## Elementary Linear Algebra 7th Edition

(a) A basis for the solution space is \left[\begin{aligned}2 \\1\\0\\ 0 \end{aligned}\right], \left[\begin{aligned} 3 \\0\\4 \\ 1 \end{aligned}\right] . (b) The dimension of the solution space is $2$.
The coefficient matrix is given by $$\left[ \begin {array}{cccc} 2&4&3&-6\\ 1&2&2&-5 \\ 3&6&5&-11\end {array} \right] .$$ The reduced row echelon form is $$\left[ \begin {array}{cccc} 1&2&0&3\\ 0&0&1&-4 \\ 0&0&0&0\end {array} \right] .$$ The corresponding system is \begin{aligned} x_1 +2x_2- 3x_4 &=0\\ x_3-4x_4 &=0\\ \end{aligned}. The solution of the above system is $x_1=- 2s+3t$,$x_2=s$,$x_3=4t$, $x_4=t$. This means that the solution space of $Ax = 0$ consists of the vectors on the form x= \left[\begin{aligned} x_1\\ x_2\\x_3\\x_4 \end{aligned}\right]= \left[\begin{aligned} 2s+3t\\s\\4t\\ t \end{aligned}\right] = s\left[\begin{aligned}2 \\1\\0\\ 0 \end{aligned}\right]+t\left[\begin{aligned} 3 \\0\\4 \\ 1 \end{aligned}\right] . (a) A basis for the solution space is \left[\begin{aligned}2 \\1\\0\\ 0 \end{aligned}\right], \left[\begin{aligned} 3 \\0\\4 \\ 1 \end{aligned}\right] . (b) The dimension of the solution space is $2$.