Question

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 A be the \({\bf{1}} \times {\bf{5}}\) matrix \(\left( {\begin{array}{*{20}{c}}{\bf{2}}&{\bf{5}}&{ - {\bf{3}}}&{\bf{0}}&{\bf{6}}\end{array}} \right)\). Note that \({\bf{Nul}}\,\,A\) is in \({\mathbb{R}^{\bf{5}}}\). Let \(H = {\bf{Nul}}\,\,A\).

Short answer

\(f\left( {{x_1},{x_2},{x_3},{x_4},{x_5}} \right) = 2{x_1} + 5{x_2} - 3{x_3} + 6{x_5}\) and \(d = 0\)

Step-by-step solution

Write the matrix equation

The matrix equation can be written as follows:

\(\begin{array}{c}A{\bf{x}} = 0\\\left( {\begin{array}{*{20}{c}}2&5&{ - 3}&0&6\end{array}} \right)\left( {\begin{array}{*{20}{c}}{{x_1}}\\{{x_2}}\\{{x_3}}\\{{x_4}}\\{{x_5}}\end{array}} \right) = 0\end{array}\)

Write the equation using matrix multiplication

The matrix equation is shown below:

\(\begin{array}{c}2{x_1} + 5{x_2} - 3{x_3} + 0 + 6{x_5} = 0\\2{x_1} + 5{x_2} - 3{x_3} + 6{x_5} = 0\end{array}\)

So, \(f\left( {{x_1},{x_2},{x_3},{x_4},{x_5}} \right) = 2{x_1} + 5{x_2} - 3{x_3} + 6{x_5}\) and \(d = 0\).