2. suppose a is the matrix \n$a = \\begin{bmatrix} 0 & 1 & 2 & 2 \\\\ 0 & 3 & 8 & 7 \\\\ 0 & 0 & 4 & 2…

Explanation:

Step1: Write the augmented matrix for (Ax = 0)

The matrix (A) is (\begin{bmatrix}0&1&2&2\0&3&8&7\0&0&4&2\end{bmatrix}), and for (Ax = 0), the augmented matrix (\left[\begin{array}{cccc|c}0&1&2&2&0\0&3&8&7&0\0&0&4&2&0\end{array}\right]) is considered. First, perform row operations. Subtract 3 times the first row from the second row: (R_2 = R_2 - 3R_1). [ \begin{bmatrix}0&1&2&2&0\0&0&2&1&0\0&0&4&2&0\end{bmatrix} ] Then, subtract 2 times the second row from the third row: (R_3 = R_3 - 2R_2). [ \begin{bmatrix}0&1&2&2&0\0&0&2&1&0\0&0&0&0&0\end{bmatrix} ] Now, divide the second row by 2: (R_2=\frac{1}{2}R_2), we get (\begin{bmatrix}0&1&2&2&0\0&0&1&\frac{1}{2}&0\0&0&0&0&0\end{bmatrix}).

Step2: Express variables in terms of free variables

Let (x_1 = t) (free variable, since the first column has no leading 1) and (x_4 = s) (free variable, since the fourth column has no leading 1). From the second row: (x_3+\frac{1}{2}x_4 = 0\implies x_3=-\frac{1}{2}s). From the first row: (x_2 + 2x_3+2x_4 = 0). Substitute (x_3 = -\frac{1}{2}s) into it: (x_2+2\left(-\frac{1}{2}s\right)+2s = 0\implies x_2 - s + 2s=0\implies x_2=-s).

Step3: Write the general solution and find special solutions

The general solution is (x=\begin{bmatrix}x_1\x_2\x_3\x_4\end{bmatrix}=\begin{bmatrix}t\-s\-\frac{1}{2}s\s\end{bmatrix}=t\begin{bmatrix}1\0\0\0\end{bmatrix}+s\begin{bmatrix}0\-1\-\frac{1}{2}\1\end{bmatrix}). To make the solution simpler (eliminate fractions), we can multiply the second vector by 2 (since scalar multiples of special solutions are still special solutions). So the special solutions are (\mathbf{v_1}=\begin{bmatrix}1\0\0\0\end{bmatrix}) and (\mathbf{v_2}=\begin{bmatrix}0\-2\-1\2\end{bmatrix}) (or the non - scaled version (\begin{bmatrix}0\-1\-\frac{1}{2}\1\end{bmatrix}), but the integer - valued one is more convenient). The nullspace of (A) is the set of all linear combinations of the special solutions of (Ax = 0), i.e., (\text{Null}(A)=\text{span}\left{\begin{bmatrix}1\0\0\0\end{bmatrix},\begin{bmatrix}0\-2\-1\2\end{bmatrix}\right}) (or equivalently (\text{span}\left{\begin{bmatrix}1\0\0\0\end{bmatrix},\begin{bmatrix}0\-1\-\frac{1}{2}\1\end{bmatrix}\right})).

Answer:

The special solutions to (Ax = 0) are (\boldsymbol{\begin{bmatrix}1\0\0\0\end{bmatrix}}) and (\boldsymbol{\begin{bmatrix}0\-2\-1\2\end{bmatrix}}) (or (\boldsymbol{\begin{bmatrix}0\-1\-\frac{1}{2}\1\end{bmatrix}})). The nullspace of (A) is the set of all linear combinations of these special solutions, meaning it is a two - dimensional subspace of (\mathbb{R}^4) spanned by the special solutions of (Ax = 0).