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 8E from Chapter 5.4 from Lay's Linear Algebra and Its Applications, 5th Edition.
Problem 8E
Chapter:
Problem:
Let = {b1, b2, b3} be a basis for a vector space V. Find T(3b1 - 4b2) when T is a linear transformation from V to V whose matrix relative to is.
Step-by-Step Solution
Given Information
We are given with a transformation matrix defined by \[ [ T ] _ { \mathcal { B } } = \left[ \begin{array} { r r r } { 0 } & { - 6 } & { 1 } \\ { 0 } & { 5 } & { - 1 } \\ { 1 } & { - 2 } & { 7 } \end{array} \right] \] We have to find the transformation of $3{{\bf{b}}_1} - 4{{\bf{b}}_2}$
Step-1: The vector form
The vector form is given by: \[ \left[ \left( 3 b _ { 1 } - 4 b _ { 2 } + 0 b _ { 3 } \right) \right] _ { \mathrm { B } } = \left[ \begin{array} { c } { 3 } \\ { - 4 } \\ { 0 } \end{array} \right] \]
Step-2: The required Transformation
\[\begin{array}{l} \left[ {T\left( {3{b_1} - 4{b_2}} \right)} \right] = \left[ {\begin{array}{*{20}{c}} 0&{ - 6}&1\\ 0&5&{ - 1}\\ 1&{ - 2}&7 \end{array}} \right]\left[ {\begin{array}{*{20}{c}} 3\\ { - 4}\\ 0 \end{array}} \right]\\ = \left[ {\begin{array}{*{20}{c}} {0 + 24 + 0}\\ {0 - 20 + 0}\\ {3 + 8 + 0} \end{array}} \right]\\ = \left[ {\begin{array}{*{20}{c}} {24}\\ { - 20}\\ {11} \end{array}} \right] \end{array}\] Therefore,
\[ T \left( 3 b _ { 1 } - 4 b _ { 2 } \right) = 24 b _ { 1 } - 20 b _ { 2 } + 11 b _ { 3 } \]
We are given with a transformation matrix defined by \[ [ T ] _ { \mathcal { B } } = \left[ \begin{array} { r r r } { 0 } & { - 6 } & { 1 } \\ { 0 } & { 5 } & { - 1 } \\ { 1 } & { - 2 } & { 7 } \end{array} \right] \] We have to find the transformation of $3{{\bf{b}}_1} - 4{{\bf{b}}_2}$
Step-1: The vector form
The vector form is given by: \[ \left[ \left( 3 b _ { 1 } - 4 b _ { 2 } + 0 b _ { 3 } \right) \right] _ { \mathrm { B } } = \left[ \begin{array} { c } { 3 } \\ { - 4 } \\ { 0 } \end{array} \right] \]
Step-2: The required Transformation
\[\begin{array}{l} \left[ {T\left( {3{b_1} - 4{b_2}} \right)} \right] = \left[ {\begin{array}{*{20}{c}} 0&{ - 6}&1\\ 0&5&{ - 1}\\ 1&{ - 2}&7 \end{array}} \right]\left[ {\begin{array}{*{20}{c}} 3\\ { - 4}\\ 0 \end{array}} \right]\\ = \left[ {\begin{array}{*{20}{c}} {0 + 24 + 0}\\ {0 - 20 + 0}\\ {3 + 8 + 0} \end{array}} \right]\\ = \left[ {\begin{array}{*{20}{c}} {24}\\ { - 20}\\ {11} \end{array}} \right] \end{array}\] Therefore,
\[ T \left( 3 b _ { 1 } - 4 b _ { 2 } \right) = 24 b _ { 1 } - 20 b _ { 2 } + 11 b _ { 3 } \]