# Thread: reduce row echelon question

1. ## reduce row echelon question

Let the reduced row echelon form of $A$ be

$\left(\begin{array}{ccccc} 1 & 0 & 2 & 0 & -2\\ \\0 & 1 & -5 & 0 & -3\\ \\0 & 0 & 0 & 1 & 6\end{array}\right)$

Determini $A$ if the first second and fourth columns of $A$ are

$\left(\begin{array}{c} 1 \\ \\ - 1 \\ \\3 \end{array}\right), \left(\begin{array}{c} 0 \\ \\ - 1 \\ \\1 \end{array}\right) and \left(\begin{array}{c} 1 \\ \\ - 2 \\ \\0 \end{array}\right)$

would this just mean that you substitute the first second and fourth and perform row operations?

2. Originally Posted by lllll
Let the reduced row echelon form of $A$ be

$\left(\begin{array}{ccccc} 1 & 0 & 2 & 0 & -2\\ \\0 & 1 & -5 & 0 & -3\\ \\0 & 0 & 0 & 1 & 6\end{array}\right)$

Determini $A$ if the first second and fourth columns of $A$ are

$\left(\begin{array}{c} 1 \\ \\ - 1 \\ \\3 \end{array}\right), \left(\begin{array}{c} 0 \\ \\ - 1 \\ \\1 \end{array}\right) and \left(\begin{array}{c} 1 \\ \\ - 2 \\ \\0 \end{array}\right)$

would this just mean that you substitute the first second and fourth and perform row operations?
In the Matrix A we need to express the columns without 1 as a liner combination of the colums with ones.

for example

$c_3=2 \cdot c_1-5 \cdot c_2$ and

$c_5=-2 \cdot c_1-3 \cdot c_2 + 6 \cdot c_4$

These are called the dependency equations

You can find the original column vectors using the same equations.

so

$c_3= 2 \cdot \begin{bmatrix}1 \\ -1 \\ 3 \end{bmatrix}
-5 \cdot \begin{bmatrix}0 \\ -1 \\ 1 \end{bmatrix} =
\begin{bmatrix}2 \\ 3 \\ 1 \end{bmatrix}$