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 H be the plane in \({\mathbb{R}^{\bf{3}}}\) spanned by the rows of \(B = \left( {\begin{array}{*{20}{c}}{\bf{1}}&{\bf{4}}&{ - {\bf{5}}}\\{\bf{0}}&{ - {\bf{2}}}&{\bf{8}}\end{array}} \right)\). That is, \(H = {\bf{Row}}\,B\).

Short answer

The linear functional is \(f\left( {{x_1},{x_2},{x_3}} \right) = - 11{x_1} + 4{x_2} + {x_3}\) and \(d = 0\).

Step-by-step solution

Write the matrix equation

The matrix equation can be written as follows:

\(\begin{array}{c}B{\bf{x}} = 0\\\left( {\begin{array}{*{20}{c}}1&4&{ - 5}\\0&{ - 2}&8\end{array}} \right)\left( {\begin{array}{*{20}{c}}{{x_1}}\\{{x_2}}\\{{x_3}}\end{array}} \right) = \left( {\begin{array}{*{20}{c}}0\\0\end{array}} \right)\end{array}\)

Write the equation using matrix multiplication

The matrix equation can be simplified as shown below:

\({x_1} + 4{x_2} - 5{x_3} = 0\)

And,

\(\begin{array}{c} - 2{x_2} + 8{x_3} = 0\\{x_2} = 4{x_3}\end{array}\)

Simplifying the above equations:

\(\begin{array}{c}{x_1} + 4\left( {4{x_3}} \right) - 5{x_3} = 0\\{x_1} = - 11{x_3}\end{array}\)

Write the general solution

The general solution is:

\(\begin{array}{c}{\bf{x}} = \left( {\begin{array}{*{20}{c}}{{x_1}}\\{{x_2}}\\{{x_3}}\end{array}} \right)\\ = \left( {\begin{array}{*{20}{c}}{ - 11{x_3}}\\{4{x_3}}\\{{x_3}}\end{array}} \right)\\ = \left( {\begin{array}{*{20}{c}}{ - 11}\\4\\1\end{array}} \right){x_3}\end{array}\)

So, \(f\left( {{x_1},{x_2},{x_3}} \right) = - 11{x_1} + 4{x_2} + {x_3}\) and \(d = 0\).