Chapter 1, Problem 4CA

To determine

To arrange: The given system of linear equations according to number of solutions.

Answer to Problem 4CA

The correct order is $B<C<A$ .

Explanation of Solution

Given information: Three systems of equations.

Calculation:

For $A$ ,

Adding equations,

  $\begin{array}{l}-14x-28y+7z=-49Multiplying1ste{q}^{n}by-7\\ 14x+28y-7z=49\\ =================\\ 0=0\end{array}$

As the identity is $0=0$ , the system has infinitely many solutions.

For $B$ ,

Adding equations,

  $\begin{array}{l}-3x+3y-3z=24Multiplying2nde{q}^{n}by3\\ 3x-3y+3z=5\\ =============\\ 0=29\end{array}$

which is never true. Thus, this part has no solution.

For $C$ ,

  $\begin{array}{l}3x+3y-4z=44\\ -3x+6y+3z=33Multiplying2ndequationby3\\ =============\\ 9y-z=77\\ -4x+8y+4z=44Multiplying2ndequationby4\\ 4x-y+2z=18\\ =============\\ 7y+6z=62\end{array}$

  $\begin{array}{l}9y-z=77-(i)\\ 7y+6z=62-(ii)\end{array}$

Multiplying $(i)$ by $6$ and $(i)$ adding and $(ii)$ ,

  $\begin{array}{l}54y-6z=462\\ 7y+6z=62\\ ===========\\ 61y=524\\ y=\frac{524}{61}\end{array}$

Substituting the value of $y$ in $(ii)$ ,

  $\begin{array}{l}7(\frac{524}{61})+6z=62\\ \frac{3668}{61}+6z=62\\ 6z=\frac{62}{1}-\frac{3668}{61}\\ 6z=\frac{3782-3668}{61}\\ z=\frac{19}{61}\end{array}$

Substituting the value of $z$ in second equation,

  $\begin{array}{l}-x+2(\frac{524}{61})+\frac{19}{61}=11\\ -x+\frac{1048}{61}+\frac{19}{61}=11\\ -x=\frac{11}{1}-\frac{1067}{61}\\ -x=\frac{671-1067}{61}\\ -x=-\frac{396}{61}\\ x=\frac{396}{61}\end{array}$

System $C$ has one solution $(\frac{396}{61},\frac{524}{61},\frac{19}{61})$ .

Thus, the correct order is $B<C<A$ .