Question 8.2.2: Solve 2x1 + 6x2 + x3 = 7 x1 + 2x2 - x3 = -1 5x1 + 7x2 - 4x3 ......

We begin by interchanging the first and second rows:

\begin{aligned} x_{1}+2 x_{2}-x_{3} & =-1 \\ 2 x_{1}+6 x_{2}+x_{3} & =7 \\ 5 x_{1}+7 x_{2}-4 x_{3} & =9 . \end{aligned}

Our goal now is to eliminate x_{1} from the second and third equations. If we add to the second equation -2 times the first equation, we obtain the equivalent system

\begin{aligned} x_{1}+2 x_{2}-x_{3} & =-1 \\ 2 x_{2}+3 x_{3} & =9 \\ 5 x_{1}+7 x_{2}-4 x_{3} & =9 . \end{aligned}

By adding to the third equation -5 times the first equation, we get a new equivalent system:

\begin{aligned} x_{1}+2 x_{2}-x_{3} & =-1 \\ 2 x_{2}+3 x_{3} & =9 \\ -3 x_{2}+x_{3} & =14. \end{aligned}

We are now going to use the second equation to eliminate the variable x_{2} from the first and third equations. To simplify matters, let us multiply the second equation by \frac{1}{2} :

\begin{aligned} x_{1}+2 x_{2}-x_{3} & =-1 \\ x_{2}+\frac{3}{2} x_{3} & =\frac{9}{2} \\ -3 x_{2}+x_{3} & =14 . \end{aligned}

Adding to the first equation -2 times the second equation yields

\begin{aligned} x_{1} \quad  -4 x_{3} & =-10 \\ x_{2}+\frac{3}{2} x_{3} & =\frac{9}{2} \\ -3 x_{2}+x_{3} & =14 . \end{aligned}

Next, by adding 3 times the second equation to the third equation we get

\begin{aligned} x_{1} \quad  -4 x_{3} & =-10 \\ x_{2}+\frac{3}{2} x_{3} & =\frac{9}{2} \\ \frac{11}{2} x_{3} & =\frac{55}{2} \end{aligned}

We shall use the last equation to eliminate the variable x_{3} from the first and second equations. To this end, we multiply the third equation by \frac{2}{11} :

\begin{aligned} x_{1} \quad-4 x_{3} & =-10 \\ x_{2}+\frac{3}{2} x_{3} & =\frac{9}{2} \\ x_{3} & =5 . \end{aligned}

At this point we could use back-substitution; that is, substitute the value x_{3}=5 back into the remaining equations to determine x_{1} and x_{2}. However, by continuing with our systematic elimination, we add to the second equation -\frac{3}{2} times the third equation:

\begin{aligned} x_{1} \quad-4 x_{3} & =-10 \\ x_{2} \quad & =-3 \\ x_{3} & =5 . \end{aligned}

Finally, by adding to the first equation 4 times the third equation, we obtain

\begin{array}{crc} & x_{1} \qquad &=10 \\ & x_{2} \quad&=-3 \\ & x_{3}&=5 \text {. } \end{array}

It is now apparent that x_{1}=10, x_{2}=-3, x_{3}=5 is the solution of the original system. The answer written as the ordered triple (10,-3,5) means that the planes represented by the three equations in the system intersect at a point as in Figure 8.2.2(a).