Question 8.7.1: Use Cramer's rule to solve the system 3x1 + 2x2 + x3 = 7 x1 ......
Use Cramer’s rule to solve the system
\begin{array}{r} 3 x_{1}+2 x_{2}+x_{3}=7 \\ x_{1}-x_{2}+3 x_{3}=3 \\ 5 x_{1}+4 x_{2}-2 x_{3}=1. \end{array}
The solution requires the evaluation of four determinants:
\begin{aligned} \operatorname{det} \mathrm{A}=\left|\begin{array}{rrr} 3 & 2 & 1 \\ 1 & -1 & 3 \\ 5 & 4 & -2 \end{array}\right|=13, \quad \operatorname{det} \mathrm{A}_{1}=\left|\begin{array}{rrr} 7 & 2 & 1 \\ 3 & -1 & 3 \\ 1 & 4 & -2 \end{array}\right|=-39, \\ \operatorname{det} \mathrm{A}_{2}=\left|\begin{array}{rrr} 3 & 7 & 1 \\ 1 & 3 & 3 \\ 5 & 1 & -2 \end{array}\right|=78, \quad \operatorname{det} \mathrm{A}_{3}=\left|\begin{array}{rrr} 3 & 2 & 7 \\ 1 & -1 & 3 \\ 5 & 4 & 1 \end{array}\right|=52. \end{aligned}
Thus, (6) gives
x_1=\frac{\operatorname{det} A _1}{\operatorname{det} A }, \quad x_2=\frac{\operatorname{det} A _2}{\operatorname{det} A }, \ldots, \quad x_n=\frac{\operatorname{det} A _n}{\operatorname{det} A } (6)
x_{1}=\frac{\operatorname{det} \mathrm{A}_{1}}{\operatorname{det} \mathrm{A}}=-3, \quad x_{2}=\frac{\operatorname{det} \mathrm{A}_{2}}{\operatorname{det} \mathrm{A}}=6, \quad x_{3}=\frac{\operatorname{det} \mathrm{A}_{3}}{\operatorname{det} \mathrm{A}}=4.