Question

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

Short answer

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

Step-by-step solution

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.

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