Notes: `2x - y + 3z = 24, 2y - z = 14, 7x - 5y = 6.` Use matricies to solve the system of equations (if possible). Use Gauss-Jordan elimination.

Monday, December 8, 2014

$\displaystyle{2}{x}-{y}+{3}{z}={24},{2}{y}-{z}={14},{7}{x}-{5}{y}={6}.$ Use matricies to solve the system of equations (if possible). Use Gauss-Jordan elimination.

$\displaystyle{2}{x}-{y}+{3}{z}={24}$

$\displaystyle{2}{y}-{z}={14}$

$\displaystyle{7}{x}-{5}{y}={6}$

Write the equations as,

$\displaystyle{\left[\matrix{{2}&-{1}&{3}&{\mid}{24}\\{0}&{2}&-{1}&{\mid}{14}\\{7}&-{5}&{0}&{\mid}{6}}\right]}$

Make the pivot in the first column by dividing the First row by 2,

$\displaystyle{\left[\matrix{{1}&-\frac{{1}}{{2}}&\frac{{3}}{{2}}&{\mid}{12}\\{0}&{2}&-{1}&{\mid}{14}\\{7}&-{5}&{0}&{\mid}{6}}\right]}$

Multiply the first row by 7,

$\displaystyle{\left[\matrix{{7}&-\frac{{7}}{{2}}&\frac{{21}}{{2}}&{\mid}{84}\\{0}&{2}&-{1}&{\mid}{14}\\{7}&-{5}&{0}&{\mid}{6}}\right]}$

Subtract the 1st row from the 3rd row and restore it,

$\displaystyle{\left[\matrix{{1}&-\frac{{1}}{{2}}&\frac{{3}}{{2}}&{\mid}{12}\\{0}&{2}&-{1}&{\mid}{14}\\{0}&-\frac{{3}}{{2}}&-\frac{{21}}{{2}}&{\mid}-{78}}\right]}$

Make the pivot in the second column by dividing the 2nd row by 2,

$\displaystyle{\left[\matrix{{1}&-\frac{{1}}{{2}}&\frac{{3}}{{2}}&{\mid}{12}\\{0}&{1}&-\frac{{1}}{{2}}&{\mid}{7}\\{0}&-\frac{{3}}{{2}}&-\frac{{21}}{{2}}&{\mid}-{78}}\right]}$

Multiply the 2nd row by -1/2,

$\displaystyle{\left[\matrix{{1}&-\frac{{1}}{{2}}&\frac{{3}}{{2}}&{\mid}{12}\\{0}&-\frac{{1}}{{2}}&\frac{{1}}{{4}}&{\mid}-\frac{{7}}{{2}}\\{0}&-\frac{{3}}{{2}}&-\frac{{21}}{{2}}&{\mid}-{78}}\right]}$

subtract the 2nd row from the 1st row and restore it,

...

$\displaystyle{2}{x}-{y}+{3}{z}={24}$

$\displaystyle{2}{y}-{z}={14}$

$\displaystyle{7}{x}-{5}{y}={6}$

Write the equations as,

$\displaystyle{\left[\matrix{{2}&-{1}&{3}&{\mid}{24}\\{0}&{2}&-{1}&{\mid}{14}\\{7}&-{5}&{0}&{\mid}{6}}\right]}$

Make the pivot in the first column by dividing the First row by 2,

$\displaystyle{\left[\matrix{{1}&-\frac{{1}}{{2}}&\frac{{3}}{{2}}&{\mid}{12}\\{0}&{2}&-{1}&{\mid}{14}\\{7}&-{5}&{0}&{\mid}{6}}\right]}$

Multiply the first row by 7,

$\displaystyle{\left[\matrix{{7}&-\frac{{7}}{{2}}&\frac{{21}}{{2}}&{\mid}{84}\\{0}&{2}&-{1}&{\mid}{14}\\{7}&-{5}&{0}&{\mid}{6}}\right]}$

Subtract the 1st row from the 3rd row and restore it,

$\displaystyle{\left[\matrix{{1}&-\frac{{1}}{{2}}&\frac{{3}}{{2}}&{\mid}{12}\\{0}&{2}&-{1}&{\mid}{14}\\{0}&-\frac{{3}}{{2}}&-\frac{{21}}{{2}}&{\mid}-{78}}\right]}$

