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Q20E

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Found in: Page 437

### Linear Algebra and its Applications

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 $$d = 0$$.

See the step by step solution

## Step 1: Write the matrix equation

The matrix equation can be written as follows:

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

## Step 2: Write the equation using matrix multiplication

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:

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

## Step 3: Write the general solution

The general solution is:

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

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