Chapter 8, Problem 30RE

To determine

The solution of the system of linear equation based on gauss Jordon Elimination method.

Answer to Problem 30RE

It is reduced row-echelon form and the corresponding solutions are,

  $x=\frac{31}{42},y=\frac{5}{14},z=\frac{13}{84}$

Explanation of Solution

Given information:

The given matrix is shown below,

  $\{\begin{array}{c}4x+4y+4z=5\\ 4x-2y-8z=1\\ 5x+3y+8z=6\end{array}$

Formula used:

The gauss Jordan elimination method is used.

Calculation:

Write augmented matrix for system,

  $\left[\begin{array}{ccccc}4& 4& 4& ⋮& 5\\ 4& -2& -8& ⋮& 1\\ 5& 3& 8& ⋮& 6\end{array}\right]$

Perform row operation $-R_3+R_1$

  $\left[\begin{array}{ccccc}-1& 1& -4& ⋮& -1\\ 4& -2& -8& ⋮& 1\\ 5& 3& 8& ⋮& 3\end{array}\right]$

Perform row operation $-R_1$

  $\left[\begin{array}{ccccc}1& -1& 4& ⋮& 1\\ 4& -2& -8& ⋮& 1\\ 5& 3& 8& ⋮& 6\end{array}\right]$

Perform row operation $-4R_1+R_2$

  $\left[\begin{array}{ccccc}1& -1& 4& ⋮& 1\\ 0& 2& -24& ⋮& -3\\ 5& 3& 8& ⋮& 6\end{array}\right]$

Perform row operation $-5R_1+R_3$

  $\left[\begin{array}{ccccc}1& -1& 4& ⋮& 1\\ 0& 2& -24& ⋮& -3\\ 0& 8& -12& ⋮& 1\end{array}\right]$

Perform row operation $\frac{1}{2}{R}_1$

  $\left[\begin{array}{ccccc}1& -1& 4& ⋮& 1\\ 0& 1& -12& ⋮& -\frac{3}{2}\\ 0& 8& 12& ⋮& 1\end{array}\right]$

Perform row operation $-8R_2+R_3$

  $\left[\begin{array}{ccccc}1& -1& 4& ⋮& 1\\ 0& 1& -12& ⋮& -\frac{3}{2}\\ 0& 0& 84& ⋮& 13\end{array}\right]$

Perform row operation $\frac{1}{84}{R}_3$

  $\left[\begin{array}{ccccc}1& -1& 4& ⋮& 1\\ 0& 1& -12& ⋮& -\frac{3}{2}\\ 0& 0& 1& ⋮& \frac{13}{84}\end{array}\right]$

Perform row operation $12R_2+R_3$

  $\left[\begin{array}{ccccc}1& -1& 4& ⋮& 1\\ 0& 1& 0& ⋮& \frac{5}{14}\\ 0& 0& 1& ⋮& \frac{13}{84}\end{array}\right]$

Perform row operation $R_2+R_1$

  $\left[\begin{array}{ccccc}1& 0& 4& ⋮& \frac{19}{14}\\ 0& 1& 0& ⋮& \frac{5}{14}\\ 0& 0& 1& ⋮& \frac{13}{84}\end{array}\right]$

Perform row operation $-4R_3+R_1$

  $\left[\begin{array}{ccccc}1& 0& 0& ⋮& \frac{31}{42}\\ 0& 1& 0& ⋮& \frac{5}{14}\\ 0& 0& 1& ⋮& \frac{13}{84}\end{array}\right]$

This is reduced row-echelon form and the corresponding solutions are,

  $x=\frac{31}{42},y=\frac{5}{14},z=\frac{13}{84}$

Conclusion:

It is reduced row-echelon form and the corresponding solutions are,

  $x=\frac{31}{42},y=\frac{5}{14},z=\frac{13}{84}$