Answer to Question #346039 in Linear Algebra for Beza

Question #346039

Find the solution x1+2x2-3x3+x4=0

3x1-x2+5x-3x4=0

2x1+x2+x4=0


1
Expert's answer
2022-05-30T15:15:00-0400

Write down the augmented matrix


"A=\\begin{pmatrix}\n 1 & 2 & -3 & 1 &| & 0 \\\\\n 3 & -1 & 5 & -1 &| & 0 \\\\\n 2 & 1 & 0 & 1 &| & 0 \\\\\n\\end{pmatrix}"

"R_2=R_2-3R_1"


"\\begin{pmatrix}\n 1 & 2 & -3 & 1 &| & 0 \\\\\n 0 & -7 & 14 & -4 &| & 0 \\\\\n 2 & 1 & 0 & 1 &| & 0 \\\\\n\\end{pmatrix}"

"R_3=R_3-2R_1"


"\\begin{pmatrix}\n 1 & 2 & -3 & 1 &| & 0 \\\\\n 0 & -7 & 14 & -4 &| & 0 \\\\\n 0 & -3 & 6 & -1 &| & 0 \\\\\n\\end{pmatrix}"

"R_2=-R_2\/7"


"\\begin{pmatrix}\n 1 & 2 & -3 & 1 &| & 0 \\\\\n 0 & 1 & -2 & 4\/7 &| & 0 \\\\\n 0 & -3 & 6 & -1 &| & 0 \\\\\n\\end{pmatrix}"

"R_1=R_1-2R_2"


"\\begin{pmatrix}\n 1 & 0 & 1 & -1\/7 &| & 0 \\\\\n 0 & 1 & -2 & 4\/7 &| & 0 \\\\\n 0 & -3 & 6 & -1 &| & 0 \\\\\n\\end{pmatrix}"

"R_3=R_3+3R_2"


"\\begin{pmatrix}\n 1 & 0 & 1 & -1\/7 &| & 0 \\\\\n 0 & 1 & -2 & 4\/7 &| & 0 \\\\\n 0 & 0 & 0 & 5\/7 &| & 0 \\\\\n\\end{pmatrix}"

"R_3=7R_3\/5"


"\\begin{pmatrix}\n 1 & 0 & 1 & -1\/7 &| & 0 \\\\\n 0 & 1 & -2 & 4\/7 &| & 0 \\\\\n 0 & 0 & 0 & 1 &| & 0 \\\\\n\\end{pmatrix}"


"R_1=R_1+R_3\/7"


"\\begin{pmatrix}\n 1 & 0 & 1 & 0 &| & 0 \\\\\n 0 & 1 & -2 & 4\/7 &| & 0 \\\\\n 0 & 0 & 0 & 1 &| & 0 \\\\\n\\end{pmatrix}"

"R_2=R_2-4R_3\/7"


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

Take "x_3=t, t\\in \\R."

Then "x_1=-t, x_2=2t, x_3=t, x_4=0, t\\in \\R"



"\\{-t, 2t, t, 0\\}, t\\in \\R"


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