Chapter 7.3, Problem 104E

To determine

To find: values of a, b and c (if possible) such that the system of linear equations has (a) a unique solution (b) no solution and (c) an infinite number of solutions.

Answer to Problem 104E

(a) If a + b + c=0, will have a unique solution.

(b) If $a+b+c\ne 0$, will have no solution.

(c) Never can have an infinity number of solutions.

Explanation of Solution

Given information: The given system of linear equation is,

  $\begin{array}{c}x+y=2\\ y+z=2\\ x+z=2\\ ax+by+cz=0\end{array}$

Calculation:

The given system of linear equation is,

  $\begin{array}{c}x+y=2\\ y+z=2\\ x+z=2\\ ax+by+cz=0\end{array}$

The augmented matrix of the given system of linear equations is,

  $\left[\begin{array}{cccc}1& 1& 0& 2\\ 0& 1& 1& 2\\ 1& 0& 1& 2\\ a& b& c& 0\end{array}\right]$ ---------------(1)

Use Gauss Jordan elimination. Doing $R_3-R_1\to R_3,R_4-a{R}_1\to R_4$ , got

  $\left[\begin{array}{cccc}1& 1& 0& 2\\ 0& 1& 1& 2\\ 0& -1& 1& 0\\ 0& b-c& c& -2a\end{array}\right]$ -----------(2)

Doing $R_2+R_3\to R_3,(b-a)R_3+R_4\to R_4$

  $\left[\begin{array}{cccc}1& 1& 0& 2\\ 0& 1& 1& 2\\ 0& 0& 2& 2\\ 0& 0& c+b-a& -2a\end{array}\right]$ ---------(3)

Doing $R_3/2\to R_3$ gives

  $\left[\begin{array}{cccc}1& 1& 0& 2\\ 0& 1& 1& 2\\ 0& 0& 1& 1\\ 0& 0& c+b-a& -2a\end{array}\right]$ ---------(4)

Doing $(c+b-a)R_3-R_4\to R_4$

  $\left[\begin{array}{cccc}1& 1& 0& 2\\ 0& 1& 1& 1\\ 0& 0& 1& 1\\ 0& 0& 0& a+b+c\end{array}\right]$ ---------(5)

Doing $R_2-R_3\to R_3.$

  $\left[\begin{array}{cccc}1& 0& 0& 2\\ 0& 1& 0& 1\\ 0& 0& 1& 1\\ 0& 0& 0& a+b+c\end{array}\right]$ ---------(6)

Finally ,doing $R_1-R_2\to R_1.$ obtain the reduced row echelon matrix,

  $\left[\begin{array}{cccc}1& 0& 0& 1\\ 0& 1& 0& 1\\ 0& 0& 1& 1\\ 0& 0& 0& a+b+c\end{array}\right]$ ---------(7)

Then, by back substitution. $x=1,y=1,z=1,a+b+c=0.$

Then, the system

(d) If a + b + c=0, will have a unique solution.

(e) If $a+b+c\ne 0$, will have no solution.

(f) Never can have an infinity number of solutions.