## Linear Algebra and Its Applications (5th Edition)

a. no, 3 b. infinitely many c. yes, $v_{1}+v_{2}=w$
a. The set $S=\{v_{1},v_{2},v_{3}\}$ contains the three given vectors. $w$ is not equal to any of them, so $w\not\in S$. b. Span$S=\{av_{1}+bv_{2}+cv_{3}, a,b,c\in \mathbb{R}\}$ has infinitely many vectors (there are infinitely many ways of choosing a,b, and c) c. If $w\in$Span$S$, then there exist $a,b,c\in \mathbb{R}$ such that $w=av_{1}+bv_{2}+cv_{3}.$ $a, b,$ and $c$ are solutions (if they exist) to a linear system with the augmented matrix $\left[\begin{array}{lllll} 1 & 2 & 4 & | & 3\\ 0 & 1 & 2 & | & 1\\ -1 & 3 & 6 & | & 2 \end{array}\right]\left\{\begin{array}{l} .\\ .\\ \leftarrow R_{3}+R_{1}. \end{array}\right.$ $\left[\begin{array}{lllll} 1 & 2 & 4 & | & 3\\ 0 & 1 & 2 & | & 1\\ 0 & 5 & 10 & | & 5 \end{array}\right]\left\{\begin{array}{l} .\\ .\\ \leftarrow (\frac{1}{5}R_{3}). \end{array}\right.$ $\left[\begin{array}{lllll} 1 & 2 & 4 & | & 3\\ 0 & 1 & 2 & | & 1\\ 0 & 1 & 2 & | & 1 \end{array}\right]\left\{\begin{array}{l} .\\ .\\ \leftarrow R_{3}-R_{2}. \end{array}\right.$ $\left[\begin{array}{lllll} 1 & 2 & 4 & | & 3\\ 0 & 1 & 2 & | & 1\\ 0 & 0 & 0 & | & 0 \end{array}\right]$ (consistent, solutions exist) We take any $c\in \mathbb{R},$ back substituing, $b=1-2c$ back substituing, $a=3-2(1-2c)-4c=1,$ So, there exist a,b,c such that $w\in$Span$S$. (For example, c=0$\Rightarrow v_{1}+v_{2}=w$)