Chapter 10.3, Problem 51E

To determine

The solutions of given system of linear equations are to be estimated:

  $\begin{array}{l}x-y+w=0\\ 3x-z+2w=0\\ x-4y+z+2w=0\end{array}$

Answer to Problem 51E

The solutions eqns. $\begin{array}{l}x-y+w=0\\ 3x-z+2w=0\\ x-4y+z+2w=0\end{array}$ has infinitely solutions

Explanation of Solution

Given:

  $\begin{array}{l}x-y+w=0\\ 3x-z+2w=0\\ x-4y+z+2w=0\end{array}$

Calculations:

The eqns are given as:

  $x-y+w=0$.......(1)

  $3x-z+2w=0$.......(2)

  $x-4y+z+2w=0$.......(3)

Take two system of equation to find new two eq.:

System 1:

  $x-4y+z+2w=0$ and $3x-z+2w=0$

Take the subtraction:

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

  $-2x-4y+2z=0$.......(4)

System 2:

  $x-4y+z+2w=0$ and $x-y+w=0$

Multiply by $2$ : $2x-2y+2w=0$

Now, take subtraction:

  $2x-2y+2w-\left(x-4y+z+2w\right)=0$

  $x+2y+z=0$

  $\times 2$

  $2x+4y+2z=0$.......(5)

Now, need to solve the system for two new eqns by addition:

  $\begin{array}{l}-2x-4y+2z+2x+4y+2z=0\\ 4z=0\\ z=0\end{array}$

Put the value of $\left(z=0\right)$ in eq. (4) and (5):

  $\begin{array}{l}-2x-4y+2z=0\\ -2x-4y+2\left(0\right)=0\end{array}$

  $2x+4y=0$.......(7)

And

  $\begin{array}{l}2x+4y+2z=0\\ 2x+4y+2\left(0\right)=0\end{array}$

  $2x+4y=0$.......(8)

Now, from eq. (7) and (8):

Both equations are same which implies that for this system of linear equations contain infinite number of solutions.

So, the solutions of given set of eqns. is $x=-2,y=1,z=0,w=3$ .

Conclusion:

Hence, the infinitely solutions are available for the given set of system.