Is \{(1,4, –6), (1,5, 8), (2, 1, 1), (0, 1,0)\}

interdicoxd 2022-01-06 Answered
Is \(\displaystyle{\left\lbrace{\left({1},{4},–{6}\right)},{\left({1},{5},{8}\right)},{\left({2},{1},{1}\right)},{\left({0},{1},{0}\right)}\right\rbrace}\) a linearly independent subset of \(\displaystyle{R}^{{3}}\)?

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Expert Answer

usumbiix
Answered 2022-01-07 Author has 1667 answers
Consider the set of Vectors
\(\displaystyle{S}={\left\lbrace{\left({1},{4},{6}\right)},{\left({1},{5},{8}\right)},{\left({2},{1},{1}\right)},{\left({0},{1},{0}\right)}\right\rbrace}\)
Our objective is to determine Whether the set S is linear independent subset of Vector Space \(\displaystyle{R}^{{3}}\)
No, S is not linearly independent subset of Vector Space \(\displaystyle{R}^{{3}}\)
We know,
Dimension of Finite Dimensional Vector Space be n.
And Basis the vector space be B
So, Using Converse of Replacement Theorem
\(\displaystyle{\left|{B}\right|}={n}\)
Let linear independent subset of Vector Space be L, it contains m elements
And Generating subset of Vector Space be G
So, Using Replacement Theorem
\(\displaystyle{\left|{L}\right|}\le{\left|{G}\right|}\)
So, \(\displaystyle{\left|{L}\right|}\le{n}={\left|{B}\right|}\)
Hence, \(\displaystyle{\left|{S}\right|}={4}{>}\dim{\left({R}^{{3}}\right)}={3}\)
So,
\(\displaystyle{S}={\left\lbrace{\left({1},{4},{6}\right)},{\left({1},{5},{8}\right)},{\left({2},{1},{1}\right)},{\left({0},{1},{0}\right)}\right\rbrace}\) is not linearly independent.
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