 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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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.