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 1.9 from Lay's Linear Algebra and Its Applications, 5th Edition.
Problem 1E
Chapter:
Problem:
In Exercises 1-10, assume that T is a linear transformation. Find the standard matrix of T...
Step-by-Step Solution
Step 1
Let the Vectors are:\[{{\bf{e}}_{\bf{1}}} = \left[ {\begin{array}{*{20}{c}}0\\1\end{array}} \right];\,\,\,{{\bf{e}}_{\bf{2}}} = \left[ {\begin{array}{*{20}{c}}1\\0\end{array}} \right]\]Given Transformation :\[T:{{\rm{R}}^2} \to {{\rm{R}}^4}::T\left( {{{\bf{e}}_1}} \right) = \left[ {\begin{array}{*{20}{c}}3\\1\\3\\1\end{array}} \right],{\mkern 1mu} {\mkern 1mu} :T\left( {{{\bf{e}}_2}} \right) = \left[ {\begin{array}{*{20}{c}}{ - 5}\\2\\0\\0\end{array}} \right]{\mkern 1mu} \]We have to find the transformation matrix for the above transformation.
Step 2: The Transformation Matrix
The Transformation Matrix is the matrix A such that:\[T\left( {\bf{x}} \right) = A\left( {\bf{x}} \right)\]So: \[\begin{array}{l}A = \left[ {\begin{array}{*{20}{c}}{T\left( {{{\bf{e}}_{\bf{1}}}} \right)}&{T\left( {{{\bf{e}}_2}} \right)}\end{array}} \right]\\\\ = \left[ {\begin{array}{*{20}{c}}{T\left( {\begin{array}{*{20}{c}}0\\1\end{array}} \right)}&{T\left( {\begin{array}{*{20}{c}}1\\0\end{array}} \right)}\end{array}} \right]\\\\ = \left[ {\begin{array}{*{20}{c}}{\begin{array}{*{20}{c}}3\\1\\3\\1\end{array}}&{\begin{array}{*{20}{c}}{ - 5}\\2\\0\\0\end{array}}\end{array}} \right]\end{array}\]
ANSWER
\[A = \left[ {\begin{array}{*{20}{c}}{\begin{array}{*{20}{c}}3\\1\\3\\1\end{array}}&{\begin{array}{*{20}{c}}{ - 5}\\2\\0\\0\end{array}}\end{array}} \right]\]
Let the Vectors are:\[{{\bf{e}}_{\bf{1}}} = \left[ {\begin{array}{*{20}{c}}0\\1\end{array}} \right];\,\,\,{{\bf{e}}_{\bf{2}}} = \left[ {\begin{array}{*{20}{c}}1\\0\end{array}} \right]\]Given Transformation :\[T:{{\rm{R}}^2} \to {{\rm{R}}^4}::T\left( {{{\bf{e}}_1}} \right) = \left[ {\begin{array}{*{20}{c}}3\\1\\3\\1\end{array}} \right],{\mkern 1mu} {\mkern 1mu} :T\left( {{{\bf{e}}_2}} \right) = \left[ {\begin{array}{*{20}{c}}{ - 5}\\2\\0\\0\end{array}} \right]{\mkern 1mu} \]We have to find the transformation matrix for the above transformation.
Step 2: The Transformation Matrix
The Transformation Matrix is the matrix A such that:\[T\left( {\bf{x}} \right) = A\left( {\bf{x}} \right)\]So: \[\begin{array}{l}A = \left[ {\begin{array}{*{20}{c}}{T\left( {{{\bf{e}}_{\bf{1}}}} \right)}&{T\left( {{{\bf{e}}_2}} \right)}\end{array}} \right]\\\\ = \left[ {\begin{array}{*{20}{c}}{T\left( {\begin{array}{*{20}{c}}0\\1\end{array}} \right)}&{T\left( {\begin{array}{*{20}{c}}1\\0\end{array}} \right)}\end{array}} \right]\\\\ = \left[ {\begin{array}{*{20}{c}}{\begin{array}{*{20}{c}}3\\1\\3\\1\end{array}}&{\begin{array}{*{20}{c}}{ - 5}\\2\\0\\0\end{array}}\end{array}} \right]\end{array}\]
ANSWER
\[A = \left[ {\begin{array}{*{20}{c}}{\begin{array}{*{20}{c}}3\\1\\3\\1\end{array}}&{\begin{array}{*{20}{c}}{ - 5}\\2\\0\\0\end{array}}\end{array}} \right]\]