Answer to Question #202600 in Linear Algebra for Amna

Question #202600

Evaluate the following system of equation by Gaussian Elimination method

x1+5x2+2x3=9

x1+x2+7x3=6

-3x2+4x3=-2



1
Expert's answer
2021-06-03T17:01:50-0400

Augmented matrix


"A=\\begin{pmatrix}\n 1 & 5 & 2 & \\ \\ 9\\\\\n 1 & 1 & 7 & \\ \\ 6\\\\\n 0 & -3 & 4 & \\ \\ -2\\\\\n\\end{pmatrix}"

"R_2=R_2-R_1"


"\\begin{pmatrix}\n 1 & 5 & 2 & \\ \\ 9\\\\\n 0 & -4 & 5 & \\ \\ -3\\\\\n 0 & -3 & 4 & \\ \\ -2\\\\\n\\end{pmatrix}"

"R_2=-\\dfrac{R_2}{4}"


"\\begin{pmatrix}\n 1 & 5 & 2 & \\ \\ 9\\\\\n 0 & 1 & -5\/4 & \\ \\ 3\/4\\\\\n 0 & -3 & 4 & \\ \\ -2\\\\\n\\end{pmatrix}"

"R_1=R_1-5R_2"


"\\begin{pmatrix}\n 1 & 0 & 33\/4 & \\ \\ 21\/4\\\\\n 0 & 1 & -5\/4 & \\ \\ 3\/4\\\\\n 0 & -3 & 4 & \\ \\ -2\\\\\n\\end{pmatrix}"

"R_3=R_3+3R_2"


"\\begin{pmatrix}\n 1 & 0 & 33\/4 & \\ \\ 21\/4\\\\\n 0 & 1 & -5\/4 & \\ \\ 3\/4\\\\\n 0 & 0 & 1\/4 & \\ \\ 1\/4\\\\\n\\end{pmatrix}"

"R_3=4R_3"


"\\begin{pmatrix}\n 1 & 0 & 33\/4 & \\ \\ 21\/4\\\\\n 0 & 1 & -5\/4 & \\ \\ 3\/4\\\\\n 0 & 0 & 1 & \\ \\ 1\\\\\n\\end{pmatrix}"

"R_1=R_1-\\dfrac{33R_3}{4}"


"\\begin{pmatrix}\n 1 & 0 & 0 & \\ \\ -3\\\\\n 0 & 1 & -5\/4 & \\ \\ 3\/4\\\\\n 0 & 0 & 1 & \\ \\ 1\\\\\n\\end{pmatrix}"

"R_2=R_2+\\dfrac{5R_3}{4}"


"\\begin{pmatrix}\n 1 & 0 & 0 & \\ \\ -3\\\\\n 0 & 1 & 0 & \\ \\ 2\\\\\n 0 & 0 & 1 & \\ \\ 1\\\\\n\\end{pmatrix}"


"x_1=-3, x_2=2, x_3=1"

"(-3,2,1)"



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