Chapter 7.3, Problem 108E

a

To determine

To Find: Value of $a,b,c$

Answer to Problem 108E

If $a+b+c=0$ , will have a unique solution

Explanation of Solution

Given information:

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

Formula used: When forming the augmented matrix of a system. Begin by vertically aligning the variables in the equations and using zeros for the coefficients of missing variables

Calculation:

The augmented matrix is $\left[\begin{array}{cccc}1& 1& 0& 2\\ 0& 1& 1& 2\\ 1& 0& 1& 2\\ a& b& c& 0\end{array}\right]$

Doing

  $\begin{array}{l}{R}_3-R_1\to R_3\\ R_4-a{R}_1\to R_4\end{array}$

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

Doing

  $\begin{array}{l}{R}_3+R_2\to R_3\\ R_4+(b-a)R_3\to R_4\\ \end{array}$

we obtain

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

Now doing $\frac{R_3}{2}\to R_3$ gives

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

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

obtain

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

And now doing $R_1-R_2\to R_1$

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

We have $x=1,y=1,z=1,a+b+c=0$

If $a+b+c=0$ will have a unique solution

b

To determine

To Find: Value of $a,b,c$

Answer to Problem 108E

a. $a+b+c\ne 0$ will have no solution

Explanation of Solution

Given information:

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

Formula used: When forming the augmented matrix of a system. Begin by vertically aligning the variables in the equations and using zeros for the coefficients of missing variables

Calculation:

The augmented matrix is $\left[\begin{array}{cccc}1& 1& 0& 2\\ 0& 1& 1& 2\\ 1& 0& 1& 2\\ a& b& c& 0\end{array}\right]$

Doing

  $\begin{array}{l}-R_1+R_3\to R_3\\ R_4-a{R}_1\to R_4\end{array}$

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

Doing

  $\begin{array}{l}{R}_2+R_3\to R_3\\ R_4+(b-a)R_4\to R_4\end{array}$

we obtain

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

Now doing $\frac{R_3}{2}\to R_3$

gives

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

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

obtain

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

And now doing $R_1-R_2\to R_1$

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

We have $x=1,y=1,z=1,a+b+c=0$

If $a+b+c\ne 0$ will have no solution

c

To determine

To Find: Value of Value of a,b ,c

Answer to Problem 108E

Never can have an infinite number of solutions

Explanation of Solution

Given information:

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

Formula used: When forming the augmented matrix of a system. Begin by vertically aligning the variables in the equations and using zeros for the coefficients of missing variables

Calculation:

The augmented matrix is $\left[\begin{array}{cccc}1& 1& 0& 2\\ 0& 1& 1& 2\\ 1& 0& 1& 2\\ a& b& c& 0\end{array}\right]$

Doing $R_3-R_1\to R_3,R_4-a{R}_1\to R_4$

  $\begin{array}{l}-R_1+R_3\to R_3\\ R_4-a{R}_1\to R_4\end{array}$

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

Doing

  $\begin{array}{l}{R}_2+R_3\to R_3\\ R_4+(b-a)R_4\to R_4\end{array}$

we obtain

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

Now doing $\frac{R_3}{2}\to R_3$

gives

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

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

obtain

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

And now doing $R_1-R_2\to R_1$

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

We have $x=1,y=1,z=1,a+b+c=0$

Never can have an infinite number of solutions