Question: In Exercises 15-20, write a formula for a linear functional f and specify a number d, so that \(\left( {f:d} \right)\) the hyperplane H described in the exercise.

Let H be the plane in \({\mathbb{R}^{\bf{3}}}\) spanned by the rows of \(B = \left( {\begin{array}{*{20}{c}}{\bf{1}}&{\bf{3}}&{\bf{5}}\\{\bf{0}}&{\bf{2}}&{\bf{4}}\end{array}} \right)\). That is, \(H = {\bf{Row}}\,B\). (Hint: How is H is related to Nul B?see section 6.1.)

The matrix equation can be written as follows:

\(\begin{array}{c}B{\bf{x}} = 0\\\left( {\begin{array}{*{20}{c}}1&3&5\\0&2&4\end{array}} \right)\left( {\begin{array}{*{20}{c}}{{x_1}}\\{{x_2}}\\{{x_3}}\end{array}} \right) = \left( {\begin{array}{*{20}{c}}0\\0\end{array}} \right)\end{array}\)

The matrix equation can be simplified as shown below:

\({x_1} + 3{x_2} + 5{x_3} = 0\)

And,

\(\begin{array}{c}2{x_2} + 4{x_3} = 0\\{x_2} = - 2{x_3}\end{array}\)

Simplifying the above equations:

\(\begin{array}{c}{x_1} + 3\left( { - 2{x_3}} \right) + 5{x_3} = 0\\{x_1} = {x_3}\end{array}\)

The general solution is:

\(\begin{array}{c}{\bf{x}} = \left( {\begin{array}{*{20}{c}}{{x_1}}\\{{x_2}}\\{{x_3}}\end{array}} \right)\\ = \left( {\begin{array}{*{20}{c}}{{x_3}}\\{ - 2{x_3}}\\{{x_3}}\end{array}} \right)\\ = \left( {\begin{array}{*{20}{c}}1\\{ - 2}\\1\end{array}} \right){x_3}\end{array}\)

So, \(f\left( {{x_1},{x_2},{x_3}} \right) = {x_1} - 2{x_2} + {x_3}\) and \(d = 0\).