Respuesta :
Answer:
The system has unique solution.
[tex]\therefore x=-\frac57[/tex], [tex]y=\frac57[/tex] and [tex]z=-\frac{30}{7}[/tex]
Step-by-step explanation:
Given system of equation is
2x+3y+z=5
2x+9y+0.z=5
12x+y+3z=5
The augmented matrix is
[tex]\left[\begin{array}{ccc}2&3&1\\2&9&0\\12&1&3\end{array}\right|\left\begin{array}{c}5\\5\\5\end{array}\right][/tex]
Subtract row 1 from row 2 ([tex]R_2=R_2-R_1[/tex])
[tex]\left[\begin{array}{ccc}2&3&1\\0&6&-1\\12&1&3\end{array}\right|\left\begin{array}{c}5\\0\\5\end{array}\right][/tex]
subtract row 1 multiplied by 6 from row 3 ([tex]R_3=R_3-6R_1[/tex])
[tex]\left[\begin{array}{ccc}2&3&1\\0&6&-1\\0&-17&-3\end{array}\right|\left\begin{array}{c}5\\0\\-25\end{array}\right][/tex]
Divide row 1 by 2 ([tex]R_1=\frac{R_1}2[/tex])
[tex]\left[\begin{array}{ccc}1&\frac32&\frac12\\0&6&-1\\0&-17&-3\end{array}\right|\left\begin{array}{c}\frac52\\0\\-25\end{array}\right][/tex]
Divide row 2 by 6 [tex](R_2=\frac{R_2}6)[/tex]
[tex]\left[\begin{array}{ccc}1&\frac32&\frac12\\ \\0&1&\frac{-1}{6}\\ \\0&-17&-3\end{array}\right|\left\begin{array}{c}\frac52\\ \\0\\ \\-25\end{array}\right][/tex]
Subtract row 2 multiplied by [tex]\frac32[/tex] from row 1 [tex](R_1=R_1-\frac32 R_2)[/tex]
[tex]\left[\begin{array}{ccc}1&0&\frac34\\ \\0&1&\frac{-1}{6}\\ \\0&-17&-3\end{array}\right|\left\begin{array}{c}\frac52\\ \\0\\ \\-25\end{array}\right][/tex]
Add row 2 multiplied by 17 to row 3 [tex](R_3=R_3+17R_2)[/tex]
[tex]\left[\begin{array}{ccc}1&0&\frac34\\ \\0&1&-\frac{1}{6}\\ \\0&0&-\frac{35}{6}\end{array}\right|\left\begin{array}{c}\frac52\\ \\0\\ \\-25\end{array}\right][/tex]
Multiply row 3 by [tex]-\frac{6}{35}[/tex] [tex](R_3=-\frac6{35} R_3)[/tex]
[tex]\left[\begin{array}{ccc}1&0&\frac34\\ \\0&1&-\frac{1}{6}\\ \\0&0&1\end{array}\right|\left\begin{array}{c}\frac52\\ \\0\\ \\\frac{30}7\end{array}\right][/tex]
Subtract row 3 multiplied by [tex]\frac34[/tex] from row 1 [tex](\ R_1=R_1-\frac34R_3)[/tex]
[tex]\left[\begin{array}{ccc}1&0&0\\ \\0&1&-\frac{1}{6}\\ \\0&0&1\end{array}\right|\left\begin{array}{c}-\frac57\\ \\0\\ \\\frac{30}7\end{array}\right][/tex]
Add row 3 multiplied by [tex]\frac16[/tex] to row 2 [tex](R_2=R_2+\frac16R_3)[/tex]
[tex]\left[\begin{array}{ccc}1&0&0\\ \\0&1&0\\ \\0&0&1\end{array}\right|\left\begin{array}{c}-\frac57\\ \\\frac57\\ \\\frac{30}7\end{array}\right][/tex]
[tex]\therefore x=-\frac57[/tex], [tex]y=\frac57[/tex] and [tex]z=-\frac{30}{7}[/tex]
Unique solution: The system has one specific solution.
The system has unique solution.
