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Found in: Page 191

### 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

# 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 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

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