In April 2025
Answer to Question #229491 in Algebra for Ann
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
"\\displaystyle \nx + 2y +z=2\n\\\\5x-6y=2\n\\\\2x-z=8\n\\\\\\text{We can write this in matrix form}\n\\\\\\begin{pmatrix}\n1&2&1\\\\\n5&-6&0\\\\\n2&0&-1\n\\end{pmatrix}\\begin{pmatrix}\nx\\\\y\\\\z\n\\end{pmatrix}=\\begin{pmatrix}\n2\\\\2\\\\8\n\\end{pmatrix}\n\\\\\\text{Next,we use the Cramer's rule to to solve the system of equations}\n\\\\detA = 1\\begin{vmatrix}\n5&-6\\\\\n2&0\n\\end{vmatrix}-1\\begin{vmatrix}\n1&2\\\\\n5&-6\n\\end{vmatrix}=28\n\\\\\\delta_x=\\begin{vmatrix}\n2&2&1\\\\\n2&-6&0\\\\\n8&0&-1\n\\end{vmatrix}=64\n\\\\\\delta_y=\\begin{vmatrix}\n1&2&1\\\\\n5&2&0\\\\\n2&8&-1\n\\end{vmatrix}=44\n\\\\\\delta_z=\\begin{vmatrix}\n1&2&2\\\\\n5&-6&2\\\\\n2&0&8\n\\end{vmatrix}=-96\n\\\\\\text{Therefore using Kramer's rule we have that, $\\overline{x}=\\frac{64}{28}=2.29$}\n\\\\\\overline{y} = \\frac{44}{28}=1.57\n\\\\\\overline{z} = \\frac{-96}{28} =-3.42\n\\\\\\text{Hence, the values of x,y and z are 2, 2 and -3 approximately}"
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