Q:

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

Accepted Solution

A:
Answer:System of equations: [tex]x_1+2x_2+2x_3=6\\2x_1+x_2+x_3=6\\x_1+x_2+3x_3=6[/tex]Augmented matrix: [tex]\left[\begin{array}{cccc}1&2&2&6\\2&1&1&6\\1&1&3&6\end{array}\right][/tex]Reduced Row Echelon matrix: [tex]\left[\begin{array}{cccc}1&2&2&6\\0&1&1&2\\0&0&1&1\end{array}\right][/tex]Step-by-step explanation:Convert the system into an augmented matrix: [tex]\left[\begin{array}{cccc}1&2&2&6\\2&1&1&6\\1&1&3&6\end{array}\right][/tex]For notation, R_n is the new nth row and r_n the unchanged one. 1. Operations: [tex]R_2=-2r_1+r_2\\R_3=-r_1+r_3[/tex]Resulting matrix:[tex]\left[\begin{array}{cccc}1&2&2&6\\0&-3&-3&-6\\0&-1&1&0\end{array}\right][/tex]2. Operations: [tex]R_2=-\frac{1}{3}r_2[/tex]Resulting matrix:[tex]\left[\begin{array}{cccc}1&2&2&6\\0&1&1&2\\0&-1&1&0\end{array}\right][/tex]3. Operations: [tex]R_3=r_2+r_3[/tex]Resulting matrix:[tex]\left[\begin{array}{cccc}1&2&2&6\\0&1&1&2\\0&0&2&2\end{array}\right][/tex]4. Operations: [tex]R_3=\frac{1}{2}r_3[/tex]Resulting matrix:[tex]\left[\begin{array}{cccc}1&2&2&6\\0&1&1&2\\0&0&1&1\end{array}\right][/tex]