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

@@ -36,8 +36,17 @@ $$\begin{bmatrix}
 3 & 6 & 2 \\ h & 12 & k
 \end{bmatrix} \sim \begin{bmatrix}
 h & 2h & \frac{2h}{3} \\ h & 12 & k
-\end{bmatrix} \sim \begin{bmatrix}
+\end{bmatrix} \sim \left[\begin{array}{cc|c}
 h & 2h & \frac{2h}{3} \\ 0 & 12 - 2h & k - \frac{2h}{3}
-\end{bmatrix} $$
+\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$$
+This means that there are no solutions when $$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]
+$$