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 3E from Chapter 2.2 from Lay's Linear Algebra and Its Applications, 5th Edition.
Problem 3E
Chapter:
Problem:
Find the inverses of the matrices in Exercises...
Step-by-Step Solution
Step 1
We are given with following Matrix\[A = \left[ {\begin{array}{*{20}{c}}8&5\\{ - 7}&{ - 5}\end{array}} \right]\]We have to find the inverse of the above matrix, A using the definition given in Theorem 4 of the textbook.
As per the theorem, the inverse of a matrix $A = \left[ {\begin{array}{*{20}{l}}a&b\\c&d\end{array}} \right]$ is given by the following equations (with the condition that the matrix A is invertible and $ad - bc \ne 0$):
$${A^{ - 1}} = \dfrac{1}{{ad - bc}}\left[ {\begin{array}{*{20}{c}}d&{ - b}\\{ - c}&a\end{array}} \right]$$
Step 2: Find the Inverse
Substitute the elements of the matrix in above formula for matrix inverse: \[\begin{array}{l}{A^{ - 1}} = \dfrac{1}{{( - 5)}}\left[ {\begin{array}{*{20}{c}}{ - 5}&{ - 5}\\7&8\end{array}} \right]\\\\ = \dfrac{1}{{8( - 5) - ( - 7)5}}\left[ {\begin{array}{*{20}{c}}{ - 5}&{ - 5}\\7&8\end{array}} \right]\\\\ = \dfrac{1}{{ - 5}}\left[ {\begin{array}{*{20}{c}}{ - 5}&{ - 5}\\7&8\end{array}} \right]\\\\ = \left[ {\begin{array}{*{20}{c}}1&1\\{ - \dfrac{7}{5}}&{\dfrac{{ - 8}}{5}}\end{array}} \right]\end{array}\]
ANSWER
\[{A^{ - 1}} = \left[ {\begin{array}{*{20}{c}}1&1\\{ - \dfrac{7}{5}}&{\dfrac{{ - 8}}{5}}\end{array}} \right]\]
We are given with following Matrix\[A = \left[ {\begin{array}{*{20}{c}}8&5\\{ - 7}&{ - 5}\end{array}} \right]\]We have to find the inverse of the above matrix, A using the definition given in Theorem 4 of the textbook.
As per the theorem, the inverse of a matrix $A = \left[ {\begin{array}{*{20}{l}}a&b\\c&d\end{array}} \right]$ is given by the following equations (with the condition that the matrix A is invertible and $ad - bc \ne 0$):
$${A^{ - 1}} = \dfrac{1}{{ad - bc}}\left[ {\begin{array}{*{20}{c}}d&{ - b}\\{ - c}&a\end{array}} \right]$$
Step 2: Find the Inverse
Substitute the elements of the matrix in above formula for matrix inverse: \[\begin{array}{l}{A^{ - 1}} = \dfrac{1}{{( - 5)}}\left[ {\begin{array}{*{20}{c}}{ - 5}&{ - 5}\\7&8\end{array}} \right]\\\\ = \dfrac{1}{{8( - 5) - ( - 7)5}}\left[ {\begin{array}{*{20}{c}}{ - 5}&{ - 5}\\7&8\end{array}} \right]\\\\ = \dfrac{1}{{ - 5}}\left[ {\begin{array}{*{20}{c}}{ - 5}&{ - 5}\\7&8\end{array}} \right]\\\\ = \left[ {\begin{array}{*{20}{c}}1&1\\{ - \dfrac{7}{5}}&{\dfrac{{ - 8}}{5}}\end{array}} \right]\end{array}\]
ANSWER
\[{A^{ - 1}} = \left[ {\begin{array}{*{20}{c}}1&1\\{ - \dfrac{7}{5}}&{\dfrac{{ - 8}}{5}}\end{array}} \right]\]