College Algebra (10th Edition)

$x=8/7, \ y=5/7, \ z=17/7$ or $(8/7, \ 5/7, \ 17/7)$
Writing the system in matrix form, $AX=B,$ the solution is $X=A^{-1}B$. Here, $A=\left[\begin{array}{rrr}{3}&{3}&{1}\\{1}&{2}&{1}\\{2}&{-1}&{1}\end{array}\right]$, as in exercise $40$. There, we found $A^{-1}=\left[\begin{array}{rrr} {\displaystyle \frac{3}{7}}&{-\displaystyle \frac{4}{7}}&{\displaystyle \frac{1}{7}}\\ {\displaystyle \frac{1}{7}}&{\displaystyle \frac{1}{7}}&{-\displaystyle \frac{2}{7}}\\ {-\displaystyle \frac{5}{7}}&{\displaystyle \frac{9}{7}}&{\displaystyle \frac{3}{7}}\end{array}\right]$ With $B=\left[\begin{array}{l} 8\\ 5\\ 4 \end{array}\right]$, the solution, $X=A^{-1}B$ equals $X=\left[\begin{array}{l} \frac{1}{7}(24-20+4)\\ \frac{1}{7}(8+5-8)\\ \frac{1}{7}(-40+45+12) \end{array}\right]=\left[\begin{array}{l} 8/7\\ 5/7\\ 17/7 \end{array}\right]$ $x=8/7, \ y=5/7, \ z=17/7$ or $(8/7, \ 5/7, \ 17/7)$