Answer to Question #237087 in Linear Algebra for lavanya

Question #237087

Show that the set 𝑆 = {(1, 0, 1)(1,1, 0)(−1, 0, −1)} is linearly dependent in 𝑉3(𝑅)


1
Expert's answer
2021-09-14T11:25:00-0400

Let "\\vec v_1=\\begin{pmatrix}\n 1\\\\\n 0 \\\\\n1\n\\end{pmatrix}, \\vec v_2=\\begin{pmatrix}\n 1\\\\\n 1 \\\\\n0\n\\end{pmatrix}" and "\\vec v_3=\\begin{pmatrix}\n -1\\\\\n 0 \\\\\n-1\n\\end{pmatrix}."

We have to determine whether or not we can find real numbers "a_1, a_2, a_3" which are not all zero, such that


"a_1\\vec v_1+a_2\\vec v_2+a_3\\vec v_3=\\vec 0"

Substitute


"a_1\\begin{pmatrix}\n 1\\\\\n 0 \\\\\n1\n\\end{pmatrix}+a_2\\begin{pmatrix}\n 1\\\\\n 1 \\\\\n0\n\\end{pmatrix}+a_3\\begin{pmatrix}\n -1\\\\\n 0 \\\\\n-1\n\\end{pmatrix}=\\begin{pmatrix}\n 0\\\\\n 0 \\\\\n0\n\\end{pmatrix}"

Augmented matrix


"A=\\begin{pmatrix}\n 1 & 1 & -1 & & 0 \\\\\n 0 & 1 & 0 & & 0 \\\\\n 1 & 0 & -1 & & 0 \\\\\n\\end{pmatrix}"

"R_3=R_3-R_1"


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

"R_1=R_1-R_2"


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

"R_3=R_3+R_2"


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


If we take "a_3=1," then the solution is "(a_1, a_2, a_3)=(1, 0, 1)"

So this set of equations has a non–zero solution.

Therefore, set "\ud835\udc46 = \\{(1, 0, 1)(1,1, 0)(\u22121, 0, \u22121)\\}" is a linearly dependent in "V^3."



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