Answer to Question #114211 in Linear Algebra for Stacey

Question

can the system
3xy-4y-z=5
3x+2y-z=0
x-y+z=1
be solved by using Cramer's rule? if not, why? if yes, solve it.

Answer 1

$\begin{vmatrix} 3 & -4 & -1 \\ 3 & 2 & -1 \\ 1 & -1 &1 \end{vmatrix} =6+4+3+2+12-3=24$

$\begin{vmatrix} 5 & -4 & -1 \\ 0 & 2 & -1 \\ 1 & -1 &1 \end{vmatrix} =10+4+0+2+0-5=11$

$\begin{vmatrix} 3 & 5 & -1 \\ 3 & 0 & -1 \\ 1 & 1 &1 \end{vmatrix} =0-3-5+0-15+3=-20$

$\begin{vmatrix} 3 & -4 & 5 \\ 3 & 2 & 0 \\ 1 & -1 &1 \end{vmatrix} =6-15+0-10+12-0=-7$

Hence, $x=\frac{11}{24}, y=-\frac{20}{24}, z=-\frac{7}{24}$