Question

Use Cramer's Rule to solve the system of linear equations.

xa + 2xc = 6
-3xa + 4xb + 6xc = 30
-xa - 2xb + 3xc = 8

Answer 1

$\begin{gathered} \Delta=\left|\begin{array}{ccc} x & 0 & 2 x \\ -3 x & 4 x & 6 x \\ -x & -2 x & 3 x \end{array}\right|=x\left(12 x^{2}+12 x^{2}\right)-0+2 x\left(6 x^{2}+4 x^{2}\right) \\ =24 x^{3}+20 x^{3}=44 x^{3} \\ \Delta_{1}=\left|\begin{array}{ccc} 6 & 0 & 2 x \\ 30 & 4 x & 6 x \\ 8 & -2 x & 3 x \end{array}\right|=6\left(12 x^{2}+12 x^{2}\right)-0+2 x(-60 x-32 x) \\ =144 x^{2}-184 x=4 x(36 x-46) \\\Delta_{2}=\left|\begin{array}{ccc} x & 6 & 2 x \\ -3 x & 30 & 6 x \\ -x & 8 & 3 x \end{array}\right| \\ =x(90 x-48 x)-6\left(-9 x^{2}+6 x^{2}\right)+2 x(-24 x+30 x) \\ =42 x+18 x^{2}+12 x^{2}=42 x+30 x^{2}=2 x(21+15 x) \\ \Delta_{3}=\left|\begin{array}{ccc} x & 0 & 6 \\ -3 x & 4 x & 30 \\ -x & -2 x & 8 \end{array}\right|=x(32 x+60 x)-0+6\left(6 x^{2}+4 x^{2}\right) \\ =92 x^{2}+60 x^{2}=152 x^{2} \end{gathered}$

$Now, a=\frac{\Delta_{1}}{\Delta}=\frac{4 x(36 x-46)}{44 x^{3}}=\frac{(36 x-46)}{11 x^{2}},\\b=\frac{\Delta_{2}}{\Delta}=\frac{2 x(21+15 x)}{44 x^{3}}= \frac{(21+15 x)}{22 x^{2}},\\ c=\frac{\Delta_{3}}{\Delta}=\frac{152 x^{2}}{44 x^{3}}=\frac{38}{11 x}$