Answer to Question #229491 in Algebra for Ann

Question

Use Cramer's rule to solve the unknown of the system: x + 2y =2 - z,3x - 6y = 2 - 2x,2x =8 + z

i.x =2, y=1/3, z=-2

ii.x =-3,y =2, z=-1/2

iii.x =4,y =-1/2, z =-3

iv.x =3, y =1/2, z=-2

Answer 1

$\displaystyle x + 2y +z=2 \\5x-6y=2 \\2x-z=8 \\\text{We can write this in matrix form} \\\begin{pmatrix} 1&2&1\\ 5&-6&0\\ 2&0&-1 \end{pmatrix}\begin{pmatrix} x\\y\\z \end{pmatrix}=\begin{pmatrix} 2\\2\\8 \end{pmatrix} \\\text{Next,we use the Cramer's rule to to solve the system of equations} \\detA = 1\begin{vmatrix} 5&-6\\ 2&0 \end{vmatrix}-1\begin{vmatrix} 1&2\\ 5&-6 \end{vmatrix}=28 \\\delta_x=\begin{vmatrix} 2&2&1\\ 2&-6&0\\ 8&0&-1 \end{vmatrix}=64 \\\delta_y=\begin{vmatrix} 1&2&1\\ 5&2&0\\ 2&8&-1 \end{vmatrix}=44 \\\delta_z=\begin{vmatrix} 1&2&2\\ 5&-6&2\\ 2&0&8 \end{vmatrix}=-96 \\\text{Therefore using Kramer's rule we have that, $\overline{x}=\frac{64}{28}=2.29$} \\\overline{y} = \frac{44}{28}=1.57 \\\overline{z} = \frac{-96}{28} =-3.42 \\\text{Hence, the values of x,y and z are 2, 2 and -3 approximately}$