## Elementary Linear Algebra 7th Edition

(a) $S$ spans $R^3$. (b) $S$ is linearly independent set of vectors. (c) $W$ is a basis for $R^3$.
Let $S$ be given by $$S=\{(4,0,1),(0,-3,2),(5,10,0)\}.$$ (a) Consider $u\in R^3$ such that $$u=(x,y,z)=a(4,0,1)+b(0,-3,2)+c(5,10,0), \quad a,b,c\in R.$$ Which yields the following system of equations \begin{align*} 4a+5c&=x\\ -3b+10c&=y\\ a+2b&=z. \end{align*} The coefficient matrix $$\left[ \begin {array}{ccc} 4&0&5\\ 0&-3&10\\ 1&2\end {array} \right]$$ has non zero determinant and hence there exist a unique solution for the above system and hence one can calculate $a,b,c$. Therefore, $S$ spans $R^3$. (b) Assume that $$a(4,0,1)+b(0,-3,2)+c(5,10,0)=(0,0,0), \quad a,b,c\in R.$$ Which yields the following system of equations \begin{align*} a+11b+2c&=0\\ -5a+6b+3c&=0\\ 4a-b+5c&=0. \end{align*} The coefficient matrix $$\left[ \begin {array}{ccc} 4&0&5\\ 0&-3&10\\ 1&2\end {array} \right]$$ has non zero determinant and hence there exist a unique solution for the above system, that is, the trivial solution $a=0,b=0,c=0$. Hence $S$ is linearly independent set of vectors. (c) Since $W$ is linearly independent set and spans $R^3$, then it is a basis for $R^3$.