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 7E from Chapter 7.1 from Lay's Linear Algebra and Its Applications, 5th Edition.
Problem 7E
Chapter:
Problem:
Determine which of the matrices in Exercises 7–12 are orthogonal. If orthogonal, find the inverse.
Step-by-Step Solution
Given Information
We are given with a matrix: \[ A = \left[ \begin{array} { c c } { 0.6 } & { 0.8 } \\ { 0.8 } & { - 0.6 } \end{array} \right] \] We have to check if the matrix is orthogonal or not.
Step-1:
The matrix is said to be orthogonal if, \[ A A ^ { T } = A ^ { T } A = I \]
Step-2: Find the transpose of A
\[ A ^ { T } = \left[ \begin{array} { c c } { 0.6 } & { 0.8 } \\ { 0.8 } & { - 0.6 } \end{array} \right] \]
Step-3: The product of A and $A^T$
\[ \begin{aligned} A A ^ { T } & = \left[ \begin{array} { c c } { 0.6 } & { 0.8 } \\ { 0.8 } & { - 0.6 } \end{array} \right] \left[ \begin{array} { c c } { 0.6 } & { 0.8 } \\ { 0.8 } & { - 0.6 } \end{array} \right] \\ & = \left[ \begin{array} { c c } { ( 0.6 ) ( 0.6 ) + ( 0.8 ) ( 0.8 ) } & { ( 0.6 ) ( 0.8 ) + ( 0.8 ) ( - 0.6 ) } \\ { ( 0.8 ) ( 0.6 ) + ( - 0.6 ) ( 0.8 ) } & { ( 0.8 ) ( 0.8 ) + ( - 0.6 ) ( - 0.6 ) } \end{array} \right] \\ & = \left[ \begin{array} { c c } { 0.36 + 0.64 } & { 0.48 - 0.48 } \\ { 0.36 + 0.64 } & { 0.48 - 0.48 } \end{array} \right] \\ & = \left[ \begin{array} { c c } { 1 } & { 0 } \\ { 0 } & { 1 } \end{array} \right] \\ & = I \end{aligned} \] Therefore,
The matrix is orthogonal
Step-4: The inverse of matrix A
For an orthogonal matrix, the inverse is equal to its transpose
Therefore,
\[ A ^ { - 1 } = A ^ { T } = \left[ \begin{array} { c c } { 0.6 } & { 0.8 } \\ { 0.8 } & { - 0.6 } \end{array} \right] \]
We are given with a matrix: \[ A = \left[ \begin{array} { c c } { 0.6 } & { 0.8 } \\ { 0.8 } & { - 0.6 } \end{array} \right] \] We have to check if the matrix is orthogonal or not.
Step-1:
The matrix is said to be orthogonal if, \[ A A ^ { T } = A ^ { T } A = I \]
Step-2: Find the transpose of A
\[ A ^ { T } = \left[ \begin{array} { c c } { 0.6 } & { 0.8 } \\ { 0.8 } & { - 0.6 } \end{array} \right] \]
Step-3: The product of A and $A^T$
\[ \begin{aligned} A A ^ { T } & = \left[ \begin{array} { c c } { 0.6 } & { 0.8 } \\ { 0.8 } & { - 0.6 } \end{array} \right] \left[ \begin{array} { c c } { 0.6 } & { 0.8 } \\ { 0.8 } & { - 0.6 } \end{array} \right] \\ & = \left[ \begin{array} { c c } { ( 0.6 ) ( 0.6 ) + ( 0.8 ) ( 0.8 ) } & { ( 0.6 ) ( 0.8 ) + ( 0.8 ) ( - 0.6 ) } \\ { ( 0.8 ) ( 0.6 ) + ( - 0.6 ) ( 0.8 ) } & { ( 0.8 ) ( 0.8 ) + ( - 0.6 ) ( - 0.6 ) } \end{array} \right] \\ & = \left[ \begin{array} { c c } { 0.36 + 0.64 } & { 0.48 - 0.48 } \\ { 0.36 + 0.64 } & { 0.48 - 0.48 } \end{array} \right] \\ & = \left[ \begin{array} { c c } { 1 } & { 0 } \\ { 0 } & { 1 } \end{array} \right] \\ & = I \end{aligned} \] Therefore,
The matrix is orthogonal
Step-4: The inverse of matrix A
For an orthogonal matrix, the inverse is equal to its transpose
Therefore,
\[ A ^ { - 1 } = A ^ { T } = \left[ \begin{array} { c c } { 0.6 } & { 0.8 } \\ { 0.8 } & { - 0.6 } \end{array} \right] \]