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Linear Algebra and its Applications
Found in: Page 191
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

In Exercise 3, find the vector x determined by the given coordinate vector \({\left( x \right)_{\rm B}}\) and the given basis \({\rm B}\).

3. \({\rm B} = \left\{ {\left( {\begin{array}{*{20}{c}}{\bf{1}}\\{ - {\bf{4}}}\\{\bf{3}}\end{array}} \right),\left( {\begin{array}{*{20}{c}}{\bf{5}}\\{\bf{2}}\\{ - {\bf{2}}}\end{array}} \right),\left( {\begin{array}{*{20}{c}}{\bf{4}}\\{ - {\bf{7}}}\\{\bf{0}}\end{array}} \right)} \right\},{\left( x \right)_{\rm B}} = \left( {\begin{array}{*{20}{c}}{\bf{3}}\\{\bf{0}}\\{ - {\bf{1}}}\end{array}} \right)\)

Vector \(x = \left( {\begin{array}{*{20}{c}}{ - 1}\\{ - 5}\\9\end{array}} \right)\)

See the step by step solution

Step by Step Solution

Step 1: Use the definition

The coordinates of x relative to basis \({\rm B} = \left\{ {{b_{\bf{1}}},{b_{\bf{2}}},...,{b_n}} \right\}\) are the weights \({c_{\bf{1}}},{c_{\bf{2}}},...,{c_n}\), such that \(x = {c_{\bf{1}}}{b_{\bf{1}}} + {c_{\bf{2}}}{b_{\bf{2}}} + ... + {c_n}{b_n}\). Then, \({\left( x \right)_{\rm B}} = \left( {\begin{array}{*{20}{c}}{{c_1}}\\{{c_2}}\\ \vdots \\{{c_n}}\end{array}} \right)\).

Step 2: Find x

By the above definition, you get

\[\begin{array}{c}x = 3\left[ {\begin{array}{*{20}{c}}1\\{ - 4}\\3\end{array}} \right] + 0\left[ {\begin{array}{*{20}{c}}5\\2\\{ - 2}\end{array}} \right] + \left( { - 1} \right)\left[ {\begin{array}{*{20}{c}}4\\{ - 7}\\0\end{array}} \right]\\ = \left[ {\begin{array}{*{20}{c}}3\\{ - 12}\\9\end{array}} \right] + 0 + \left[ {\begin{array}{*{20}{c}}{ - 4}\\7\\0\end{array}} \right]\\ = \left[ {\begin{array}{*{20}{c}}{3 - 4}\\{ - 12 + 7}\\9\end{array}} \right]\\x = \left[ {\begin{array}{*{20}{c}}{ - 1}\\{ - 5}\\9\end{array}} \right].\end{array}\]

Step 3: Draw a conclusion

Hence, vector \(x = \left( {\begin{array}{*{20}{c}}{ - 1}\\{ - 5}\\9\end{array}} \right)\).


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