Blog: `2x + 2y - z = 2, x - 3y + z = -28, -x + y = 14.` Use matricies to solve the system of equations (if possible). Use Gauss-Jordan elimination.

Monday, June 22, 2015

$\displaystyle{2}{x}+{2}{y}-{z}={2},{x}-{3}{y}+{z}=-{28},-{x}+{y}={14}.$ Use matricies to solve the system of equations (if possible). Use Gauss-Jordan elimination.

$\displaystyle{2}{x}+{2}{y}-{z}={2}$

$\displaystyle{x}-{3}{y}+{z}=-{28}$

$\displaystyle-{x}+{y}={14}$

The equations can be written as,

$\displaystyle{\left[\matrix{{2}&{2}&-{1}&{\mid}{2}\\{1}&-{3}&{1}&{\mid}-{28}\\-{1}&{1}&{0}&{\mid}{14}}\right]}$

R2 $\displaystyle\to$ R1+R2+R3

$\displaystyle{\left[\matrix{{2}&{2}&-{1}&{\mid}{2}\\{2}&{0}&{0}&{\mid}-{12}\\-{1}&{1}&{0}&{\mid}{14}}\right]}$

R1 $\displaystyle\to$ R1-R2

$\displaystyle{\left[\matrix{{0}&{2}&-{1}&{\mid}{14}\\{2}&{0}&{0}&{\mid}-{12}\\-{1}&{1}&{0}&{\mid}{14}}\right]}$

R3 $\displaystyle\to$ 2R3+R2

$\displaystyle{\left[\matrix{{0}&{2}&-{1}&{\mid}{14}\\{2}&{0}&{0}&{\mid}-{12}\\{0}&{2}&{0}&{\mid}{16}}\right]}$

R1 $\displaystyle\to$ R1-R3

$\displaystyle{\left[\matrix{{0}&{0}&-{1}&{\mid}-{2}\\{2}&{0}&{0}&{\mid}-{12}\\{0}&{2}&{0}&{\mid}{16}}\right]}$

Now,

$\displaystyle-{z}=-{2}$

$\displaystyle\therefore{z}={2}$

$\displaystyle{2}{x}=-{12}$

$\displaystyle{x}=-\frac{{12}}{{2}}=-{6}$

$\displaystyle{2}{y}={16}$

$\displaystyle{y}=\frac{{16}}{{2}}={8}$

So, the solutions are x=-6,y=8,z=2

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