Chapter 7, Problem 48RE

a.

To determine

To Find: the values of a,b and c such that the system of linear equations have unique solution.

Answer to Problem 48RE

The given system have unique solution if $a=-b-c$

Explanation of Solution

Given:

The given system of linear equations is $\{\begin{array}{c}x+y=2\\ y+z=2\\ x+z=2\\ ax+by+cz=0\end{array}$

Concept Used:

The given system can be solve using Row Echelon form

Calculation:

Write the augmented matrix A for the given system of equations

  $A=\left[\begin{array}{ccc}1& 1& 0\\ 0& 1& 1\\ 1& 0& 1\\ a& b& c\end{array}|\begin{array}{c}2\\ 2\\ 2\\ 0\end{array}\right]$

Reduce it to Row Echelon form

  $\begin{array}{l}{R}_3\to R_3-R_1,R_4\to R_4-a{R}_1\\ \left[\begin{array}{ccc}1& 1& 0\\ 0& 1& 1\\ 0& -1& 1\\ 0& b-a& c\end{array}|\begin{array}{c}2\\ 2\\ 0\\ -2a\end{array}\right]\\ R_3\to R_3+R_2,R_4\to R_4-\left(b-a\right)R_2\\ \left[\begin{array}{ccc}1& 1& 0\\ 0& 1& 1\\ 0& 0& 2\\ 0& 0& a-b+c\end{array}|\begin{array}{c}2\\ 2\\ 2\\ -2b\end{array}\right]\\ R_3\to \frac{1}{2}{R}_3\\ \left[\begin{array}{ccc}1& 1& 0\\ 0& 1& 1\\ 0& 0& 1\\ 0& 0& a-b+c\end{array}|\begin{array}{c}2\\ 2\\ 1\\ -2b\end{array}\right]\\ R_4\to R_4-\left(a-b+c\right)R_3\\ \left[\begin{array}{ccc}1& 1& 0\\ 0& 1& 1\\ 0& 0& 1\\ 0& 0& 0\end{array}|\begin{array}{c}2\\ 2\\ 1\\ -a-b-c\end{array}\right]\end{array}$

The reduced Roe Echelon for of the matrix A is

  $B=\left[\begin{array}{ccc}1& 1& 0\\ 0& 1& 1\\ 0& 0& 1\\ 0& 0& 0\end{array}|\begin{array}{c}2\\ 2\\ 1\\ -a-b-c\end{array}\right]$

The given system have a unique solution if $-a-b-c=0$ or $a=-b-c$

Conclusion:

Hence the system have a unique solution if $-a-b-c=0$ or $a=-b-c$

b.

To determine

To Find: the values of a,b and c such that the system of linear equations have no solutions

Answer to Problem 48RE

The given system have no solution if $a+b+c\ne 0$

Explanation of Solution

Given:

The given system of linear equations is $\{\begin{array}{c}x+y=2\\ y+z=2\\ x+z=2\\ ax+by+cz=0\end{array}$

Concept Used:

The given system can be solve using Row Echelon form

Calculation:

Write the augmented matrix A for the given system of equations

  $A=\left[\begin{array}{ccc}1& 1& 0\\ 0& 1& 1\\ 1& 0& 1\\ a& b& c\end{array}|\begin{array}{c}2\\ 2\\ 2\\ 0\end{array}\right]$

