Question

1. forward elimination changes ( amathbf{x} = mathbf{b} ) to a row reduced ( rmathbf{x} = mathbf{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 (mathbf{d})?

Explanation:

Step1: Identify pivot/free variables

From the complete solution, $x_1$ is a pivot variable; $x_2, x_3$ are free variables.

Step2: Construct reduced row echelon R

Reduced row echelon form $R$ has leading 1s in pivot columns, 0s above/below pivots:

$$R =
LATEXBLOCK0
$$

Step3: Calculate vector d

Substitute particular solution $\mathbf{x}_p =

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

$ into $R\mathbf{x} = \mathbf{d}$:

$$\mathbf{d} = R\mathbf{x}_p =
LATEXBLOCK2

LATEXBLOCK3
=
LATEXBLOCK4
$$

Answer:

The reduced row echelon matrix $R$ is

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

and $\mathbf{d} =

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

$