Chapter 10.2, Problem 31E

To determine

To calculate:The solution of linear equation

Answer to Problem 31E

  $x,y,z=\left(3-t,-3+2t,t\right)$

The equation is linear.

Explanation of Solution

Given information:

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

Calculation:

Naming the given equations as,

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

Eliminate $x$ from equation $\left(2\right)$ ,

Multiplying equation $\left(1\right)$ by $-1$ ,

  $\begin{array}{l}\left(x+y-z=0\right)\times \left(-1\right)\\ \left(x+y-z\times \left(-1\right)\right)=0\times \left(-1\right)\\ \left(x\times \left(-1\right)\right)+\left(y\times \left(-1\right)\right)-\left(z\times \left(-1\right)\right)=0\\ -1x+\left(-1y\right)-\left(-1z\right)=0\end{array}$

  $-x-y+z=0$

Naming it as equation $\left(4\right)$ ,

  $-x-y+z=\mathrm{0........}\left(4\right)$

Adding equation $\left(2\right)$ and equation $\left(4\right)$ ,

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

Naming it as equation $.\left(5\right)$ ,

  $y-2z=-\mathrm{3........}\left(5\right)$

Eliminate $x$ from equation $\left(3\right)$ ,

Multiplying equation $\left(1\right)$ by $-2$ ,

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

Naming it as equation $\left(6\right)$ ,

  $-2x-2y+2z=\mathrm{0........}\left(6\right)$

Adding equation $\left(3\right)$ and equation $\left(6\right)$ ,

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

Multiplying equation $.\left(5\right)$ by $-1$ ,

  $\begin{array}{l}\left(y-2z=-3\right)\times \left(-1\right)\\ \left(y-2z\times \left(-1\right)\right)=-3\times \left(-1\right)\\ \left(y\times \left(-1\right)\right)-\left(2z\times \left(-1\right)\right)=3\\ -y-\left(-2z\right)=3\\ -y+2z=3\end{array}$

Naming it as equation $\left(7\right)$ ,

  $-y+2z=\mathrm{3........}\left(7\right)$

Adding equation $.\left(5\right)$ and equation $\left(7\right)$ ,

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

The equation $.\left(5\right)$ is true and hence dropping it from the system,

Now there are only two equations.

Using these two equations to solve $x$ & $y$ in terms of $z$ ,

As $z$ can take any value So there are infinitely many solutions.

Using the equation $.\left(5\right)$ ,

  $y-2z=-3$

Rearranging the above equation,

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

Solving for $x$ in terms of $z$ ,

Using equation $\left(1\right)$ ,

  $x+y-z=0$

Substituting the value $y=-3+2z$ in above equation,

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

Assuming $z=t$ ,

Where in $t$ is a real number.

Substitute $z=t$ in $y=-3+2z$ ,

  $\begin{array}{l}y=-3+2z\\ y=-3+2t\end{array}$

Substitute $z=t$ in $x=3-z$ ,

  $\begin{array}{l}x=3-z\\ x=3-t\end{array}$

Therefore, the final solution is $x,y,z=\left(3-t,-3+2t,t\right)$