Reduce it to Row Echelon form

  $\begin{array}{l}{R}_3\to R_3-R_1,R_4\to R_4-a{R}_1\\ \left[\begin{array}{ccc}1& 1& 0\\ 0& 1& 1\\ 0& -1& 1\\ 0& b-a& c\end{array}|\begin{array}{c}2\\ 2\\ 0\\ -2a\end{array}\right]\\ R_3\to R_3+R_2,R_4\to R_4-\left(b-a\right)R_2\\ \left[\begin{array}{ccc}1& 1& 0\\ 0& 1& 1\\ 0& 0& 2\\ 0& 0& a-b+c\end{array}|\begin{array}{c}2\\ 2\\ 2\\ -2b\end{array}\right]\\ R_3\to \frac{1}{2}{R}_3\\ \left[\begin{array}{ccc}1& 1& 0\\ 0& 1& 1\\ 0& 0& 1\\ 0& 0& a-b+c\end{array}|\begin{array}{c}2\\ 2\\ 1\\ -2b\end{array}\right]\\ R_4\to R_4-\left(a-b+c\right)R_3\\ \left[\begin{array}{ccc}1& 1& 0\\ 0& 1& 1\\ 0& 0& 1\\ 0& 0& 0\end{array}|\begin{array}{c}2\\ 2\\ 1\\ -a-b-c\end{array}\right]\end{array}$

The reduced Roe Echelon for of the matrix A is

  $B=\left[\begin{array}{ccc}1& 1& 0\\ 0& 1& 1\\ 0& 0& 1\\ 0& 0& 0\end{array}|\begin{array}{c}2\\ 2\\ 1\\ -a-b-c\end{array}\right]$

The given system have no solution if $-a-b-c\ne 0\Rightarrow a+b+c\ne 0$

Conclusion:

Hence the system have no solution if $a+b+c\ne 0$

c.

To determine

To Find: the values of a,b and c such that the system of linear equations have infinitely many solutions.

Answer to Problem 48RE

There exist no such value of a,b and c such that the system have infinitely many solutions

Explanation of Solution

Given:

The given system of linear equations is $\{\begin{array}{c}x+y=2\\ y+z=2\\ x+z=2\\ ax+by+cz=0\end{array}$

Concept Used:

The given system can be solve using Row Echelon form

Calculation:

Write the augmented matrix A for the given system of equations

  $A=\left[\begin{array}{ccc}1& 1& 0\\ 0& 1& 1\\ 1& 0& 1\\ a& b& c\end{array}|\begin{array}{c}2\\ 2\\ 2\\ 0\end{array}\right]$

Reduce it to Row Echelon form

  $\begin{array}{l}{R}_3\to R_3-R_1,R_4\to R_4-a{R}_1\\ \left[\begin{array}{ccc}1& 1& 0\\ 0& 1& 1\\ 0& -1& 1\\ 0& b-a& c\end{array}|\begin{array}{c}2\\ 2\\ 0\\ -2a\end{array}\right]\\ R_3\to R_3+R_2,R_4\to R_4-\left(b-a\right)R_2\\ \left[\begin{array}{ccc}1& 1& 0\\ 0& 1& 1\\ 0& 0& 2\\ 0& 0& a-b+c\end{array}|\begin{array}{c}2\\ 2\\ 2\\ -2b\end{array}\right]\\ R_3\to \frac{1}{2}{R}_3\\ \left[\begin{array}{ccc}1& 1& 0\\ 0& 1& 1\\ 0& 0& 1\\ 0& 0& a-b+c\end{array}|\begin{array}{c}2\\ 2\\ 1\\ -2b\end{array}\right]\\ R_4\to R_4-\left(a-b+c\right)R_3\\ \left[\begin{array}{ccc}1& 1& 0\\ 0& 1& 1\\ 0& 0& 1\\ 0& 0& 0\end{array}|\begin{array}{c}2\\ 2\\ 1\\ -a-b-c\end{array}\right]\end{array}$

The reduced Roe Echelon for of the matrix A is

  $B=\left[\begin{array}{ccc}1& 1& 0\\ 0& 1& 1\\ 0& 0& 1\\ 0& 0& 0\end{array}|\begin{array}{c}2\\ 2\\ 1\\ -a-b-c\end{array}\right]$

There exist no such value of a,b and c such that the system have infinitely many solutions

Conclusion:

There exist no such value of a,b and c such that the system have infinitely many solutions