Question

For problem (1) and (2), use Gauss-Jordan reduction to transform the augmented matrix of each system to RREF. Use it to discuss the solutions of the system (i.e., no solutions, a unique solution,or infinitely many solutions).(1)2x+ 3y+z= 2x+ 9y= 12x+y+ 3z= 5

Answer 1

Answer:

The system has unique solution.

$\therefore x=-\frac57$, $y=\frac57$ and  $z=-\frac{30}{7}$

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

$\left[\begin{array}{ccc}2&3&1\\2&9&0\\12&1&3\end{array}\right|\left\begin{array}{c}5\\5\\5\end{array}\right]$

Subtract row 1 from row 2  ($R_2=R_2-R_1$)

$\left[\begin{array}{ccc}2&3&1\\0&6&-1\\12&1&3\end{array}\right|\left\begin{array}{c}5\\0\\5\end{array}\right]$

subtract row 1 multiplied by 6 from row 3   ($R_3=R_3-6R_1$)

$\left[\begin{array}{ccc}2&3&1\\0&6&-1\\0&-17&-3\end{array}\right|\left\begin{array}{c}5\\0\\-25\end{array}\right]$

Divide row 1 by 2     ($R_1=\frac{R_1}2$)

$\left[\begin{array}{ccc}1&\frac32&\frac12\\0&6&-1\\0&-17&-3\end{array}\right|\left\begin{array}{c}\frac52\\0\\-25\end{array}\right]$

Divide row 2 by 6   $(R_2=\frac{R_2}6)$

$\left[\begin{array}{ccc}1&\frac32&\frac12\\ \\0&1&\frac{-1}{6}\\ \\0&-17&-3\end{array}\right|\left\begin{array}{c}\frac52\\ \\0\\ \\-25\end{array}\right]$

Subtract row 2 multiplied by $\frac32$ from row 1 $(R_1=R_1-\frac32 R_2)$

$\left[\begin{array}{ccc}1&0&\frac34\\ \\0&1&\frac{-1}{6}\\ \\0&-17&-3\end{array}\right|\left\begin{array}{c}\frac52\\ \\0\\ \\-25\end{array}\right]$

Add row 2 multiplied by 17 to row 3 $(R_3=R_3+17R_2)$

$\left[\begin{array}{ccc}1&0&\frac34\\ \\0&1&-\frac{1}{6}\\ \\0&0&-\frac{35}{6}\end{array}\right|\left\begin{array}{c}\frac52\\ \\0\\ \\-25\end{array}\right]$

Multiply row 3 by $-\frac{6}{35}$   $(R_3=-\frac6{35} R_3)$

$\left[\begin{array}{ccc}1&0&\frac34\\ \\0&1&-\frac{1}{6}\\ \\0&0&1\end{array}\right|\left\begin{array}{c}\frac52\\ \\0\\ \\\frac{30}7\end{array}\right]$

Subtract row 3 multiplied by $\frac34$ from row 1 $(\ R_1=R_1-\frac34R_3)$

$\left[\begin{array}{ccc}1&0&0\\ \\0&1&-\frac{1}{6}\\ \\0&0&1\end{array}\right|\left\begin{array}{c}-\frac57\\ \\0\\ \\\frac{30}7\end{array}\right]$

Add row 3 multiplied by $\frac16$ to row 2  $(R_2=R_2+\frac16R_3)$

$\left[\begin{array}{ccc}1&0&0\\ \\0&1&0\\ \\0&0&1\end{array}\right|\left\begin{array}{c}-\frac57\\ \\\frac57\\ \\\frac{30}7\end{array}\right]$

$\therefore x=-\frac57$, $y=\frac57$ and  $z=-\frac{30}{7}$

Unique solution: The system has one specific solution.

The system has unique solution.