Blog: `-x + y - z = -14, 2x - y + z = 21, 3x + 2y + z = 19` Use matricies to solve the system of equations (if possible). Use Gauss-Jordan elimination.

Wednesday, March 4, 2009

$\displaystyle-{x}+{y}-{z}=-{14},{2}{x}-{y}+{z}={21},{3}{x}+{2}{y}+{z}={19}$ Use matricies to solve the system of equations (if possible). Use Gauss-Jordan elimination.

The augmented matrix is

$\displaystyle{\left[\matrix{-{1}&{1}&-{1}&-{14}\\{2}&-{1}&{1}&{21}\\{3}&{2}&{1}&{19}}\right]}$

On applying $\displaystyle{R}_{{1}}-\gt{R}_{{1}}+{R}_{{2}}$ we get (means changing 1st row as the sum of of first and second row)

$\displaystyle{\left[\matrix{{1}&{0}&{0}&{7}\\{2}&-{1}&{1}&{21}\\{3}&{2}&{1}&{19}}\right]}$

On applying $\displaystyle{R}_{{2}}-\gt{R}_{{2}}-{2}{R}_{{1}}$ we get

$\displaystyle{\left[\matrix{{1}&{0}&{0}&{7}\\{0}&-{1}&{1}&{7}\\{3}&{2}&{1}&{19}}\right]}$

On applying $\displaystyle{R}_{{2}}-\gt-{R}_{{2}}$ and $\displaystyle{R}_{{3}}-\gt{R}_{{3}}-{3}{R}_{{1}}$ we get

$\displaystyle{\left[\matrix{{1}&{0}&{0}&{7}\\{0}&{1}&-{1}&-{7}\\{0}&{2}&{1}&-{2}}\right]}$

On applying $\displaystyle{R}_{{3}}-\gt{R}_{{3}}-{2}{R}_{{2}}$ we get

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

On applying $\displaystyle{R}_{{3}}-\gt\frac{{{R}_{{3}}}}{{3}}$ we get

$\displaystyle{\left[\matrix{{1}&{0}&{0}&{7}\\{0}&{1}&-{1}&-{7}\\{0}&{0}&{1}&{4}}\right]}$

Hence the given system of equations is equivalent to the following system of equations

$\displaystyle{x}={7}$

$\displaystyle{y}-{z}=-{7}$ and

$\displaystyle{z}={4}$

$\displaystyle\therefore$ the solution set is

$\displaystyle{x}={7},{y}=-{3},{z}={4}$

No comments:

Post a Comment

Subscribe to: Post Comments (Atom)