What is the next step in Gauss-Jordan Elimination for this augmented matrix?

[karush]

complete
$$\left[
\begin{array}{rrrr|r}
1& -6& 4& 0&-1\\
0& 2& -7& 0&4\\
0& 0& 1& 2&-3\\
0& 0& 4& 1&2\
\end{array}\right]$$
ok assume next step is $r_2/2$ and $r_4/4$ introducing fractions

 

[cbarker1]

Yes for $r_2/2$. But no for $r_4/4$. You need to multiply by the value from above and below by the given row value. So for instance, you multiple by -4 on fourth row and then add them to the 5th row.

 

[karush]

how about $r_2/2$.and $r_4-r_3(- 4)$

 

[cbarker1]

complete
$$\left[
\begin{array}{rrrr|r}
1& -6& 4& 0&-1\\
0& 2& -7& 0&4\\
0& 0& 1& 2&-3\\
0& 0& 4& 1&2\
\end{array}\right]$$
ok assume next step is $r_2/2$ and $r_4/4$ introducing fractions

What is the complete question for this exercise?

It should be -7R4+R3=> R3. Then it follow that the R4= {0,0,1,2} and R3= {0,0,0,-14}, where {} means the row entries.

 

[karush]

Consider each matrix in Exercises 5 and 6 as the augmented matrix of a linear system. State in words the next two elementary row operations that should be performed in the process of solving the system.

so it looks like the idea is to get the zeros triangle a complete solve would be complicalted

 

[cbarker1]

Where is the zeros triangle needs to be on the bottom of the matrix or the top of the matrix?

 

[karush]

bottom already has the zeros except one

symbolab answer

$$\begin{bmatrix}1&0&0&0&28\\ 0&1&0&0&\frac{11}{2}\\ 0&0&1&0&1\\ 0&0&0&1&-2\end{bmatrix}$$

 

[cbarker1]

I see, Gauss-Jordan Elimination.