Chapter 8, Problem 29RE

To determine

To solve: the system of linear equations.

Answer to Problem 29RE

The solution does not exist.

Explanation of Solution

Given information:

Given equations

  $\begin{array}{l}-x+y+2z=1\\ 2x+3y+z=-2\\ 5x+4y+2z=4\end{array}$

Calculation:

SO, the augmented matrix is:

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

To apply Gauss-Jorden elimination, this has to be converted to the reduced row echelon form by the elementary row transformation.

Perform $R_2\to R_2+2R_2$ and $R_3\to R_3+5R_1$

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

Perform $R_1\to -R_1$ and $R_2\to \frac{1}{5}{R}_2$

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

Perform $R_3\to R_3-6R_2$

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

So, the system of equation becomes:

  $\begin{array}{c}x-y-2z=-1\\ y+z=\frac{4}{5}\\ 0x+0y+0z=\frac{6}{5}\end{array}$

The second equation is invalid for any value of $x,y$ and $z$ .

Hence, the solution does not exist.