Blog: `x + 2y - 3z = -28, 4y + 2z = 0, -x + y - z = -5` Use matricies to solve the system of equations. Use Gaussian elimination with...

Wednesday, June 12, 2013

$\displaystyle{x}+{2}{y}-{3}{z}=-{28},{4}{y}+{2}{z}={0},-{x}+{y}-{z}=-{5}$ Use matricies to solve the system of equations. Use Gaussian elimination with...

$\displaystyle{x}+{2}{y}-{3}{z}=-{28}$

$\displaystyle{4}{y}+{2}{z}={0}$

$\displaystyle-{x}+{y}-{z}=-{5}$

The equations in the matrix form can be written as,

$\displaystyle{\left[\matrix{{1}&{2}&-{3}&-{28}\\{0}&{4}&{2}&{0}\\-{1}&{1}&-{1}&-{5}}\right]}$

Add Row 1 and Row 3

$\displaystyle{\left[\matrix{{1}&{2}&-{3}&-{28}\\{0}&{4}&{2}&{0}\\{0}&{3}&-{4}&-{33}}\right]}$

Multiply Row 2 by 2 and Add it to Row 3

$\displaystyle{\left[\matrix{{1}&{2}&-{3}&-{28}\\{0}&{4}&{2}&{0}\\{0}&{11}&{0}&-{33}}\right]}$

Now the equations can be written as,

$\displaystyle{x}+{2}{y}-{3}{z}=-{28}$ ----- equation 1

$\displaystyle{4}{y}+{2}{z}={0}$ ------ equation 2

$\displaystyle{11}{y}=-{33}$ ----- equation 3

From equation 3,

$\displaystyle{y}=-\frac{{33}}{{11}}=-{3}$

Substitute back y in equation 2,

$\displaystyle{4}{\left(-{3}\right)}+{2}{z}={0}$

$\displaystyle-{12}+{2}{z}={0}$

$\displaystyle{2}{z}={12}$

$\displaystyle{z}=\frac{{12}}{{2}}={6}$

substitute back y and z in equation 1,

$\displaystyle{x}+{2}{\left(-{3}\right)}-{3}{\left({6}\right)}=-{28}$

$\displaystyle{x}-{6}-{18}=-{28}$

$\displaystyle{x}=-{28}+{18}+{6}$

$\displaystyle{x}=-{4}$

So the solutions are x=-4, y=-3, z=6

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