## Elementary Linear Algebra 7th Edition

(a) A basis for the solution space is \left[\begin{aligned}2\\1 \\0 \end{aligned}\right], \left[\begin{aligned}-3\\0\\1 \end{aligned}\right]. (b) The dimension of the solution space is $2$.
The coefficient matrix is given by $$\left[ \begin {array}{ccc} 1&-2&3\\ -3&6&-9 \end {array} \right] .$$ The reduced row echelon form is $$\left[ \begin {array}{ccc} 1&-2&3\\ 0&0&0 \end {array} \right] .$$ The corresponding system is \begin{aligned} x -2y+3z&=0\\ \end{aligned}. The solution of the above system is $x=2s-3t$,$y=s$, $z=t$. This means that the solution space of $Ax = 0$ consists of the vectors on the form x= \left[\begin{aligned} x\\ y\\z \end{aligned}\right]= \left[\begin{aligned}2s-3t\\s \\t \end{aligned}\right] =s\left[\begin{aligned}2\\1 \\0 \end{aligned}\right] + t \left[\begin{aligned}-3\\0\\1 \end{aligned}\right] . (a) A basis for the solution space is \left[\begin{aligned}2\\1 \\0 \end{aligned}\right], \left[\begin{aligned}-3\\0\\1 \end{aligned}\right]. (b) The dimension of the solution space is $2$.