Question

Use the row operations tool to solve the following system of equations, obtaining the solutions in fraction form.

Answer 1

Answer:

$x=-\frac{7}{26}, y=\frac{37}{26}, z=-\frac{21}{13}$

Step-by-step explanation:

The matrix representation of the system of linear equations is:

$\left(\begin{array}{ccccc}15&13&4&\vdots&8\\11&13&9&\vdots&1\\3&5&7&\vdots&-5\\8&8&2&\vdots&6\\7&5&2&\vdots&2\end{array}\right)$

First apply the following row operations:

$\begin{array}{c}R_{2}\to R_{2}+(-\frac{11}{15})R_{1}\\R_{3}\to R_{3}+(-\frac{1}{5})R_{1}\\R_{4}\to R_{4}+(-\frac{8}{15})R_{1}\\R_{5}\to R_{5}+(-\frac{7}{15})R_{1}\end{array}$

The resulting matrix is:

$\left(\begin{array}{ccccc}15&13&4&\vdots&8\\0&\frac{52}{15}&\frac{91}{15}&\vdots&-\frac{73}{15}\\0&\frac{12}{5}&\frac{31}{5}&\vdots&-\frac{33}{5}\\0&\frac{16}{15}&-\frac{2}{15}&\vdots&\frac{26}{15}\\0&-\frac{16}{15}&\frac{2}{15}&\vdots&-\frac{26}{15}\end{array}\right)$

Then apply  the row operations:

$\begin{array}{c}R_{3}\to R_{3}+(-\frac{12}{5}\cdot \frac{15}{52})R_{2}\\R_{4}\to R_{4}+(-\frac{-16}{15}\cdot \frac{15}{52})R_{2}\\R_{5}\to R_{5}+(\frac{16}{15}\cdot \frac{15}{52})R_{2}\end{array}$

The resulting matrix is:

$\left(\begin{array}{ccccc}15&13&4&\vdots&8\\0&\frac{52}{15}&\frac{91}{15}&\vdots&-\frac{73}{15}\\0&0&2&\vdots&-\frac{42}{13}\\0&0&-2&\vdots&\frac{42}{13}\\0&0&2&\vdots&-\frac{42}{13}\end{array}\right)$

Now apply the row operations:

$\begin{array}{c} R_{4}\to R_{4}+R_{3}\\R_{5}\to R_{5}+(-1)R_{3}\end{array}$

The resulting matrix is:

$\left(\begin{array}{ccccc}15&13&4&\vdots&8\\0&\frac{52}{15}&\frac{91}{15}&\vdots&-\frac{73}{15}\\0&0&2&\vdots&-\frac{42}{13}\\0&0&0&\vdots&0\\0&0&0&\vdots&0\end{array} \right)$

The equivalent linear system associated to this matrix is

$\begin{cases}15x+13y+4z=8\\\frac{52}{15}y+\frac{91}{15}z=-\frac{73}{15}\\2z=-\frac{42}{13}\end{cases}$

To Solve this last system is very simple by substitution. The solutions are:

$x=-\frac{7}{26}, y=\frac{37}{26}, z=-\frac{21}{13}$