2. suppose ( a ) is the matrix 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 already in row - echelon form, and for (Ax=0), the augmented matrix ([A|0]) is the same as (A) since the right - hand side is all zeros. So we have the system of linear equations: [ \begin{cases} 0x_1 + 1x_2+2x_3 + 2x_4=0\ 0x_1 + 3x_2+8x_3 + 7x_4=0\ 0x_1 + 0x_2+4x_3 + 2x_4=0 \end{cases} ]

Step2: Solve the third equation for (x_2) (wait, no, the third equation is (4x_3 + 2x_4=0), solve for (x_3) in terms of (x_4))

From (4x_3+2x_4 = 0), we can divide both sides by 2: (2x_3+x_4=0), then (x_3=-\frac{1}{2}x_4). Let (x_4 = t) (where (t\in\mathbb{R}) is a free variable), so (x_3=-\frac{1}{2}t).

Step3: Substitute (x_3) into the first equation to solve for (x_2)

The first equation is (x_2 + 2x_3+2x_4=0). Substitute (x_3 = -\frac{1}{2}t) and (x_4=t) into it: (x_2+2\left(-\frac{1}{2}t\right)+2t = 0) (x_2 - t + 2t=0) (x_2+t = 0), so (x_2=-t)

Step4: Substitute (x_2) and (x_3) into the second equation to check (and also, note that (x_1) is a free variable? Wait, no, in the matrix (A), the first column has all zeros, so (x_1) is a free variable. Let (x_1 = s) (where (s\in\mathbb{R}) is a free variable), (x_4=t).

Wait, let's re - express the variables. Let (x_1 = s) and (x_4=t) (two free variables since the rank of (A) is 3, and the number of variables (n = 4), so the dimension of the nullspace is (n - \text{rank}(A)=4 - 3=1)? Wait, no, wait the matrix (A) has 4 columns. Let's do row operations on (A) to get it to reduced row - echelon form. Row 2: (R_2-3R_1): (R_2 = R_2-3R_1), so the second row becomes (\begin{bmatrix}0&3 - 3\times1&8 - 3\times2&7 - 3\times2\end{bmatrix}=\begin{bmatrix}0&0&2&1\end{bmatrix}) Row 3: (R_3) is (\begin{bmatrix}0&0&4&2\end{bmatrix}), then (R_3 - 2R_2): (4 - 2\times2=0), (2-2\times1 = 0), so (R_3-2R_2=\begin{bmatrix}0&0&0&0\end{bmatrix}) Row 1: (R_1) is (\begin{bmatrix}0&1&2&2\end{bmatrix}), (R_1-2R_3/4)? Wait, better to do (R_1 - R_2): (R_1-R_2=\begin{bmatrix}0&1&0&1\end{bmatrix}) Now the reduced row - echelon form of (A) is: [ \begin{bmatrix} 0&1&0&1\ 0&0&2&1\ 0&0&0&0 \end{bmatrix} ] Divide the second row by 2: [ \begin{bmatrix} 0&1&0&1\ 0&0&1&\frac{1}{2}\ 0&0&0&0 \end{bmatrix} ] So the system is: (x_2+x_4 = 0) (x_3+\frac{1}{2}x_4=0) (x_1) is free (let (x_1 = s)), (x_4) is free (let (x_4=t)) From (x_2+x_4 = 0), we have (x_2=-t) From (x_3+\frac{1}{2}x_4 = 0), we have (x_3=-\frac{1}{2}t) So the general solution of (Ax = 0) is: (x=\begin{bmatrix}x_1\x_2\x_3\x_4\end{bmatrix}=\begin{bmatrix}s\-t\-\frac{1}{2}t\t\end{bmatrix}=s\begin{bmatrix}1\0\0\0\end{bmatrix}+t\begin{bmatrix}0\-1\-\frac{1}{2}\1\end{bmatrix}) To eliminate the fraction, we can let (t = 2r) (where (r\in\mathbb{R})). Then (x_2=-2r), (x_3=-r), (x_4 = 2r) So the general solution is (x=s\begin{bmatrix}1\0\0\0\end{bmatrix}+r\begin{bmatrix}0\-2\-1\2\end{bmatrix}) The special solution (when we set the free variables one - by - one to 1 and others to 0) When (s = 1) and (r = 0), we get the vector (\begin{bmatrix}1\0\0\0\end{bmatrix}) When (s = 0) and (r = 1), we get the vector (\begin{bmatrix}0\-2\-1\2\end{bmatrix}) Wait, actually, the nullspace of (A) is the set of all solutions to (Ax = 0). The nullspace is a subspace of (\mathbb{R}^4). The special solutions are the vectors that form a basis for the nullspace. From the reduced row - echelon form, the pivot columns are the second and third columns (since the leading 1s are in the second and third rows, second and third columns). So the free variables are (x_1) and (x_4). Let (x_1 = 1), (x_4=0): Then from (x_2+x_4 = 0), (x_2 = 0); from (x_3+\frac{1}{2}x_4=0), (x_3 = 0). So one special solution is (\begin{bmatrix}1\0\0\0\end{bmatrix}) Let (x_1 = 0), (x_4 = 2) (to eliminate the fraction): Then from (x_2+x_4=0), (x_2=-2); from (x_3+\frac{1}{2}x_4 = 0), (x_3=-1). So the other special solution is (\begin{bmatrix}0\-2\-1\2\end{bmatrix}) The nullspace of (A) is the set of all linear combinations of these two special solutions. In other words, (\text{null}(A)=\left{s\begin{bmatrix}1\0\0\0\end{bmatrix}+r\begin{bmatrix}0\-2\-1\2\end{bmatrix}:s,r\in\mathbb{R}\right}), which is a two - dimensional subspace of (\mathbb{R}^4) (since the dimension of the nullspace is (n-\text{rank}(A)=4 - 2=2), because the rank of (A) is 2, as there are two non - zero rows in the reduced row - echelon form).

Answer:

The special solutions are (\boldsymbol{\begin{bmatrix}1\0\0\0\end{bmatrix}}) and (\boldsymbol{\begin{bmatrix}0\-2\-1\2\end{bmatrix}}) (or scalar multiples thereof). The nullspace of (A) is the set of all linear combinations of these two special solutions, i.e., (\text{null}(A)=\left{s\begin{bmatrix}1\0\0\0\end{bmatrix}+r\begin{bmatrix}0\-2\-1\2\end{bmatrix}:s,r\in\mathbb{R}\right}), a two - dimensional subspace of (\mathbb{R}^4) consisting of all vectors (\begin{bmatrix}x_1\x_2\x_3\x_4\end{bmatrix}) such that (x_2=-2x_4/2=-x_4) (wait, no, from the equations (x_2=-x_4) when we set (x_1 = 0) and (x_3=-\frac{1}{2}x_4)) and (x_1) is arbitrary.