Answer
(a) A basis for the solution space is
$$ \left[\begin{aligned}-1 \\-\frac{3}{2} \\1\\0\\0 \end{aligned}\right], \left[\begin{aligned} 2 \\ -8 \\0\\1\\0 \end{aligned}\right],\left[\begin{aligned}-1 \\ -4 \\0\\0\\1 \end{aligned}\right] .$$
(b) The dimension of the solution space is $3$.
Work Step by Step
The coefficient matrix is given by
$$
\left[ \begin {array}{ccccc} 1&3&2&22&13\\ 1&0&1&-2
&1\\ 3&6&5&42&27\end {array} \right]
.
$$
The reduced row echelon form is
$$
\left[ \begin {array}{ccccc} 1&0&1&-2&1\\ 0&1&\frac{1}{3}&8
&4\\ 0&0&0&0&0\end {array} \right]
.
$$
The corresponding system is
$$
\begin{aligned}
x_1 +x_3- 2x_4+x_5&=0\\
x_2+\frac{1}{3}x_3+8x_4+4x_5&=0\\
\end{aligned}.
$$
The solution of the above system is $x_1=-r+2s-t$,$x_2=-\frac{3}{2}r-8s-4t$,$x_3=r$, $x_4=s$, $x_5=t$. This means that the solution space of $Ax = 0 $ consists of the vectors on the form
$$x= \left[\begin{aligned} x_1\\ x_2\\x_3\\x_4 \\x_5 \end{aligned}\right]= \left[\begin{aligned}-r+2s-t\\-\frac{3}{2}r-8s-4t\\r\\s\\t \end{aligned}\right] = r\left[\begin{aligned}-1 \\-\frac{3}{2} \\1\\0\\0 \end{aligned}\right]+s\left[\begin{aligned} 2 \\ -8 \\0\\1\\0 \end{aligned}\right]+t\left[\begin{aligned}-1 \\ -4 \\0\\0\\1 \end{aligned}\right] .$$
(a) A basis for the solution space is
$$ \left[\begin{aligned}-1 \\-\frac{3}{2} \\1\\0\\0 \end{aligned}\right], \left[\begin{aligned} 2 \\ -8 \\0\\1\\0 \end{aligned}\right],\left[\begin{aligned}-1 \\ -4 \\0\\0\\1 \end{aligned}\right] .$$
(b) The dimension of the solution space is $3$.
