Elementary Linear Algebra 7th Edition

(a) A basis for the solution space is \left[\begin{aligned}-4 \\-1 \\1\\0 \end{aligned}\right],\left[\begin{aligned} -3 \\-\frac{2}{3} \\0\\1 \end{aligned}\right] . (b) The dimension of the solution space is $2$.
The coefficient matrix is given by $$\left[ \begin {array}{cccc} 3&3&15&11\\ 1&-3&1&1 \\ 2&3&11&8\end {array} \right] .$$ The reduced row echelon form is $$\left[ \begin {array}{cccc} 1&0&4&3\\ 0&1&1&\frac{2}{3} \\ 0&0&0&0\end {array} \right] .$$ The corresponding system is \begin{aligned} x_1 +4x_3+3x_4&=0\\ x_2+ x_3+\frac{2}{3}x_4&=0\\ \end{aligned}. The solution of the above system is $x_1=-4s-3t$,$x_2=- s-\frac{2}{3}t$,$x_3=s$, $x_4=t$. 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}-4s-3t\\-s-\frac{2}{3}t \\s\\t \end{aligned}\right] =s\left[\begin{aligned}-4 \\-1 \\1\\0 \end{aligned}\right]+t\left[\begin{aligned} -3 \\-\frac{2}{3} \\0\\1 \end{aligned}\right] . (a) A basis for the solution space is \left[\begin{aligned}-4 \\-1 \\1\\0 \end{aligned}\right],\left[\begin{aligned} -3 \\-\frac{2}{3} \\0\\1 \end{aligned}\right] . (b) The dimension of the solution space is $2$.