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Linear Algebra and its Applications
Found in: Page 437
Linear Algebra and its Applications

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

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Short Answer

Question: Let \({{\bf{p}}_{\bf{1}}} = \left( {\begin{array}{*{20}{c}}{\bf{2}}\\{ - {\bf{3}}}\\{\bf{1}}\\{\bf{2}}\end{array}} \right)\), \({{\bf{p}}_{\bf{2}}} = \left( {\begin{array}{*{20}{c}}{\bf{1}}\\{\bf{2}}\\{ - {\bf{1}}}\\{\bf{3}}\end{array}} \right)\), \({{\bf{n}}_{\bf{1}}} = \left( {\begin{array}{*{20}{c}}{\bf{1}}\\{\bf{2}}\\{\bf{4}}\\{\bf{2}}\end{array}} \right)\), and \({{\bf{n}}_{\bf{2}}} = \left( {\begin{array}{*{20}{c}}{\bf{2}}\\{\bf{3}}\\{\bf{1}}\\{\bf{5}}\end{array}} \right)\), let \({H_{\bf{1}}}\) be the hyperplane in \({\mathbb{R}^{\bf{4}}}\) through \({{\bf{p}}_{\bf{1}}}\) with normal \({{\bf{n}}_{\bf{1}}}\), and let \({H_{\bf{2}}}\) be the hyperplane through \({{\bf{p}}_{\bf{2}}}\) with normal \({{\bf{n}}_{\bf{2}}}\). Give an explicit description of \({H_{\bf{1}}} \cap {H_{\bf{2}}}\). (Hint: Find a point p in \({H_{\bf{1}}} \cap {H_{\bf{2}}}\) and two linearly independent vectors \({{\bf{v}}_{\bf{1}}}\) and \({{\bf{v}}_{\bf{2}}}\) that span a subspace parallel to the 2-dimensional flat \({H_{\bf{1}}} \cap {H_{\bf{2}}}\).)

\({H_1} \cap {H_2} = \left\{ {{\bf{x}}:{\bf{x}} = {\bf{p}} + {x_3}{{\bf{v}}_1} + {x_4}{{\bf{v}}_2}} \right\}\)

See the step by step solution

Step by Step Solution

Step 1: Find the values of \({d_{\bf{1}}}\) and \({d_{\bf{2}}}\)

Find \({d_1}\) for \({H_1}\).

\(\begin{array}{c}{d_1} = {{\bf{n}}_1} \cdot {{\bf{p}}_1}\\ = 1\left( 2 \right) + 2\left( { - 3} \right) + 4\left( 1 \right) + 2\left( 2 \right)\\ = 4\end{array}\)

Find \({d_2}\) for \({H_2}\).

\(\begin{array}{c}{d_2} = {{\bf{n}}_2} \cdot {{\bf{p}}_2}\\ = 2\left( 2 \right) + 3\left( 2 \right) + 1\left( { - 1} \right) + 5\left( 3 \right)\\ = 22\end{array}\)

Step 2: Write the augmented matrix

The system of equations is:

\(\left( {\begin{array}{*{20}{c}}1&2&4&2\end{array}} \right){\bf{x}} = 4\)

\(\left( {\begin{array}{*{20}{c}}2&3&1&5\end{array}} \right){\bf{x}} = 22\)

The augmented matrix can be written as shown below:

\(\left( {\begin{array}{*{20}{c}}1&2&4&2&4\\2&3&1&5&{22}\end{array}} \right) = \left( {\begin{array}{*{20}{c}}1&0&{ - 10}&4&{32}\\0&1&7&{ - 1}&{ - 14}\end{array}} \right)\)

Step 3: Write the general solution

The general solution is:

\(\begin{array}{c}x = \left( {\begin{array}{*{20}{c}}{32}\\{ - 14}\\0\\0\end{array}} \right) + {x_3}\left( {\begin{array}{*{20}{c}}{10}\\{ - 7}\\1\\0\end{array}} \right) + {x_4}\left( {\begin{array}{*{20}{c}}{ - 4}\\1\\0\\1\end{array}} \right)\\ = p + {x_3}{{\bf{v}}_1} + {x_4}{{\bf{v}}_2}\end{array}\)


\(p = \left( {\begin{array}{*{20}{c}}{32}\\{ - 14}\\0\\0\end{array}} \right)\), \({{\bf{v}}_1} = \left( {\begin{array}{*{20}{c}}{10}\\{ - 7}\\1\\0\end{array}} \right)\) and \({{\bf{v}}_2} = \left( {\begin{array}{*{20}{c}}{ - 4}\\1\\0\\1\end{array}} \right)\)

Therefore, \({H_1} \cap {H_2} = \left\{ {{\bf{x}}:{\bf{x}} = {\bf{p}} + {x_3}{{\bf{v}}_1} + {x_4}{{\bf{v}}_2}} \right\}\).

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