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Found in: Page 36

### Linear Algebra With Applications

Book edition 5th
Author(s) Otto Bretscher
Pages 442 pages
ISBN 9780321796974

# a. Using technology, generate a random ${\mathbf{4}}{\mathbf{×}}{\mathbf{3}}$ matrix A. (The entries may be either single-digit integers or numbers between 0 and 1, depending on the technology you are using.) Find ${\mathbf{rref}}\left(A\right)$. Repeat this experiment a few times. b. What does the reduced row-echelon form of most ${\mathbf{4}}{\mathbf{×}}{\mathbf{3}}$matrices look like? Explain.

a. A $3×4$ matrix $A$ has $rref\left(A\right)=\left[\begin{array}{ccc}1& 0& -1\\ 0& 1& 2\\ 0& 0& 0\\ 0& 0& 0\end{array}\right]$

b. The reduced row-echelon form of matrix has leading 1’s with zero elements below the leading 1’s and non-zero elements above the leading 1’s.

See the step by step solution

## Step 1: Consider the matrix.

The reduced row-echelon form of a matrix has number of leading 1’s in each row and is denoted as $rref\left(A\right)=\left[\begin{array}{ccc}1& \dots & a\\ 0& \ddots & ⋮\\ 0& 0& 1\end{array}\right]$.

The matrix is,

$A=\left[\begin{array}{ccc}1& 2& 3\\ 4& 5& 6\\ 7& 8& 9\\ 1& 2& 3\end{array}\right]$

## Step 2: Find the reduced row-echelon form.

Consider the matrix

$A=\left[\begin{array}{ccc}1& 2& 3\\ 4& 5& 6\\ 7& 8& 9\\ 1& 2& 3\end{array}\right]$

$=A=\left[\begin{array}{ccc}7& 8& 9\\ 0& \frac{6}{7}& \frac{12}{7}\\ 0& 0& 0\\ 0& 0& 0\end{array}\right]$

$rref\left(A\right)\left[\begin{array}{ccc}1& 0& -1\\ 0& 1& 2\\ 0& 0& 0\\ 0& 0& 0\end{array}\right]$

localid="1659343693826" $Therefore,rref\left(A\right)\left[\begin{array}{ccc}1& 0& -1\\ 0& 1& 2\\ 0& 0& 0\\ 0& 0& 0\end{array}\right]$.

## Step 3: Explanation for the reduced row-echelon form.

A matrix in reduced row echelon form is used to solve systems of linear equations. There are four prerequisites for the reduced row echelon form:

• The number 1 is the first non-zero integer in the first row (the leading entry).
• The second row begins with the number 1, which is more to the right than the first row's leading item. The number 1 must be further to the right in each consecutive row.
• Each row's first item must be the sole non-zero number in its column.
• Any rows that are not zero are pushed to the bottom of the matrix.

Consider the reduced row-echelon form.

$rref\left(A\right)\left[\begin{array}{ccc}1& 0& -1\\ 0& 1& 2\\ 0& 0& 0\\ 0& 0& 0\end{array}\right]$

The reduced row-echelon form of matrix has leading 1’s with zero elements below the leading 1’s and non-zero elements above the leading 1’s.

a. $rref\left(A\right)\left[\begin{array}{ccc}1& 0& -1\\ 0& 1& 2\\ 0& 0& 0\\ 0& 0& 0\end{array}\right]$