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Q20E

Expert-verifiedFound in: Page 437

Book edition
5th

Author(s)
David C. Lay, Steven R. Lay and Judi J. McDonald

Pages
483 pages

ISBN
978-03219822384

**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 column space of the matrix \(B = \left( {\begin{array}{*{20}{c}}{\bf{1}}&{\bf{0}}\\{\bf{5}}&{\bf{2}}\\{ - {\bf{4}}}&{ - {\bf{4}}}\end{array}} \right)\). That is, \(H = {\bf{Col}}\,B\).(Hint: How is \({\bf{Col}}\,B\)related to Nul \({B^T}\)? See section 6.1)**

The linear functional is \(f\left( {{x_1},{x_2},{x_3}} \right) = - 6{x_1} + 2{x_2} + {x_3}\) and \(d = 0\).

The **matrix equation** can be written as follows:

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

The matrix equation can be simplified as shown below:

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

And,

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

Simplifying the above equations:

\(\begin{array}{c}{x_1} + 5\left( {2{x_3}} \right) - 4{x_3} = 0\\{x_1} = - 6{x_3}\end{array}\)

The **general solution** is:

\(\begin{array}{c}{\bf{x}} = \left( {\begin{array}{*{20}{c}}{{x_1}}\\{{x_2}}\\{{x_3}}\end{array}} \right)\\ = \left( {\begin{array}{*{20}{c}}{ - 6{x_3}}\\{2{x_3}}\\{{x_3}}\end{array}} \right)\\ = \left( {\begin{array}{*{20}{c}}{ - 6}\\2\\1\end{array}} \right){x_3}\end{array}\)

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

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