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

The system has unique solution.

`\therefore x=-\frac57`,
`y=\frac57` and
`z=-(30)/(7)`

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&(-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&(-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&-(1)/(6)\\ \\0&0&-(35)/(6)\end{array}\right|\left\begin{array}{c}\frac52\\ \\0\\ \\-25\end{array}\right]`

Multiply row 3 by
`-(6)/(35)`
`(R_3=-\frac6{35} R_3)`

`\left[\begin{array}{ccc}1&0&\frac34\\ \\0&1&-(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&-(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=-(30)/(7)`

Unique solution: The system has one specific solution.

The system has unique solution.