## Elementary Linear Algebra 7th Edition

(a) A basis for the solution space is \left[\begin{aligned}-2\\1 \\0 \end{aligned}\right] . (b) The dimension of the solution space is $1$.
The coefficient matrix is given by $$\left[ \begin {array}{ccc} 1&2&-4\\ -3&-6&-12 \end {array} \right] .$$ The reduced row echelon form is $$\left[ \begin {array}{ccc} 1&2&0\\ 0&0&1 \end {array} \right] .$$ The corresponding system is \begin{aligned} x +2y&=0\\ z&=0\\ \end{aligned}. The solution of the above system is $x=-2t$,$y=t$, $z=0$. 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}-2t\\t \\0 \end{aligned}\right] =t\left[\begin{aligned}-2\\1 \\0 \end{aligned}\right] . (a) A basis for the solution space is \left[\begin{aligned}-2\\1 \\0 \end{aligned}\right] . (b) The dimension of the solution space is $1$.