Chapter 1.4, Problem 35E

a.

To determine

The values for a,b and c so that system has no solution.

Answer to Problem 35E

System has no solution if $a=b=c=-1$ .

Explanation of Solution

Given information: The system of three equations are given:

  $\begin{array}{l}x+y+z=2\\ ax+by+cz=10\\ x-2y+z=4\end{array}$

Calculation: : The system of three equations are given:

  $\begin{array}{l}x+y+z=2\\ ax+by+cz=10\\ x-2y+z=4\end{array}$

Choose $a=b=c=-1$ and the system has no solution. because equations are:

  $\begin{array}{l}x+y+z=2(i)\\ -x-y-z=10(ii)\\ x-2y+z=4(iii)\end{array}$

Add equation $(i)\mathrm{and}(ii)$ we get:

  $\begin{array}{l}x+y+z=2\\ -x-y-z=10\\ \overline{0=12}\end{array}$

This is false equation therefore it has no solution.

Therefore value of $a=b=c=-1$ .

b.

To determine

The values for a,b and c so that system has exactly one solution.

Answer to Problem 35E

System has exactly one solution if $a=1\mathrm{and}b=c=0$ .

Explanation of Solution

Given information: The system of three equations are given:

  $\begin{array}{l}x+y+z=2\\ ax+by+cz=10\\ x-2y+z=4\end{array}$

Calculation: The system of three equations are given:

  $\begin{array}{l}x+y+z=2\\ ax+by+cz=10\\ x-2y+z=4\end{array}$

Choose $a=1\mathrm{and}b=c=0$ and the system has one solution. because equations are:

  $\begin{array}{l}x+y+z=2(i)\\ x=10(ii)\\ x-2y+z=4(iii)\end{array}$

Subtract equation $(iii)\mathrm{from}(i)$ we get:

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

Put $x=10\mathrm{and}y=\frac{-2}{3}$ in equation $(i)$ we get:

  $\begin{array}{l}10-\frac{2}{3}+z=2\\ z=2+\frac{2}{3}-10\\ z=\frac{6+2-30}{3}=-\frac{22}{3}\end{array}$

Therefore solution is $(x,y,z)=(10,-\frac{2}{3},-\frac{22}{3})$ .

c.

To determine

The values for a,b and c so that system has infinitely many solutions.

Answer to Problem 35E

System has many solutions if $a=b=c=5$ .

Explanation of Solution

Given information: The system of three equations are given:

  $\begin{array}{l}x+y+z=2\\ ax+by+cz=10\\ x-2y+z=4\end{array}$

Calculation: The system of three equations are given:

  $\begin{array}{l}x+y+z=2\\ ax+by+cz=10\\ x-2y+z=4\end{array}$

Choose $a=b=c=5$ and the system has many solutions. Because equations are:

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

Equation $(i)$ multiplies by $5$ and add $(i)\mathrm{and}(ii)$ we get:

  $\begin{array}{l}5x+5y+5z=10\\ 5x+5y+5z=10\\ ----\\ \overline{0=0}\end{array}$

Therefore system has many solutions.