Chapter 9.9, Problem 25WE

To determine

To find: whether there is any solution and also find the general expression for the solution.

Answer to Problem 25WE

The general expressions for the solutions of the given system is $\{(0,2z,z)\}.$

Explanation of Solution

Given:

  $x-2y+4z=0$

  $3x-y+2z=0$

  $x+3y-6z=0$

Calculation:

Let us first transform the given system of equations into a triangular system.

Multiply first equation by $-3$ and add the result to the second equation.

  $\begin{array}{l}3x-y+2z=0\\ -3(x-2y+4z=0)\\ 5y-10z=0\end{array}$$\begin{array}{l}x-2y+4z=0\\ 5y-10z=0\\ x+3y-6z=0\end{array}$

Then, multiply the first equation by $-1$ and add the result to the third equation.

  $\begin{array}{l}x+3y-6z=0\\ -1(x-2y+4z=0)\\ 5y-10z=0\end{array}$$\begin{array}{l}Resultingsystem\\ x-2y+4z=0\\ 5y-10z=0\\ -5y+10z=0\end{array}$  $(1)$

In $(1)$ , subtract the third equation from the second equation.

  $\begin{array}{l}5y-10z=0\\ -5y+10z=0\\ 0=0\end{array}$$\begin{array}{l}Resultingsystem\\ x-2y+4z=0\\ 5y-10z=0\\ 0=0\end{array}$  $(2)$

The third equation in the resulting system shows that the given system is dependent and has infinitely many solutions. The solutions for $x$ and $y$ can be determined in terms of $z$ .

In $(2)$, the value of $y$ can be obtained by solving the second equation.

  $\begin{array}{l}5y-10z=0\\ 5y=10z\\ y=\frac{10z}{5}\\ =2z\end{array}$

Substitute $2z$ for the value of $y$ in the first equation and solve for $x$ .

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

Therefore, the general expressions for the solutions of the given system is $\{(0,2z,z)\}.$

Conclusion:

The general expressions for the solutions of the given system is $\{(0,2z,z)\}.$