Notes: `2x + 6y = 16, 2x + 3y = 7` Use matricies to solve the system of equations. Use Gaussian elimination with back-substitution.

Friday, February 27, 2015

$\displaystyle{2}{x}+{6}{y}={16}$

$\displaystyle{2}{x}-{3}{y}={7}$

Write the system of equations as an augmented matrix.

$\displaystyle{\left[\matrix{{2}&{6}{\mid}{16}\\{2}&-{3}{\mid}{7}}\right]}$

Multiply the second line by -1.

$\displaystyle{\left[\matrix{{2}&{6}{\mid}{16}\\-{2}&{3}{\mid}-{7}}\right]}$

Eliminate the first column.

$\displaystyle{\left[\matrix{{2}&{6}{\mid}{16}\\{0}&{9}{\mid}{9}}\right]}$

$\displaystyle{9}{y}={9}$

$\displaystyle{y}={1}$

$\displaystyle{2}{x}+{6}{y}={16}$

$\displaystyle{2}{x}+{6}{\left({1}\right)}={16}$

$\displaystyle{2}{x}={10}$

$\displaystyle{x}={5}$

The solution set is (5, 1).

$\displaystyle{2}{x}+{6}{y}={16}$

$\displaystyle{2}{x}-{3}{y}={7}$

Write the system of equations as an augmented matrix.

$\displaystyle{\left[\matrix{{2}&{6}{\mid}{16}\\{2}&-{3}{\mid}{7}}\right]}$

Multiply the second line by -1.

$\displaystyle{\left[\matrix{{2}&{6}{\mid}{16}\\-{2}&{3}{\mid}-{7}}\right]}$

Eliminate the first column.

$\displaystyle{\left[\matrix{{2}&{6}{\mid}{16}\\{0}&{9}{\mid}{9}}\right]}$

$\displaystyle{9}{y}={9}$

$\displaystyle{y}={1}$

$\displaystyle{2}{x}+{6}{y}={16}$

$\displaystyle{2}{x}+{6}{\left({1}\right)}={16}$

$\displaystyle{2}{x}={10}$

$\displaystyle{x}={5}$

The solution set is (5, 1).

Notes