Elementary Linear Algebra 7th Edition

(a) $S$ does not span $R^3$. (b) $S$ is linearly dependent set of vectors. (c) $S$ is not a basis for $R^3$.
Let $S$ be given by $$S=\left\{\left(-\frac{1}{2}, \frac{3}{4},-1\right),(5,2,3),(-4,6,-8)\right\}.$$ (a) Consider $u\in R^3$ such that $$u=(x,y,z)=a\left(-\frac{1}{2}, \frac{3}{4},-1\right)+b(5,2,3)+c(-4,6,-8), \quad a,b,c\in R.$$ Which yields the following system of equations \begin{align*} -\frac{1}{2}a+5b-4c&=x\\ \frac{3}{4}a+2b+6c&=y\\ -a+3b-8&=z. \end{align*} The coefficient matrix $$\left[ \begin {array}{ccc} -\frac{1}{2}&5&-4\\ \frac{3}{4}&2&6\\-1&3&-8\end {array} \right]$$ has zero determinant and hence there is no solution for the above system and hence, $S$ does not span $R^3$. (b) Assume that $$a\left(-\frac{1}{2}, \frac{3}{4},-1\right)+b(5,2,3)+c(-4,6,-8)=(0,0,0), \quad a,b,c\in R.$$ Which yields the following system of equations \begin{align*} -\frac{1}{2}a+5b-4c&=0\\ \frac{3}{4}a+2b+6c&=0\\ -a+3b-8&=0. \end{align*} The coefficient matrix $$\left[ \begin {array}{ccc} -\frac{1}{2}&5&-4\\ \frac{3}{4}&2&6\\-1&3&-8\end {array} \right]$$ has zero determinant and hence there are non trivial solutions for the above system, that is, $S$ is linearly dependent set of vectors. (c) Since $S$ is not linearly independent set and does not span $R^3$, then it is not a basis for $R^3$.