Chapter 11, Problem 1CA

(a)

To determine

To write: a linear system which supports the friend’s claim.

Answer to Problem 1CA

Thus, the solution of this system is $\left(1,0,1\right)$ .

Explanation of Solution

Given:

  $\begin{array}{l}3x+y+3z=6x+y+z=24x-2y+4z=8\\ x-y+z=2\text{ 2}x+y+z=4\text{ 3}x+y+9z=12\end{array}$

Calculation:

One such linear system with exactly one solution is

  $\begin{array}{l}3x+y+9z=12\mathrm{....}\left(1\right)\hfill \\ x+y+z=2\mathrm{.......}\left(2\right)\hfill \\ x-y+z=2\mathrm{......}\left(3\right)\hfill \end{array}$

Now to prove that this system has exactly one solution,

Add $-1$ times equation (2) to equation (1) to eliminate variable y and thus obtain a new equation

  $\begin{array}{l}3x+y+9z=12\\ \underset{\_}{-x-y-z=-2}\\ 2x+8z=10\mathrm{....}\left(4\right)\end{array}$

Add equation (2) and equation (3) to eliminate variable y and thus obtain a new equation

  $\begin{array}{l}x+y+z=2\\ \underset{\_}{x-y+z=2}\\ 2x+2z=4\mathrm{....}\left(5\right)\end{array}$

Now, solve the new linear system consisting of equation (4) and (5) for both its variables.

For that, add $-1$ times equation (5) from equation (4)

  $\begin{array}{l}2x+8z=10\\ \underset{\_}{-2x-2z=-4}\\ \text{ 6}z=6\\ \Rightarrow z=1\end{array}$

Substituting it in equation (5), to obtain value of $x$

  $\begin{array}{l}2x+2\left(1\right)=4\hfill \\ 2x=2\hfill \\ x=1\hfill \end{array}$

Now, substitute values of $xandz$ in equation (2)

  $\begin{array}{l}\begin{array}{l}x+y+z=2\hfill \\ 1+y+1=2\hfill \\ y=2-2\hfill \end{array}\\ y=0\end{array}$

Thus, the solution of this system is $\left(1,0,1\right)$ .

Conclusion:

The solution of this system is $\left(1,0,1\right)$

(b)

To determine

To write: a linear system which does not support the friend’s claim.

Answer to Problem 1CA

Because, it is obtained the identity $0=0$ , the system has infinitely many solutions.

Explanation of Solution

Calculation:

One such linear system that does not have exactly one solution is

  $\begin{array}{l}\begin{array}{l}3x+y+3z=6\mathrm{...}\left(1\right)\hfill \\ x+y+z=2\mathrm{...}\left(2\right)\hfill \end{array}\\ 4x-2y+4z=8\mathrm{...}\left(3\right)\end{array}$

Now to prove that this system has exactly one solution,

Add $-1$ times equation (2) to equation (1) to eliminate variable $y$ and thus obtain a new equation

  $\begin{array}{l}3x+y+3z=6\hfill \\ \underset{\_}{-x-y-z=-2}\hfill \\ 2x+2z=4\mathrm{...}\left(4\right)\hfill \end{array}$

Add $-2$ time equation (2) and equation (3) to eliminate variable y and thus obtain a new equation

  $\begin{array}{l}\begin{array}{l}-2x-2y-2z=-4\hfill \\ \underset{\_}{4x-2y+4z=8}\hfill \end{array}\\ 2x+2z=4\mathrm{...}\left(5\right)\end{array}$

Now, solve the new linear system consisting of equation (4) and (5) for both its variables.

For that, add $-1$ times equation (5) from equation (4)

  $\begin{array}{l}\begin{array}{l}2x+2z=4\hfill \\ \underset{\_}{-2x-2z=-4}\hfill \end{array}\\ 0=0\end{array}$

Because, it is obtained the identity $0=0$ , the system has infinitely many solutions.

Conclusion:

Because, it is obtained the identity $0=0$ , the system has infinitely many solutions.