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### Linear Algebra and its Applications

Book edition 5th
Author(s) David C. Lay, Steven R. Lay and Judi J. McDonald
Pages 483 pages
ISBN 978-03219822384

# Construct a $$2 \times 3$$ matrix $$A$$, not in echelon form, such that the solution of $$Ax = 0$$ is a line in $${\mathbb{R}^3}$$.

The matrix that is not in the echelon form is \left( {\begin{aligned}{*{20}{c}}1&2&1\\1&5&2\end{aligned}} \right).

See the step by step solution

## Step 1: Determine the condition for the linear solution of $$Ax = 0$$

A solution set is a line that has one free variable in the system. In a coefficient matrix of $$2 \times 3$$, two columns should be pivot columns.

## Step 2: Construct matrix $$A$$ that is not in the echelon form

Take the example \left( {\begin{aligned}{*{20}{c}}1&2& * \\0&3& * \end{aligned}} \right). Fill any value in the entries of column 3 to obtain in an echelon matrix. Perform one row replacement operation on the second row to construct a matrix that is not in the echelon form.

\left( {\begin{aligned}{*{20}{c}}1&2&1\\0&3&1\end{aligned}} \right) \sim \left( {\begin{aligned}{*{20}{c}}1&2&1\\1&5&2\end{aligned}} \right)

Thus, the matrix which is not in the echelon form is \left( {\begin{aligned}{*{20}{c}}1&2&1\\1&5&2\end{aligned}} \right).