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The set of allhomogeneousequation solutions, \(A{\bf{x}} = 0\), is \(Nul A\), or it is called thenull space of A.

Consider the plane equation \(x + 2y + z = 0\).

In matrix form, it is\(A = \left( {\begin{array}{*{20}{c}}1&2&1\end{array}} \right)\).

The plane equation \(x + 2y + z = 0\)can also be written as\(x = - 2y - z\).

From the above equations, \(x\) corresponds to thepivot positions. So, \(x\) is the basic variable, and \(y\) and \(z\) are free variables.

Let \(y = a\), \(z = b\).

Substitute the values \(y = a\) and \(z = b\) in the equation \(x = - 2y - z\) to obtain the general solution

\(x = - 2a - b\).

Obtain the vector in the parametric form using \(x = - 2a - b\), \(y = a\), and \(z = b\).

\(\begin{array}{c}\left( {\begin{array}{*{20}{c}}x\\y\\z\end{array}} \right) = \left( {\begin{array}{*{20}{c}}{ - 2a - b}\\a\\b\end{array}} \right)\\ = \left( {\begin{array}{*{20}{c}}{ - 2a}\\a\\0\end{array}} \right) + \left( {\begin{array}{*{20}{c}}{ - b}\\0\\b\end{array}} \right)\\ = a\left( {\begin{array}{*{20}{c}}{ - 2}\\1\\0\end{array}} \right) + b\left( {\begin{array}{*{20}{c}}{ - 1}\\0\\1\end{array}} \right)\end{array}\)

It can be written as

\(\left( {\begin{array}{*{20}{c}}x\\y\\z\end{array}} \right) = y\left( {\begin{array}{*{20}{c}}{ - 2}\\1\\0\end{array}} \right) + z\left( {\begin{array}{*{20}{c}}{ - 1}\\0\\1\end{array}} \right)\).

Thebasis of vectorsis shown below:

\( \Rightarrow \left\{ {\left( {\begin{array}{*{20}{c}}{ - 2}\\1\\0\end{array}} \right),\left( {\begin{array}{*{20}{c}}{ - 1}\\0\\1\end{array}} \right)} \right\}\)

Thus, \({\bf{x}} = y\left( {\begin{array}{*{20}{c}}{ - 2}\\1\\0\end{array}} \right) + z\left( {\begin{array}{*{20}{c}}{ - 1}\\0\\1\end{array}} \right)\), and a basis for \(Nul A\) is \(\left\{ {\left( {\begin{array}{*{20}{c}}{ - 2}\\1\\0\end{array}} \right),\left( {\begin{array}{*{20}{c}}{ - 1}\\0\\1\end{array}} \right)} \right\}\).