Answer to Question #194546 in Linear Algebra for Zeeshan

Question #194546

Using gaussian elimination method find all solutions to the following system of linear equations.

2x₂ + 3x₃ + 4x₄ = 1

x₁ - 3x₂ + 4x₃ + 5x₄ = 2

-3x₁ + 10x₂ - 6x₃ -7x₄ = -4


1
Expert's answer
2021-05-19T12:58:05-0400
"\\begin{bmatrix}\n 0 & 2 & 3 & 4 & \\ \\ \\ 1 \\\\\n 1 & -3 &4 & 5 & \\ \\ \\ 2 \\\\\n-3 & 10 & -6 & -7 & \\ \\ \\ -4 \\\\\n\\end{bmatrix}"



Swap rows 1 and 2


"\\begin{bmatrix}\n 1 & -3 & 4 & 5 & \\ \\ \\ 2 \\\\\n 0 & 2 & 3 & 4 & \\ \\ \\ 1 \\\\\n-3 & 10 & -6 & -7 & \\ \\ \\ -4 \\\\\n\\end{bmatrix}"

"R_3=R_3+3R_1"


"\\begin{bmatrix}\n 1 & -3 & 4 & 5 & \\ \\ \\ 2 \\\\\n 0 & 2 & 3 & 4 & \\ \\ \\ 1 \\\\\n0 & 1 & 6 & 8 & \\ \\ \\ 2 \\\\\n\\end{bmatrix}"



"R_2=R_2\/2"


"\\begin{bmatrix}\n 1 & -3 & 4 & 5 & \\ \\ \\ 2 \\\\\n 0 & 1 & 3\/2 & 2 & \\ \\ \\ 1\/2 \\\\\n0 & 1 & 6 & 8 & \\ \\ \\ 2 \\\\\n\\end{bmatrix}"


"R_1=R_1+3R_2"


"\\begin{bmatrix}\n 1 & 0 & 17\/2 & 11 & \\ \\ \\ 7\/ 2 \\\\\n 0 & 1 & 3\/2 & 2 & \\ \\ \\ 1\/2 \\\\\n0 & 1 & 6 & 8 & \\ \\ \\ 2 \\\\\n\\end{bmatrix}"



"R_3=R_3-R_2"

"\\begin{bmatrix}\n 1 & 0 & 17\/2 & 11 & \\ \\ \\ 7\/ 2 \\\\\n 0 & 1 & 3\/2 & 2 & \\ \\ \\ 1\/2 \\\\\n0 & 0 & 9\/2 & 6 & \\ \\ \\ 3\/2 \\\\\n\\end{bmatrix}"


"R_3=2R_3\/9"


"\\begin{bmatrix}\n 1 & 0 & 17\/2 & 11 & \\ \\ \\ 7\/ 2 \\\\\n 0 & 1 & 3\/2 & 2 & \\ \\ \\ 1\/2 \\\\\n0 & 0 & 1 & 4\/3 & \\ \\ \\ 1\/3 \\\\\n\\end{bmatrix}"


"R_1=R_1-17R_3\/2"


"\\begin{bmatrix}\n 1 & 0 & 0 & -1\/3 & \\ \\ \\ 2\/ 3 \\\\\n 0 & 1 & 3\/2 & 2 & \\ \\ \\ 1\/2 \\\\\n0 & 0 & 1 & 4\/3 & \\ \\ \\ 1\/3 \\\\\n\\end{bmatrix}"


"R_2=R_2-3R_3\/2"

"\\begin{bmatrix}\n 1 & 0 & 0 & -1\/3 & \\ \\ \\ 2\/ 3 \\\\\n 0 & 1 & 0 & 0 & \\ \\ \\ 0 \\\\\n0 & 0 & 1 & 4\/3 & \\ \\ \\ 1\/3 \\\\\n\\end{bmatrix}"




"\\begin{matrix}\n x_1-\\dfrac{1}{3}x_4=\\dfrac{2}{3} \\\\ \\\\\n x_2=0 \\\\ \\\\\n x_3+\\dfrac{4}{3}x_4=\\dfrac{1}{3} \\\\ \\\\\n\\end{matrix}"

Solution set:

"\\begin{matrix}\n x_1=\\dfrac{2}{3} +\\dfrac{1}{3}x_4 \\\\ \\\\\n x_2=0 \\\\ \\\\\n x_3=\\dfrac{1}{3}-\\dfrac{4}{3}x_4 \\\\ \\\\\nx_4,\\ free \\\\ \\\\\n\\end{matrix}"

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