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

See the step by step solution

## Step 1: Write the matrix equation

The matrix equation can be written as follows:

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

## Step 2: Write the equation using matrix multiplication

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:

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

## Step 3: Write the general solution

The general solution is:

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

So, $$f\left( {{x_1},{x_2},{x_3}} \right) = {x_1} - 2{x_2} + {x_3}$$ and $$d = 0$$. ### Want to see more solutions like these? 