A Solution Method

EXAMPLE 2.1.4   Let us solve the linear system $x+ 7y + 3 z = 11, \; x + y + z = 3,$ and $4 x + 10 y - z = 13.$
Solution:

1. The above linear system and the linear system
   

   $$x+ y + z = 3 {\mbox{ Interchange the first two equations.}}$$

   $$x + 7 y + 3 z = 11$$ (2.1.2)

   $$4 x + 10 y - z = 13$$

   
   have the same set of solutions. (why?)
2. Using the $1^{\mbox{st}}$ equation, we eliminate $x$ from $2^{\mbox{nd}}$ and $3^{\mbox{rd}}$ equation to get the linear system
   

   $$x+ y + z = 3$$

   $$6 y + 2 z = 8 {\mbox{ (obtained by subtracting the first }}$$

   $${\mbox{ equation from the second equation.)}}$$

   $$6 y - 5z = 1 {\mbox{ (obtained by subtracting }} 4 {\mbox{ times the first equation}}$$

   $${\mbox{from the third equation.) }}$$ (2.1.3)

   
   This system and the system (2.1.2) has the same set of solution. (why?)
3. Using the $2^{\mbox{cd}}$ equation, we eliminate $y$ from the last equation of system (2.1.3) to get the system
   

   $$x+ y + z = 3$$

   $$6 y + 2 z = 8$$

   $$7z = 7 {\mbox{ obtained by subtracting the third equation}}$$

   $${\mbox{ from the second equation.}}$$ (2.1.4)

   
   which has the same set of solution as the system (2.1.3). (why?)
4. The system (2.1.4) and system
   

   $$x+ y + z = 3$$

   $$3 y + z = 4 {\mbox{ divide the second equation by }} 2$$

   $$z = 1 {\mbox{ divide the third equation by }} 7$$ (2.1.5)

   
   has the same set of solution. (why?)
5. Now, $z = 1$ implies $y = \displaystyle \frac{4-1}{3} = 1$ and $x = 3 - (1 +1) = 1.$ Or in terms of a vector, the set of solution is $\{ \;(x, y, z)^t \; : (x,y,z) = (1, 1, 1) \}.$