Find the basis for the subspace of R^3 spanned by S = {(81, 63, 12), (27, 21, 4), (9, 7, 1)). {(0, 0, 1), (9, 0, 0), (0, 7, 0); {(1, 0, 0), (0, 0, 1), (0, 7, 0); {(0, 0, 1), (9,7, 0); {9,0,0), (0,7, 0);

Answer with explanation:A basis of a subspace in R³ is the set of vectors that spans the subspace means the set of vectors, and is linearly  independent. The meaning of span is if any set of vectors can be written as linear combination of other vector.For example , {a, b,c}, {p,q,r} and {x,y,z} are three set of vectors.If ,{a,b,c} can be wriiten as {pm +x n, q m+y n, r m+z n}, then we say that  {a,b,c} spans the set of vectors.1. S={(81, 63, 12), (27, 21, 4), (9, 7, 1)}Writing the set of vectors in matrix form   [tex]\left[\begin{array}{ccc}81&63&12\\27&21&4\\9&7&1\end{array}\right]\\\\R_{2} \Rightarrow 3R_{2}-R_{1}\\\\R_{3} \Rightarrow 9R_{3}-R_{1}\\\\\left[\begin{array}{ccc}81&63&12\\0&0&0\\0&0&0\end{array}\right][/tex]→Basis of S=1As all the three vectors are linearly dependent.So, Basis=1.2. {(0, 0, 1), (9, 0, 0), (0, 7, 0)}If you write these three vectors in matrix form , all the three vectors are Linearly Independent.Any of the one vectors can't be written as a linear combination of otherSo, Basis =33. {(1, 0, 0), (0, 0, 1), (0, 7, 0)}All the three vectors are Linearly Independent.So, Basis=34. {(0, 0, 1), (9,7, 0)}These two vectors can't be written as a linear combination of other.So these two vectors are Independent.So,Basis=25. {9,0,0), (0,7, 0)}These two vectors can't be written as linear combination of each other.So these two vectors are Independent.So,Basis=2