Respuesta :
Answer:
[tex]x\ =\ \dfrac{209}{30}[/tex]
[tex]y\ =\ \dfrac{29}{18}[/tex]
[tex]z\ =\ \dfrac{64}{15}[/tex]
Step-by-step explanation:
Given equations are
15x + 15y + 10z = 106
5x + 15y + 25z = 135
15x + 10y - 5z = 42
The augmented matrix by using above equations can be written as
[tex]\left[\begin{array}{ccc}15&15&10\ \ |106\\5&15&25\ \ |135\\15&10&-5|42\end{array}\right][/tex]
[tex]R_1\ \rightarrow\ \dfrac{R_1}{15}[/tex]
[tex]=\ \left[\begin{array}{ccc}1&1&\dfrac{10}{15}|\dfrac{106}{15}\\5&15&25|135\\15&15&-5|42\end{array}\right][/tex]
[tex]R_1\rightarrowR_2-5R1\ and\ R_3\rightarrow\ R_3-15R_1[/tex]
[tex]=\ \left[\begin{array}{ccc}1&1&\dfrac{10}{15}|\dfrac{106}{15}\\\\0&10&\dfrac{65}{3}|\dfrac{299}{3}\\\\0&0&-15|-64\end{array}\right][/tex]
[tex]R_2\rightarrow\ \dfrac{R_2}{10}[/tex]
[tex]=\ \left[\begin{array}{ccc}1&1&\dfrac{10}{15}|\dfrac{106}{15}\\\\0&1&\dfrac{65}{30}|\dfrac{299}{30}\\\\0&0&-15|-64\end{array}\right][/tex]
[tex]R_3\rightarrow\ \dfrac{R_3}{-15}[/tex]
[tex]=\ \left[\begin{array}{ccc}1&1&\dfrac{10}{15}|\dfrac{106}{15}\\\\0&1&\dfrac{65}{30}|\dfrac{299}{30}\\\\0&0&1|\dfrac{64}{15}\end{array}\right][/tex]
[tex]R_1\rightarrow\ R_1-R_2[/tex]
[tex]=\ \left[\begin{array}{ccc}1&0&\dfrac{-3}{2}|\dfrac{17}{30}\\\\0&1&\dfrac{65}{30}|\dfrac{299}{30}\\\\0&0&1|\dfrac{64}{15}\end{array}\right][/tex]
[tex]R_1\rightarrow\ R_1+\dfrac{3}{2}R_3[/tex]
[tex]=\ \left[\begin{array}{ccc}1&0&0|\dfrac{209}{30}\\\\0&1&\dfrac{65}{30}|\dfrac{299}{30}\\\\0&0&1|\dfrac{64}{15}\end{array}\right][/tex]
[tex]R_2\rightarrow\ R_2-\dfrac{65}{30}R_3[/tex]
[tex]=\ \left[\begin{array}{ccc}1&0&\0|\dfrac{209}{30}\\\\0&1&0|\dfrac{29}{18}\\\\0&0&1|\dfrac{64}{15}\end{array}\right][/tex]
Hence, we can write from augmented matrix,
[tex]x\ =\ \dfrac{209}{30}[/tex]
[tex]y\ =\ \dfrac{29}{18}[/tex]
[tex]z\ =\ \dfrac{64}{15}[/tex]
