Blog: `3x - 2y + z = -15, -x + y + 2z = -10, x - y - 4z = 14` Use matricies to solve the system of equations. Use Gaussian elimination with...

Friday, June 14, 2013

$\displaystyle{3}{x}-{2}{y}+{z}=-{15},-{x}+{y}+{2}{z}=-{10},{x}-{y}-{4}{z}={14}$ Use matricies to solve the system of equations. Use Gaussian elimination with...

$\displaystyle{3}{x}-{2}{y}+{z}=-{15}$

$\displaystyle-{x}+{y}+{2}{z}=-{10}$

$\displaystyle{x}-{y}-{4}{z}={14}$

$\displaystyle{A}={\left[\matrix{{3}&-{2}&{1}\\-{1}&{1}&{2}\\{1}&-{1}&-{4}}\right]}$

$\displaystyle{b}={\left[\matrix{-{15}\\-{10}\\{14}}\right]}$

$\displaystyle{\left[{A}{\mid}{b}\right]}={\left[\matrix{{3}&-{2}&{1}&-{15}\\-{1}&{1}&{2}&-{10}\\{1}&-{1}&-{4}&{14}}\right]}$

Multiply 2nd Row by 3 and add Row 1

$\displaystyle{\left[\matrix{{3}&-{2}&{1}&-{15}\\{0}&{1}&{7}&-{45}\\{1}&-{1}&-{4}&{14}}\right]}$

Multiply 3rd Row by 3 and subtract it from Row 1

$\displaystyle{\left[\matrix{{3}&-{2}&{1}&-{15}\\{0}&{1}&{7}&-{45}\\{0}&{1}&{13}&-{57}}\right]}$

Subtract Row 2 from Row 3

$\displaystyle{\left[\matrix{{3}&-{2}&{1}&-{15}\\{0}&{1}&{7}&-{45}\\{0}&{0}&{6}&-{12}}\right]}$

Now the equations can be written as,

$\displaystyle{3}{x}-{2}{y}+{z}=-{15}$ ----equation 1

$\displaystyle{y}+{7}{z}=-{45}$ ------ equation 2

$\displaystyle{6}{z}=-{12}$ ------ equation 3

From equation 3,

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

Now substitute back z in equation 2,

$\displaystyle{y}+{7}{\left(-{2}\right)}=-{45}$

$\displaystyle{y}-{14}=-{45}$

$\displaystyle{y}=-{45}+{14}$

$\displaystyle{y}=-{31}$

Substitute back the value of y and z in equation 1,

$\displaystyle{3}{x}-{2}{\left(-{31}\right)}+{\left(-{2}\right)}=-{15}$

$\displaystyle{3}{x}+{62}-{2}=-{15}$

$\displaystyle{3}{x}=-{15}-{60}$

$\displaystyle{3}{x}=-{75}$

$\displaystyle{x}=-\frac{{75}}{{3}}$

$\displaystyle{x}=-{25}$

So the solution is x=-25, y=-31 and z=-2

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