Question

a. Using technology, generate a random $\mathbf{3}\mathbf{\times }\mathbf{4}$ matrix$\mathit{A}$ . (The entries may be either single-digit integers or numbers between $\mathbf{0}$and$\mathbf{1}$ , 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$\mathbf{3}\mathbf{\times }\mathbf{4}$ matrices look like? Explain.

Short answer

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

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.

Step-by-step solution

Consider the matrix.

The reduced row-echelon form of a matrix has number of leading 1’s in each row and is denoted as$rref\left(A\right)$.

$rref\left(A\right)\left[\begin{array}{ccc}1& \dots & a\\ 0& \ddots & ⋮\\ 0& 0& 1\end{array}\right]$

The matrix is,

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

Find the reduced row-echelon form.

Consider the matrix

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

$=\left[\begin{array}{cccc}9& 1& 2& 3\\ 0& \frac{49}{9}& \frac{53}{9}& \frac{19}{3}\\ 0& 0& \frac{36}{49}& \frac{72}{49}\end{array}\right]$

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

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

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.

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

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.

Final answer.

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

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.