Blog: `x + y + 4z = 5, 2x + y - z = 9` Use matricies to solve the system of equations (if possible). Use Gauss-Jordan elimination.

Sunday, March 25, 2012

$\displaystyle{x}+{y}+{4}{z}={5},{2}{x}+{y}-{z}={9}$ Use matricies to solve the system of equations (if possible). Use Gauss-Jordan elimination.

The augmented matrix is $\displaystyle{\left[\matrix{{1}&{1}&{4}&{5}\\{2}&{1}&-{1}&{9}}\right]}$

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

$\displaystyle{\left[\matrix{{1}&{1}&{4}&{5}\\{0}&{1}&{9}&{1}}\right]}$

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

$\displaystyle{\left[\matrix{{1}&{0}&-{5}&{4}\\{0}&{1}&{9}&{1}}\right]}$

The corresponding system is $\displaystyle{x}-{5}{z}={4}$

$\displaystyle{y}+{9}{z}={1}$

Let $\displaystyle{k}$ be any real number and $\displaystyle{z}={k}$ . Now substitute $\displaystyle{z}$ value in above equations and find the values of $\displaystyle{x}$ and $\displaystyle{y}$ in terms of $\displaystyle{k}$ .

$\displaystyle{x}-{5}{z}={4}$

$\displaystyle{x}={4}+{5}{z}={4}+{5}{k}$

$\displaystyle{y}+{9}{z}={1}$

$\displaystyle{y}={1}-{9}{z}={1}-{9}{k}$

Hence the solution set is $\displaystyle{\left\lbrace{4}+{5}{k},{1}-{9}{k},{k}\right\rbrace}$

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Sunday, March 25, 2012

$\displaystyle{x}+{y}+{4}{z}={5},{2}{x}+{y}-{z}={9}$ Use matricies to solve the system of equations (if possible). Use Gauss-Jordan elimination.

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