Make the pivot in the second column by dividing the 2nd row by 2,

$\displaystyle{\left[\matrix{{1}&-\frac{{1}}{{2}}&\frac{{3}}{{2}}&{\mid}{12}\\{0}&{1}&-\frac{{1}}{{2}}&{\mid}{7}\\{0}&-\frac{{3}}{{2}}&-\frac{{21}}{{2}}&{\mid}-{78}}\right]}$

Multiply the 2nd row by -1/2,

$\displaystyle{\left[\matrix{{1}&-\frac{{1}}{{2}}&\frac{{3}}{{2}}&{\mid}{12}\\{0}&-\frac{{1}}{{2}}&\frac{{1}}{{4}}&{\mid}-\frac{{7}}{{2}}\\{0}&-\frac{{3}}{{2}}&-\frac{{21}}{{2}}&{\mid}-{78}}\right]}$

subtract the 2nd row from the 1st row and restore it,

$\displaystyle{\left[\matrix{{1}&{0}&\frac{{5}}{{4}}&{\mid}\frac{{31}}{{2}}\\{0}&{1}&-\frac{{1}}{{2}}&{\mid}{7}\\{0}&-\frac{{3}}{{2}}&-\frac{{21}}{{2}}&{\mid}-{78}}\right]}$

Multiply the 2nd row by -3/2

$\displaystyle{\left[\matrix{{1}&{0}&\frac{{5}}{{4}}&{\mid}\frac{{31}}{{2}}\\{0}&-\frac{{3}}{{2}}&\frac{{3}}{{4}}&{\mid}-\frac{{21}}{{2}}\\{0}&-\frac{{3}}{{2}}&-\frac{{21}}{{2}}&{\mid}-{78}}\right]}$

subtract the 2nd row from the 3rd row and restore it,

$\displaystyle{\left[\matrix{{1}&{0}&\frac{{5}}{{4}}&{\mid}\frac{{31}}{{2}}\\{0}&{1}&-\frac{{1}}{{2}}&{\mid}{7}\\{0}&{0}&-\frac{{45}}{{4}}&{\mid}-\frac{{135}}{{2}}}\right]}$

Make the pivot in the 3rd column by dividing the 3rd row by -45/4,

$\displaystyle{\left[\matrix{{1}&{0}&\frac{{5}}{{4}}&{\mid}\frac{{31}}{{2}}\\{0}&{1}&-\frac{{1}}{{2}}&{\mid}{7}\\{0}&{0}&{1}&{\mid}{6}}\right]}$

Multiply the 3rd row by 5/4,

$\displaystyle{\left[\matrix{{1}&{0}&\frac{{5}}{{4}}&{\mid}\frac{{31}}{{2}}\\{0}&{1}&-\frac{{1}}{{2}}&{\mid}{7}\\{0}&{0}&\frac{{5}}{{4}}&{\mid}\frac{{15}}{{2}}}\right]}$

subtract the 3rd row from the first row and restore it,

$\displaystyle{\left[\matrix{{1}&{0}&{0}&{\mid}{8}\\{0}&{1}&-\frac{{1}}{{2}}&{\mid}{7}\\{0}&{0}&{1}&{\mid}{6}}\right]}$

Multiply the 3rd row by -1/2,

$\displaystyle{\left[\matrix{{1}&{0}&{0}&{\mid}{8}\\{0}&{1}&-\frac{{1}}{{2}}&{\mid}{7}\\{0}&{0}&-\frac{{1}}{{2}}&{\mid}-{3}}\right]}$

subtract the 3rd row from the 2nd row and restore it,

$\displaystyle{\left[\matrix{{1}&{0}&{0}&{\mid}{8}\\{0}&{1}&{0}&{\mid}{10}\\{0}&{0}&{1}&{\mid}{6}}\right]}$

So the solutions are x=8,y=10,z=6

Notes

Monday, December 8, 2014

$\displaystyle{2}{x}-{y}+{3}{z}={24},{2}{y}-{z}={14},{7}{x}-{5}{y}={6}.$ Use matricies to solve the system of equations (if possible). Use Gauss-Jordan elimination.

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Is Charlotte Bronte's Jane Eyre a feminist novel?

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Monday, December 8, 2014

Use matricies to solve the system of equations (if possible). Use Gauss-Jordan elimination.

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Is Charlotte Bronte&#39;s Jane Eyre a feminist novel?

$\displaystyle{2}{x}-{y}+{3}{z}={24},{2}{y}-{z}={14},{7}{x}-{5}{y}={6}.$ Use matricies to solve the system of equations (if possible). Use Gauss-Jordan elimination.

Is Charlotte Bronte's Jane Eyre a feminist novel?