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Q7E

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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 7, find the coordinate vector \({\left( x \right)_{\rm B}}\) of x relative to the given basis \({\rm B} = \left\{ {{b_{\bf{1}}},...,{b_n}} \right\}\).

7. \({b_{\bf{1}}} = \left( {\begin{array}{*{20}{c}}{\bf{1}}\\{ - {\bf{1}}}\\{ - {\bf{3}}}\end{array}} \right),{b_{\bf{2}}} = \left( {\begin{array}{*{20}{c}}{ - {\bf{3}}}\\{\bf{4}}\\{\bf{9}}\end{array}} \right),{b_{\bf{3}}} = \left( {\begin{array}{*{20}{c}}{\bf{2}}\\{ - {\bf{2}}}\\{\bf{4}}\end{array}} \right),x = \left( {\begin{array}{*{20}{c}}{\bf{8}}\\{ - {\bf{9}}}\\{\bf{6}}\end{array}} \right)\)

Coordinate vector \({\left( x \right)_{\rm B}} = \left( {\begin{array}{*{20}{c}}{ - 1}\\{ - 1}\\3\end{array}} \right)\)

See the step by step solution

Step by Step Solution

Step 1: Write the system

Note that \({\left( x \right)_{\rm B}}\)be the solution of the system.

\(\begin{array}{c}\left( {\begin{array}{*{20}{c}}{{b_1}}&{{b_2}}\end{array}} \right){\left( x \right)_{\rm B}} = x\\\left( {\begin{array}{*{20}{c}}1&{ - 3}&2\\{ - 1}&4&{ - 2}\\{ - 3}&9&4\end{array}} \right)\left( {\begin{array}{*{20}{c}}{{c_1}}\\{{c_2}}\\{{c_3}}\end{array}} \right) = \left( {\begin{array}{*{20}{c}}8\\{ - 9}\\6\end{array}} \right)\end{array}\)

Step 2: Find the reduced row echelon form

Its augmented matrix is \(\left( {\begin{array}{*{20}{c}}{{b_1}}&{{b_2}}&x\end{array}} \right) = \left( {\begin{array}{*{20}{c}}1&{ - 3}&2&8\\{ - 1}&4&{ - 2}&{ - 9}\\{ - 3}&9&4&6\end{array}} \right)\).

At row 2, add row 1 to row 2, i.e., \({R_2} \to {R_2} + {R_1}\).

Also, at row 3, multiply row 1 by 3 and add it to row 3, i.e., \({R_3} \to {R_3} + 3{R_1}\).

\( \sim \left( {\begin{array}{*{20}{c}}1&{ - 3}&2&8\\0&1&0&{ - 1}\\0&0&{10}&{30}\end{array}} \right)\)

At row 3, divide row 3 by 10.

\( \sim \left( {\begin{array}{*{20}{c}}1&{ - 3}&2&8\\0&1&0&{ - 1}\\0&0&1&3\end{array}} \right)\)

At row 1, multiply row 2 by 3 and add it to row 1, i.e., \({R_1} \to {R_1} + 2{R_2}\).

\( \sim \left( {\begin{array}{*{20}{c}}1&0&2&5\\0&1&0&{ - 1}\\0&0&1&3\end{array}} \right)\)

At row 1, multiply row 3 by 2 and subtract it from row 1, i.e., \({R_1} \to {R_1} - 2{R_3}\).

\( \sim \left( {\begin{array}{*{20}{c}}1&0&0&{ - 1}\\0&1&0&{ - 1}\\0&0&1&3\end{array}} \right)\)

This implies \({c_1} = - 1,{c_2} = - 1,\) and \({c_3} = 3\).

Step 3: Draw a conclusion

\(\begin{array}{c}{\left( x \right)_{\rm B}} = \left( {\begin{array}{*{20}{c}}{{c_1}}\\{{c_2}}\\{{c_3}}\end{array}} \right)\\ = \left( {\begin{array}{*{20}{c}}{ - 1}\\{ - 1}\\3\end{array}} \right)\end{array}\)

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