Question

Convert the following systems of equations to an augmented matrix and use Gauss-Jordan reduction to convert to an equilivalent matrix in reduced row echelon form. (Show the steps in the process of converting to G-J). You don't have to find the solution set X12x223 = 6 2a1 3 = 6 X1x23x3 = 6

Answer 1

Answer:

System of equations:

$x_1+2x_2+2x_3=6\\2x_1+x_2+x_3=6\\x_1+x_2+3x_3=6$

Augmented matrix:

$\left[\begin{array}{cccc}1&2&2&6\\2&1&1&6\\1&1&3&6\end{array}\right]$

Reduced Row Echelon matrix:

$\left[\begin{array}{cccc}1&2&2&6\\0&1&1&2\\0&0&1&1\end{array}\right]$

Step-by-step explanation:

Convert the system into an augmented matrix:

$\left[\begin{array}{cccc}1&2&2&6\\2&1&1&6\\1&1&3&6\end{array}\right]$

For notation, R_n is the new nth row and r_n the unchanged one.

1. Operations:

$R_2=-2r_1+r_2\\R_3=-r_1+r_3$

Resulting matrix:

$\left[\begin{array}{cccc}1&2&2&6\\0&-3&-3&-6\\0&-1&1&0\end{array}\right]$

2. Operations:

$R_2=-\frac{1}{3}r_2$

Resulting matrix:

$\left[\begin{array}{cccc}1&2&2&6\\0&1&1&2\\0&-1&1&0\end{array}\right]$

3. Operations:

$R_3=r_2+r_3$

Resulting matrix:

$\left[\begin{array}{cccc}1&2&2&6\\0&1&1&2\\0&0&2&2\end{array}\right]$

4. Operations:

$R_3=\frac{1}{2}r_3$

Resulting matrix:

$\left[\begin{array}{cccc}1&2&2&6\\0&1&1&2\\0&0&1&1\end{array}\right]$