Answer to Question #136020 in Analytic Geometry for Yolanda

Question

use matrices and elementary row operations to solve 5x-3y+2z=13, 2x-y-3z=1,4x-2y+4z=12

Answer 1

$\begin{aligned}\left(\begin{array}{rrr|r} 5 & -3 & 2 & 13\\ 2& -1 & -3 & 1 \\ 4 & -2 & 4 & 12\\ \end{array}\right) &{\stackrel{2R_2}{\rightarrow}} \left(\begin{array}{rrr|r} 5 & -3 & 2 & 13\\ 4& -2 & -6 & 2 \\ 4 & -2 & 4 & 12\\ \end{array}\right) \\&{\stackrel{R_3 - R_2}{\rightarrow}} \left(\begin{array}{rrr|r} 5 & -3 & 2 & 13\\ 4& -2 & -6 & 2 \\ 0 & 0 & 10 & 10 \end{array}\right) \\&{\stackrel{R_1 \times 4, R_2 \times 5}{\rightarrow}}\left(\begin{array}{rrr|r} 20 & -12 & 8 & 52\\ 20& -10 & -30& 10 \\ 0 & 0 & 10 & 10 \end{array}\right)\\&{\stackrel{R_2 - R_1}{\rightarrow}}\left(\begin{array}{rrr|r} 20 & -12 & 8& 52\\ 0& 2 & -38& -42 \\ 0 & 0 & 10 & 10 \end{array}\right) \end{aligned}\\ \textsf{By back substitution,}\\ 10z = 10, \Rightarrow z = 1\\ 2y - 38z = -42, 2y - 38(1) = -42, \\ 2y = 38 - 42, 2y = -4, y = -2\\ 20x - 12y + 8z = 52 \\ 20x - 12(-2) + 8(1) = 52 \\ 20x + 24 + 8 = 52 \\ 20x + 32 = 52\\ 20x = 20 \Rightarrow x = 1\\ \therefore x = 1, y = - 2 \hspace{0.1cm}\textsf{and}\hspace{0.1cm} z = 1 \\\textsf{are the solutions to}\\\textsf{the simultaneous equations}$