## Elementary Linear Algebra 7th Edition

(a) A basis for the solution space is \left[\begin{aligned} \frac{2}{7} \\\frac{3}{7} \\1\\ 0 \end{aligned}\right], \left[\begin{aligned}-1\\0\\0\\ 1 \end{aligned}\right] . (b) The dimension of the solution space is $2$.
The coefficient matrix is given by $$\left[ \begin {array}{cccc} 1&-3&1&1\\ 2&1&-1&2 \\ 1&4&-2&1\\ 5&-8&2&5\end {array} \right] .$$ The reduced row echelon form is $$\left[ \begin {array}{cccc} 1&0&-\frac{2}{7}&1\\ 0&1&-\frac{3}{7}&0 \\ 0&0&0&0\\ 0&0&0&0\end {array} \right] .$$ The corresponding system is \begin{aligned} x_1 -\frac{2}{7}x_3 +x_4&=0\\ x_2-\frac{3}{7}x_3 &=0\\ \end{aligned}. The solution of the above system is $x_1= \frac{2}{7}s-t$,$x_2=\frac{3}{7}s$,$x_3=s$, $x_4=t$. 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} \frac{2}{7}s-t\\\frac{3}{7}s\\s\\ t \end{aligned}\right] = s \left[\begin{aligned} \frac{2}{7} \\\frac{3}{7} \\1\\ 0 \end{aligned}\right] +t \left[\begin{aligned}-1\\0\\0\\ 1 \end{aligned}\right] . (a) A basis for the solution space is \left[\begin{aligned} \frac{2}{7} \\\frac{3}{7} \\1\\ 0 \end{aligned}\right], \left[\begin{aligned}-1\\0\\0\\ 1 \end{aligned}\right] . (b) The dimension of the solution space is $2$.