Linear Algebra and Its Applications, 5th Edition
Authors: David C. Lay, Steven R. Lay, Judi J. McDonald
ISBN-13: 978-0321982384
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See our solution for Question 1E from Chapter 4.4 from Lay's Linear Algebra and Its Applications, 5th Edition.
Problem 1E
Chapter:
Problem:
In Exercises 1-4, find the vector x determined by the given coordinate vector...
Step-by-Step Solution
Step 1
Given Basis:\[\beta = \left\{ {\left[ {\begin{array}{*{20}{r}}3\\{ - 5}\end{array}} \right],\left[ {\begin{array}{*{20}{r}}{ - 4}\\6\end{array}} \right]} \right\}\]Given Coordinate Vector:\[{[{\bf{x}}]_\beta } = \left[ {\begin{array}{*{20}{l}}5\\3\end{array}} \right]\]We have to find the vector x from the given basis $\beta$ and coordinate vector. By Theorem-7: Suppose $\beta = \left\{ {{{\bf{b}}_1}, \ldots ,{{\bf{b}}_n}} \right\}$ is a basis for V and $\bf{x}$ is in V . The coordinates of $\bf{x}$ relative to the basis $\beta$ are the weights ${c_1},{c_2}....{c_n}$ such that ${\bf{x}} = {c_1}{{\bf{b}}_1} + ,..., + {c_n}{{\bf{b}}_n}$
Step 2: The required vector $\bf{x}$
\[\begin{array}{l}{\bf{x}} = {\left[ x \right]_\beta }\beta \\ = 5\left[ {\begin{array}{*{20}{c}}3\\{ - 5}\end{array}} \right] + 3\left[ {\begin{array}{*{20}{c}}{ - 4}\\6\end{array}} \right]\\ = \left[ {\begin{array}{*{20}{c}}{15}\\{ - 25}\end{array}} \right] + \left[ {\begin{array}{*{20}{c}}{ - 12}\\{18}\end{array}} \right]\\ = \left[ {\begin{array}{*{20}{c}}{15 - 12}\\{ - 25 + 18}\end{array}} \right]\\ = \left[ {\begin{array}{*{20}{c}}3\\{ - 7}\end{array}} \right]\end{array}\]
ANSWER
\[{\bf{x}} = \left[ {\begin{array}{*{20}{c}}3\\{ - 7}\end{array}} \right]\]
Given Basis:\[\beta = \left\{ {\left[ {\begin{array}{*{20}{r}}3\\{ - 5}\end{array}} \right],\left[ {\begin{array}{*{20}{r}}{ - 4}\\6\end{array}} \right]} \right\}\]Given Coordinate Vector:\[{[{\bf{x}}]_\beta } = \left[ {\begin{array}{*{20}{l}}5\\3\end{array}} \right]\]We have to find the vector x from the given basis $\beta$ and coordinate vector. By Theorem-7: Suppose $\beta = \left\{ {{{\bf{b}}_1}, \ldots ,{{\bf{b}}_n}} \right\}$ is a basis for V and $\bf{x}$ is in V . The coordinates of $\bf{x}$ relative to the basis $\beta$ are the weights ${c_1},{c_2}....{c_n}$ such that ${\bf{x}} = {c_1}{{\bf{b}}_1} + ,..., + {c_n}{{\bf{b}}_n}$
Step 2: The required vector $\bf{x}$
\[\begin{array}{l}{\bf{x}} = {\left[ x \right]_\beta }\beta \\ = 5\left[ {\begin{array}{*{20}{c}}3\\{ - 5}\end{array}} \right] + 3\left[ {\begin{array}{*{20}{c}}{ - 4}\\6\end{array}} \right]\\ = \left[ {\begin{array}{*{20}{c}}{15}\\{ - 25}\end{array}} \right] + \left[ {\begin{array}{*{20}{c}}{ - 12}\\{18}\end{array}} \right]\\ = \left[ {\begin{array}{*{20}{c}}{15 - 12}\\{ - 25 + 18}\end{array}} \right]\\ = \left[ {\begin{array}{*{20}{c}}3\\{ - 7}\end{array}} \right]\end{array}\]
ANSWER
\[{\bf{x}} = \left[ {\begin{array}{*{20}{c}}3\\{ - 7}\end{array}} \right]\]