Chapter 10.2, Problem 47E

(a)

To determine

To prove: $\left(\frac{x_0+x_1}{2},\frac{y_0+y_1}{2},\frac{z_0+z_1}{2}\right)$ is also a solution of $\{\begin{array}{l}{a}_{1}x+b_{1}y+c_{1}z=d_1\\ a_{2}x+b_{2}y+c_{2}z=d_2\\ a_{3}x+b_{3}y+c_{3}z=d_3\end{array}$

Explanation of Solution

Proof:

Proof by contradiction: if $\left(\frac{x_0+x_1}{2},\frac{y_0+y_1}{2},\frac{z_0+z_1}{2}\right)$ is not a solution to the given system of equations, then when the values are put into the first equation, the sum will be some $d_4$ where $d_4\ne d_1$ .

Plugging the values into the first equation,

  $\begin{array}{l}\frac{a_1{x}_0+b_1{y}_0+c_1{z}_0+a_1{x}_1+b_1{y}_1+c_1{z}_1}{2}=d_4\\ \frac{d_1+a_1{x}_1+b_1{y}_1+c_1{z}_1}{2}=d_4\\ \frac{d_1+d_1}{2}=d_4\\ d_1=d_4\end{array}$

Now $\left(x_0,y_0,z_0\right)$ and $\left(x_1,y_1,z_1\right)$ are both solutions to the system of equations

The solution $\left(\frac{x_0+x_1}{2},\frac{y_0+y_1}{2},\frac{z_0+z_1}{2}\right)$ to the first is found and, using the same argument, the rest of the system of linear equations can be solved.

Conclusion:

Therefore, $\left(\frac{x_0+x_1}{2},\frac{y_0+y_1}{2},\frac{z_0+z_1}{2}\right)$ is not a solution of $\{\begin{array}{l}{a}_{1}x+b_{1}y+c_{1}z=d_1\\ a_{2}x+b_{2}y+c_{2}z=d_2\\ a_{3}x+b_{3}y+c_{3}z=d_3\end{array}$ .

(b)

To determine

To prove: if the system has two different solutions, then it has infinitely many solutions.

Explanation of Solution

Proof:

From the above proof, the way to derive new solutions for the given system is seen. Plug the new values found after using the formula back into said formula to find new, and by definition distinct, solutions.

It is continued doing this to no end, (stumbled onto one of Zero's Paradoxes), and found that there can be infinite solutions to the given system.

Conclusion:

Therefore, there can be infinite solutions to the given system.