Question

1. forward elimination changes $ax = b$ to a row reduced $rx = d$: the complete solution is

$\mathbf{x} = \

$$\begin{bmatrix} 4 \\\\ 0 \\\\ 0 \\end{bmatrix}$$

+ c_1 \

$$\begin{bmatrix} 2 \\\\ 1 \\\\ 0 \\end{bmatrix}$$

+ c_2 \

$$\begin{bmatrix} 5 \\\\ 0 \\\\ 1 \\end{bmatrix}$$

$
(a) (14 points) what is the 3 by 3 reduced row echelon matrix $r$ and what is $d$?

Explanation:

Step1: Identify solution components

The complete solution is $\mathbf{x} = \mathbf{x_p} + c_1\mathbf{x_{n1}} + c_2\mathbf{x_{n2}}$, where:
Particular solution: $\mathbf{x_p} =

$$\begin{bmatrix}4 \\ 0 \\ 0\end{bmatrix}$$

$, null space vectors: $\mathbf{x_{n1}} =

$$\begin{bmatrix}2 \\ 1 \\ 0\end{bmatrix}$$

$, $\mathbf{x_{n2}} =

$$\begin{bmatrix}5 \\ 0 \\ 1\end{bmatrix}$$

$

Step2: Find $\mathbf{d}$ from particular solution

Since $R\mathbf{x_p} = \mathbf{d}$, substitute $\mathbf{x_p}$:

$$R
LATEXBLOCK5
= \mathbf{d}$$

In reduced row echelon form, this gives $\mathbf{d} =

$$\begin{bmatrix}4 \\ 0 \\ 0\end{bmatrix}$$

$.

Step3: Derive $R$ from null space

Let $R =

$$\begin{bmatrix} r_{11} & r_{12} & r_{13} \\ r_{21} & r_{22} & r_{23} \\ r_{31} & r_{32} & r_{33} \end{bmatrix}$$

$. For null space vectors, $R\mathbf{x_n} = \mathbf{0}$:
For $\mathbf{x_{n1}}$:

$$\begin{cases} 2r_{11} + 1r_{12} + 0r_{13} = 0 \\ 2r_{21} + 1r_{22} + 0r_{23} = 0 \\ 2r_{31} + 1r_{32} + 0r_{33} = 0 \end{cases}$$

For $\mathbf{x_{n2}}$:

$$\begin{cases} 5r_{11} + 0r_{12} + 1r_{13} = 0 \\ 5r_{21} + 0r_{22} + 1r_{23} = 0 \\ 5r_{31} + 0r_{32} + 1r_{33} = 0 \end{cases}$$

Using reduced row echelon form rules (leading 1s, zeros above/below):
$r_{11}=1, r_{22}=1, r_{31}=r_{32}=r_{33}=0$. Solve for other entries:
$r_{12} = -2, r_{13} = -5; r_{21}=0, r_{23}=-5$.
Thus $R =

$$\begin{bmatrix} 1 & 0 & -2 \\ 0 & 1 & -5 \\ 0 & 0 & 0 \end{bmatrix}$$

$

Answer:

$R =

$$\begin{bmatrix} 1 & 0 & -2 \\ 0 & 1 & -5 \\ 0 & 0 & 0 \end{bmatrix}$$

$, $\mathbf{d} =

$$\begin{bmatrix} 4 \\ 0 \\ 0 \end{bmatrix}$$

$