Chapter 8.1, Problem 81E

To determine

Find the augmented matrix corresponding to the system of linear equations in reduced roe-echelon form and solve the system.

Answer to Problem 81E

The reduced row-echelon form as

  $\left[\begin{array}{cccccc}1& 0& 0& 0& ⋮& 1\\ 0& 1& 0& 0& ⋮& 0\\ 0& 0& 1& 0& ⋮& 4\\ 0& 0& 0& 1& ⋮& -2\end{array}\right]$

The solution is $x=1,y=0,z=4$ and $w=–2$ .

Explanation of Solution

Given:

The system of equations

  $\begin{array}{l}2x+y–z+2w=–6\hfill \\ 3x+4y+w=1\hfill \\ x+5y+2z+6w=–3\hfill \\ 5x+2y–z–w=3\hfill \end{array}$

Let consider the following system of equations

  $\begin{array}{l}2x+y–z+2w=–6\hfill \\ 3x+4y+w=1\hfill \\ x+5y+2z+6w=–3\hfill \\ 5x+2y–z–w=3\hfill \end{array}$

The augmented matrix for the above system of equations is

  $\left[\begin{array}{cccccc}2& 1& -1& 2& ⋮& -6\\ 3& 4& 0& 1& ⋮& 1\\ 1& 5& 2& 6& ⋮& -3\\ 5& 2& -1& -1& ⋮& 3\end{array}\right]$

Use matrix capabilities of a graphing utility or graphing calculator to write the matrix in reduced row-echelon form and get the reduced row-echelon form as

  $\left[\begin{array}{cccccc}1& 0& 0& 0& ⋮& 1\\ 0& 1& 0& 0& ⋮& 0\\ 0& 0& 1& 0& ⋮& 4\\ 0& 0& 0& 1& ⋮& -2\end{array}\right]$

When an augmented matrix is reduced to the $[I\mid B]$ form then B represents the solution set. So the solution is $x=1,y=0,z=4$ and $w=–2$ .