Chapter 7, Problem 42RE

To determine

To solve the system of linear equations by Gaussian Elimination method.

Answer to Problem 42RE

  $\boxed{x=1,y=1,z=1,w=0}$

Explanation of Solution

Given information:

Use Gaussian elimination to solve the system of linear equations and check any solutions algebraically.

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

Calculation:

Consider the given system of equations,

  $\begin{array}{l}x+4w=\mathrm{1..........}\left(1\right)\\ 3y+z-w=\mathrm{4........}\left(2\right)\\ 2y-3w=\mathrm{2........}\left(3\right)\\ 4x-y+2z=\mathrm{5.......}\left(4\right)\end{array}$

Multiply equation $\left(3\right)$ by $-1$ and add with equation $\left(2\right)$

  $\frac{\begin{array}{l}3y+z-w=4\\ -2y+3w=-2\end{array}}{y+z+2=2}$

New system of equations is.

  $\begin{array}{l}x+4w=\mathrm{1..........}\left(1\right)\\ y+z+2=\mathrm{2........}\left(2\right)\\ 2y-3w=\mathrm{2........}\left(3\right)\\ 4x-y+2z=\mathrm{5.......}\left(4\right)\end{array}$

Multiply equation $\left(2\right)$ by $-2$ and add to equation $\left(3\right)$

  $\frac{\begin{array}{l}-2y-2z-4w=-4\\ 2y-3w=2\end{array}}{-2z-7w=-2}$

Multiply equation $-2z-7w=-2$ by $-1/2$

  $z-3.5w=1$

New system of equations is.

  $\begin{array}{l}x+4w=\mathrm{1..........}\left(1\right)\\ y+z+2=\mathrm{2........}\left(2\right)\\ z-3.5w=\mathrm{1........}\left(3\right)\\ 4x-y+2z=\mathrm{5.......}\left(4\right)\end{array}$

Multiply equation $\left(1\right)$ by $-4$ and add to equation $\left(4\right)$

  $\frac{\begin{array}{l}-4x-16w=-4\\ 4x-y+2z=5\end{array}}{-y+2z-16w=1}$

Add $-y+2z-16w=1$ to equation $\left(2\right)$

  $\frac{\begin{array}{l}-y+2z-16w=1\\ y+z+2w=2\end{array}}{3z-14w=3}$

New system of equations is.

  $\begin{array}{l}x+4w=\mathrm{1..........}\left(1\right)\\ y+z+2=\mathrm{2........}\left(2\right)\\ z-3.5w=\mathrm{1........}\left(3\right)\\ 3z-14w=\mathrm{3......}\left(4\right)\end{array}$

Multiply equation $\left(3\right)$ by $-3$ and add to equation $\left(4\right)$

  $\frac{\begin{array}{l}-3z-10.5w=-3\\ 3z-14w=3\end{array}}{-24.4w=0}$

New system of equations is.

  $\begin{array}{l}x+4w=\mathrm{1..........}\left(1\right)\\ y+z+2=\mathrm{2........}\left(2\right)\\ z-3.5w=\mathrm{1........}\left(3\right)\\ w=\mathrm{0......}\left(4\right)\end{array}$

Put $w=0$ equation $\left(3\right)$

  $\begin{array}{l}z-3.5\times 0=1\\ z=1\end{array}$

Put $w=0$ , $z=1$ in equation $\left(2\right)$

  $y+z+2w=2$

  $\begin{array}{l}y+1+2\times 0=2\\ y=2-1\\ y=1\end{array}$

Put $w=0$ equation $\left(1\right)$

  $x+4w=1$

  $\begin{array}{l}x+4\times 0=1\\ x=1\end{array}$

Hence, $\boxed{x=1,y=1,z=1,w=0}$