working with systems of linear equations
 	- matrices with all constant terms of a linear system of equations are called a **augmented matrix**
-	- $$\begin{gathered} \text{Linear System} \\ a_{11}x_{1} + a_{12}x_{2} + a_{13}x_{3} = b_{1} \\ a_{21}x_{1} + a_{22}x_{2} + a_{23}x_{3} = b_{2} \\ a_{31}x_{1} + a_{32}x_{2} + a_{33}x_{3} = b_{3} \end{gathered} \quad \sim \quad \begin{gathered} \text{Augmented Matrix} \\ \begin{bmatrix}
+	- $$\begin{gathered} \text{Linear System} \\ a_{11}x_{1} + a_{12}x_{2} + a_{13}x_{3} = b_{1} \\ a_{21}x_{1} + a_{22}x_{2} + a_{23}x_{3} = b_{2} \\ a_{31}x_{1} + a_{32}x_{2} + a_{33}x_{3} = b_{3} \end{gathered} \quad \sim \quad \begin{gathered} \text{Augmented Matrix} \\ \left[\begin{array}{ccc|c}
 	a_{11} & a_{12} & a_{13} & b_{1} \\ a_{21} & a_{22} & a_{23} & b_{2} \\ a_{31} & a_{32} & a_{33} & b_{3} 
-	\end{bmatrix} \end{gathered}$$
+	\end{array}\right] \end{gathered}$$
 	- Same elementary operations can be used with augmented matrices, now it is with rows instead of equations.
 - Gaussian Elimination
 	- this is converting a matrix to **echelon form** (or row echelon form)
@@ -32,21 +32,15 @@
 
 
 To find when the system $$\begin{gathered} 3x_{1} + 6x_{2} = 2 \\ hx_{1} + 12x_{2} = k \end{gathered}$$ has no solutions
-$$\begin{bmatrix}
+$$\left[\begin{array}{cc|c}
 3 & 6 & 2 \\ h & 12 & k
-\end{bmatrix} \sim \begin{bmatrix}
+\end{array}\right] \sim \left[\begin{array}{cc|c}
 h & 2h & \frac{2h}{3} \\ h & 12 & k
-\end{bmatrix} \sim \left[\begin{array}{cc|c}
+\end{array}\right] \sim \left[\begin{array}{cc|c}
 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$$ and $$k-\frac{2h}{3} \ne 0 \to k-4 \ne 0 \to k \ne 4$$
-$$
-\left[
-\begin{array}{cccc|c}
-1 & 0 & 3 & -1 & 0 \\
-0 & 1 & 1 & -1 & 0 \\
-0 & 0 & 0 & 0 & 0 \\
-\end{array}
-\right]
-$$
+This means that there are no solutions when $$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}$$
+
+
+