Question 8.2.5: Solve the linear system in Example 2 using (a) Gaussian elim......

(a) Using row operations on the augmented matrix of the system, we obtain:

$\left(\begin{array}{rrr|r} 2 & 6 & 1 & 7 \\ 1 & 2 & -1 & -1 \\ 5 & 7 & -4 & 9 \end{array}\right) \stackrel{R_1 \leftrightarrow R_2}{\Rightarrow}\left(\begin{array}{rrr|r} 1 & 2 & -1 & -1 \\ 2 & 6 & 1 & 7 \\ 5 & 7 & -4 & 9 \end{array}\right)$

$\stackrel{\substack{-2 R_1+R_2 \\ -5 R_1+R_3}}{\Rightarrow}\left(\begin{array}{rrr|r} 1 & 2 & -1 & -1 \\ 0 & 2 & 3 & 9 \\ 0 & -3 & 1 & 14 \end{array}\right) \stackrel{\frac{1}{2} R_2}{\Rightarrow}\left(\begin{array}{rrr|r} 1 & 2 & -1 & -1 \\ 0 & 1 & \frac{3}{2} & \frac{9}{2} \\ 0 & -3 & 1 & 14 \end{array}\right)$

$\stackrel{3 R_2+R_3}{\Rightarrow}\left(\begin{array}{rrr|r} 1 & 2 & -1 & -1 \\ 0 & 1 & \frac{3}{2} & \frac{9}{2} \\ 0 & 0 & \frac{11}{2} & \frac{55}{2} \end{array}\right) \stackrel{\frac{2}{11} R_3}{\Rightarrow}\left(\begin{array}{rrr|r} 1 & 2 & -1 & -1 \\ 0 & 1 & \frac{3}{2} & \frac{9}{2} \\ 0 & 0 & 1 & 5 \end{array}\right) .$

The last matrix is in row-echelon form and represents the system

$\begin{aligned} x_1+2 x_2- \quad x_3 & =-1 \\ x_2+\frac{3}{2} x_3 & =\frac{9}{2} \\ x_3 & =5 . \end{aligned}$

Substituting $x_3=5$ into the second equation gives $x_2=-3$. Substituting both these values back into the first equation finally yields $x_1=10$.

(b) We start with the last matrix above. Since the first entries in the second and third rows are ones, we must, in turn, make the remaining entries in the second and third columns zeros:

The last matrix is in reduced row-echelon form. Bearing in mind what the matrix means in terms of equations, we see that the solution of the system is $x_1=10, x_2=-3, x_3=5.$