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 21E from Chapter 2.4 from Lay's Linear Algebra and Its Applications, 5th Edition.
Problem 21E
Chapter:
Problem:
a. Verify that A2 = I when .b. Use partitioned matrices to show that M2 = I when
Step-by-Step Solution
Given Information
We are given with a matrix: \[ A = \left[ \begin{array} { r r } { 1 } & { 0 } \\ { 3 } & { - 1 } \end{array} \right] \] We have to verify that $A ^ { 2 } = I$ and also show that $M ^ { 2 } = I$, where M is \[ M = \left[ \begin{array} { r r r r } { 1 } & { 0 } & { 0 } & { 0 } \\ { 3 } & { - 1 } & { 0 } & { 0 } \\ { 1 } & { 0 } & { - 1 } & { 0 } \\ { 0 } & { 1 } & { - 3 } & { 1 } \end{array} \right] \]
Step-1: (a) $A ^ { 2 } = I$
\[\begin{array}{l} {A^2} = A.A\\ = \left[ {\begin{array}{*{20}{c}} 1&0\\ 3&{ - 1} \end{array}} \right]\left[ {\begin{array}{*{20}{c}} 1&0\\ 3&{ - 1} \end{array}} \right]\\ = \left[ {\begin{array}{*{20}{c}} {1 + 0}&{0 + 0}\\ {3 - 3}&{0 + {{( - 1)}^2}} \end{array}} \right]\\ = \left[ {\begin{array}{*{20}{l}} 1&0\\ 0&1 \end{array}} \right]\\ = I \end{array}\]
Step-2: (b) $M ^ { 2 } = I$
Partition M \[ M = \left[ \begin{array} { c c c c c } { 1 } & { 0 } & { \vdots } & { 0 } & { 0 } \\ { 3 } & { - 1 } & { \vdots } & { 0 } & { 0 } \\ { \dots } & { \ldots } & { \dots } & { \dots } & { \ldots } \\ { 1 } & { 0 } & { \vdots } & { - 1 } & { 0 } \\ { 0 } & { 1 } & { \vdots } & { - 3 } & { 1 } \end{array} \right] \] Such that \[ M = \left[ \begin{array} { c c } { A } & { 0 } \\ { I } & { - A } \end{array} \right] \] Upon multiplying: \[\begin{array}{l} {M^2} = \left[ {\begin{array}{*{20}{c}} A&0\\ I&{ - A} \end{array}} \right]\left[ {\begin{array}{*{20}{c}} A&0\\ I&{ - A} \end{array}} \right]\\ = \left[ {\begin{array}{*{20}{c}} {{A^2} + 0}&{0 + 0}\\ {A - A}&{0 + {{( - A)}^2}} \end{array}} \right]\\ = \left[ {\begin{array}{*{20}{l}} {{A^2}}&O\\ O&{{A^2}} \end{array}} \right]\\ = \left[ {\begin{array}{*{20}{l}} I&O\\ O&I \end{array}} \right]\\ = I \end{array}\]
We are given with a matrix: \[ A = \left[ \begin{array} { r r } { 1 } & { 0 } \\ { 3 } & { - 1 } \end{array} \right] \] We have to verify that $A ^ { 2 } = I$ and also show that $M ^ { 2 } = I$, where M is \[ M = \left[ \begin{array} { r r r r } { 1 } & { 0 } & { 0 } & { 0 } \\ { 3 } & { - 1 } & { 0 } & { 0 } \\ { 1 } & { 0 } & { - 1 } & { 0 } \\ { 0 } & { 1 } & { - 3 } & { 1 } \end{array} \right] \]
Step-1: (a) $A ^ { 2 } = I$
\[\begin{array}{l} {A^2} = A.A\\ = \left[ {\begin{array}{*{20}{c}} 1&0\\ 3&{ - 1} \end{array}} \right]\left[ {\begin{array}{*{20}{c}} 1&0\\ 3&{ - 1} \end{array}} \right]\\ = \left[ {\begin{array}{*{20}{c}} {1 + 0}&{0 + 0}\\ {3 - 3}&{0 + {{( - 1)}^2}} \end{array}} \right]\\ = \left[ {\begin{array}{*{20}{l}} 1&0\\ 0&1 \end{array}} \right]\\ = I \end{array}\]
Step-2: (b) $M ^ { 2 } = I$
Partition M \[ M = \left[ \begin{array} { c c c c c } { 1 } & { 0 } & { \vdots } & { 0 } & { 0 } \\ { 3 } & { - 1 } & { \vdots } & { 0 } & { 0 } \\ { \dots } & { \ldots } & { \dots } & { \dots } & { \ldots } \\ { 1 } & { 0 } & { \vdots } & { - 1 } & { 0 } \\ { 0 } & { 1 } & { \vdots } & { - 3 } & { 1 } \end{array} \right] \] Such that \[ M = \left[ \begin{array} { c c } { A } & { 0 } \\ { I } & { - A } \end{array} \right] \] Upon multiplying: \[\begin{array}{l} {M^2} = \left[ {\begin{array}{*{20}{c}} A&0\\ I&{ - A} \end{array}} \right]\left[ {\begin{array}{*{20}{c}} A&0\\ I&{ - A} \end{array}} \right]\\ = \left[ {\begin{array}{*{20}{c}} {{A^2} + 0}&{0 + 0}\\ {A - A}&{0 + {{( - A)}^2}} \end{array}} \right]\\ = \left[ {\begin{array}{*{20}{l}} {{A^2}}&O\\ O&{{A^2}} \end{array}} \right]\\ = \left[ {\begin{array}{*{20}{l}} I&O\\ O&I \end{array}} \right]\\ = I \end{array}\]