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

Answer:

Special solutions: $\begin{bmatrix}1\0\0\0\end{bmatrix}$, $\begin{bmatrix}-\frac{1}{2}\\frac{1}{2}\1\0\end{bmatrix}$, $\begin{bmatrix}-1\-2\0\1\end{bmatrix}$ The nullspace of $A$ is the set of all linear combinations of these three special solutions, forming a 3-dimensional subspace of $\mathbb{R}^4$.

Explanation:

Step1: Identify pivot columns

Pivot columns are column 4. Free columns: 1,2,3.

Step2: Set free variables to standard basis

First, set $x_1=1, x_2=0, x_3=0$: $0x_1 + 0x_2 + 0x_3 + 4x_4 = 0 \implies x_4=0$ Solution: $\begin{bmatrix}1\0\0\0\end{bmatrix}$

Step3: Second free variable assignment

Set $x_1=0, x_2=1, x_3=0$: $0x_1 + 3x_2 + 0x_3 + 4x_4 = 0 \implies 3 + 4x_4=0 \implies x_4=-\frac{3}{4}$ $0x_1 + x_2 + 2x_3 + 2x_4 = 0 \implies 1 + 2(-\frac{3}{4}) + 0x_1=0 \implies x_1 = -\frac{1}{2}$ Solution: $\begin{bmatrix}-\frac{1}{2}\\frac{1}{2}\1\0\end{bmatrix}$ (rescaled for simplicity)

Step4: Third free variable assignment

Set $x_1=0, x_2=0, x_3=1$: $0x_1 + 0x_2 + 4x_3 + 2x_4 = 0 \implies 4 + 2x_4=0 \implies x_4=-2$ $0x_1 + 3x_2 + 8x_3 + 7x_4 = 0 \implies 8 + 7(-2) + 0x_1=0 \implies x_1=-1$ $0x_1 + x_2 + 2x_3 + 2x_4 = 0 \implies 2 + 2(-2) + 0x_1=0 \implies x_1=-1, x_2=-2$ Solution: $\begin{bmatrix}-1\-2\0\1\end{bmatrix}$

Step5: Describe nullspace

Nullspace = ${c_1\begin{bmatrix}1\0\0\0\end{bmatrix} + c_2\begin{bmatrix}-\frac{1}{2}\\frac{1}{2}\1\0\end{bmatrix} + c_3\begin{bmatrix}-1\-2\0\1\end{bmatrix} \mid c_1,c_2,c_3 \in \mathbb{R}}$, a 3D subspace of $\mathbb{R}^4$.