Linear Algebra and Its Applications, 5th Edition
Authors: David C. Lay, Steven R. Lay, Judi J. McDonald
ISBN-13: 978-0321982384
We have solutions for your book!
See our solution for Question 5E from Chapter 2.4 from Lay's Linear Algebra and Its Applications, 5th Edition.
Problem 5E
Chapter:
Problem:
In Exercises 1–9, assume that the matrices are partitioned conformably for block multiplication.
Step-by-Step Solution
Given Information
We have to find formulas for X, Y, and Z in terms of A, B, and C. We are given with partitioned matrices. \[ \left[ \begin{array} { l l } { A } & { B } \\ { C } & { 0 } \end{array} \right] \left[ \begin{array} { l l } { I } & { 0 } \\ { X } & { Y } \end{array} \right] = \left[ \begin{array} { l l } { 0 } & { I } \\ { Z } & { 0 } \end{array} \right] \]
Step-1:
\[ \left[ \begin{array} { c c } { A } & { B } \\ { C } & { 0 } \end{array} \right] \left[ \begin{array} { c c } { I } & { 0 } \\ { X } & { Y } \end{array} \right] = \left[ \begin{array} { c c } { A \cdot I + B \cdot X } & { A \cdot 0 + B \cdot Y } \\ { C \cdot I + 0 \cdot X } & { C \cdot 0 + 0 \cdot Y } \end{array} \right] \] Using the matrix properties \[\begin{array}{*{20}{l}} {I\cdot A = A}\\ {0\cdot A = 0}\\ {A + 0 = A} \end{array}\] We get \[ \begin{aligned} \left[ \begin{array} { c c } { A } & { B } \\ { C } & { 0 } \end{array} \right] \left[ \begin{array} { c c } { I } & { 0 } \\ { X } & { Y } \end{array} \right] & = \left[ \begin{array} { c c } { A \cdot I + B \cdot X } & { A \cdot 0 + B \cdot Y } \\ { C \cdot I + 0 \cdot X } & { C \cdot 0 + 0 \cdot Y } \end{array} \right] \\ & = \left[ \begin{array} { c c } { A + B X } & { B Y } \\ { C } & { 0 } \end{array} \right] \end{aligned} \]
Step-2: Equate both sides
\[ \left[ \begin{array} { c c } { A + B X } & { B Y } \\ { C } & { 0 } \end{array} \right] = \left[ \begin{array} { l l } { 0 } & { I } \\ { Z } & { 0 } \end{array} \right] \] Hence, \[ \begin{aligned} A + B X & = 0 \\ B Y & = I \\ C & = Z \end{aligned} \] From last row: $Z = C$
From second row: $BY = I \Rightarrow Y = {B^{ - 1}}$
From First row: \[\begin{array}{l} A + BX = 0\\ BX = 0 - A\\ BX = - A\\ {B^{ - 1}}(BX) = {B^{ - 1}}( - A)\\ \left( {{B^{ - 1}}B} \right)X = {B^{ - 1}}( - A)\\ \left( {{B^{ - 1}}B} \right)X = - {B^{ - 1}}A\\ IX = - {B^{ - 1}}A\\ X = - {B^{ - 1}}A \end{array}\] Therefore, the product is:
\[ \begin{array} { l } { X = - B ^ { - 1 } A } \\ { Y = B ^ { - 1 } } \\ { Z = C } \end{array} \]
We have to find formulas for X, Y, and Z in terms of A, B, and C. We are given with partitioned matrices. \[ \left[ \begin{array} { l l } { A } & { B } \\ { C } & { 0 } \end{array} \right] \left[ \begin{array} { l l } { I } & { 0 } \\ { X } & { Y } \end{array} \right] = \left[ \begin{array} { l l } { 0 } & { I } \\ { Z } & { 0 } \end{array} \right] \]
Step-1:
\[ \left[ \begin{array} { c c } { A } & { B } \\ { C } & { 0 } \end{array} \right] \left[ \begin{array} { c c } { I } & { 0 } \\ { X } & { Y } \end{array} \right] = \left[ \begin{array} { c c } { A \cdot I + B \cdot X } & { A \cdot 0 + B \cdot Y } \\ { C \cdot I + 0 \cdot X } & { C \cdot 0 + 0 \cdot Y } \end{array} \right] \] Using the matrix properties \[\begin{array}{*{20}{l}} {I\cdot A = A}\\ {0\cdot A = 0}\\ {A + 0 = A} \end{array}\] We get \[ \begin{aligned} \left[ \begin{array} { c c } { A } & { B } \\ { C } & { 0 } \end{array} \right] \left[ \begin{array} { c c } { I } & { 0 } \\ { X } & { Y } \end{array} \right] & = \left[ \begin{array} { c c } { A \cdot I + B \cdot X } & { A \cdot 0 + B \cdot Y } \\ { C \cdot I + 0 \cdot X } & { C \cdot 0 + 0 \cdot Y } \end{array} \right] \\ & = \left[ \begin{array} { c c } { A + B X } & { B Y } \\ { C } & { 0 } \end{array} \right] \end{aligned} \]
Step-2: Equate both sides
\[ \left[ \begin{array} { c c } { A + B X } & { B Y } \\ { C } & { 0 } \end{array} \right] = \left[ \begin{array} { l l } { 0 } & { I } \\ { Z } & { 0 } \end{array} \right] \] Hence, \[ \begin{aligned} A + B X & = 0 \\ B Y & = I \\ C & = Z \end{aligned} \] From last row: $Z = C$
From second row: $BY = I \Rightarrow Y = {B^{ - 1}}$
From First row: \[\begin{array}{l} A + BX = 0\\ BX = 0 - A\\ BX = - A\\ {B^{ - 1}}(BX) = {B^{ - 1}}( - A)\\ \left( {{B^{ - 1}}B} \right)X = {B^{ - 1}}( - A)\\ \left( {{B^{ - 1}}B} \right)X = - {B^{ - 1}}A\\ IX = - {B^{ - 1}}A\\ X = - {B^{ - 1}}A \end{array}\] Therefore, the product is:
\[ \begin{array} { l } { X = - B ^ { - 1 } A } \\ { Y = B ^ { - 1 } } \\ { Z = C } \end{array} \]