# Thu Feb. 23

In class we covered the definition and method for deriving identity and inverse matrices. We learned to solve systems of equations like

$\dpi{300}\inline 2x + 3y = 8$

$\dpi{300}\inline 4x - y = 10$

Using matrices

$\dpi{300}\inline \begin{bmatrix} 2 & 3 \\ 4 & -1 \end{bmatrix} \cdot \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} 8 \\ 10 \end{bmatrix}$

We would then solve the system. Multiplying on the inverse on both sides

$\dpi{300}\inline \begin{bmatrix} 2 & 3 \\ 4 & -1 \end{bmatrix}^{-1} \cdot \begin{bmatrix} 2 & 3 \\ 4 & -1 \end{bmatrix} \cdot \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} 2 & 3 \\ 4 & -1 \end{bmatrix}^{-1} \cdot \begin{bmatrix} 8 \\ 10 \end{bmatrix}$

We get

$\dpi{300}\inline \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \cdot \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} \frac{19}{7} \\ \frac{6}{7} \end{bmatrix}$

So, since $\dpi{300}\inline \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$ is the 2-by-2 identity matrix, we have

$\dpi{300}\inline \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} \frac{19}{7} \\ \frac{6}{7} \end{bmatrix}$

We also looked at solving systems of inequalities

Assignment:
6.4 on p. 331: 3, 6-10, 15, 17, 18
6.5 on p. 339: 5-9, 11-15