Differential Equations and Linear Algebra (4th Edition)

Published by Pearson
ISBN 10: 0-32196-467-5
ISBN 13: 978-0-32196-467-0

Chapter 3 - Determinants - 3.3 Cofactor Expansions - Problems - Page 234: 48

Answer

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Work Step by Step

$C_{11}=(-1)^{1+1}.\begin{vmatrix} 1 & 1 &3\\ 9 & 0& 2\\ 0 & 3 &-1 \end{vmatrix}=1.84=84$ $C_{21}=(-1)^{2+1}.\begin{vmatrix} 0 & 3 &5\\ 6 & 0 & 2\\ 0& 3 &-1 \end{vmatrix}=(-1).162=-162$ $C_{31}=(-1)^{3+1}.\begin{vmatrix} 0 & 3 &5\\ 1 & 1 & 3\\ 1 & 3 &-1 \end{vmatrix}=1.18=18$ $C_{41}=(-1)^{4+1}.\begin{vmatrix} 0 & 3 &5\\ 1 & 1 & 3\\ 9 & 0 &2 \end{vmatrix}=(-1).30=-30$ $C_{12}=(-1)^{1+2}.\begin{vmatrix} -2 & 1 &3\\ 3 & 0& 2\\ 2 & 3 &-1 \end{vmatrix}=(-1).1.2=-2$ $C_{22}=(-1)^{2+2}.\begin{vmatrix} 1 & 3 &5\\ 3 & 0 & 2\\ 2 & 3 &-1 \end{vmatrix}=1.60=60$ $C_{42}=(-1)^{4+2}.\begin{vmatrix} 1 & 3 &5\\ -2 & 1 & 3\\ 3 & 0 &2 \end{vmatrix}=1.26=26$ $C_{13}=(-1)^{1+3}.\begin{vmatrix} -2 & 1 &3\\ 3 & 9 & 2\\ 2 & 0 &-1 \end{vmatrix}=(-1).29=-29$ $C_{32}=(-1)^{2+3}.\begin{vmatrix} 1 & 3 &5\\ -2 &1 & 3\\ 2 & 3 &-1 \end{vmatrix}=(-1).(-38)=38$ $C_{23}=(-1)^{2+3}.\begin{vmatrix} 1 & 0 &5\\ 3 & 9 & 2\\ 2 & 0 &-1 \end{vmatrix}=(-1).(-99)=99$ $C_{33}=(-1)^{3+3}.\begin{vmatrix} 1 & 0 &5\\ -2 & 1 & 3\\ 2 & 0&-1 \end{vmatrix}=(-11).1=-11$ $C_{43}=(-1)^{4+3}.\begin{vmatrix} 1 & 0 &5\\ -2 & 1 & 3\\ 2 & 0 &-1 \end{vmatrix}=(-1).(-130)=130$ $C_{14}=(-1)^{4+1}.\begin{vmatrix} -2 & 1 &1\\ 3 & 9 & 0\\ 2 & 0 &3 \end{vmatrix}=(-)1.(-81)=81$ $C_{24}=(-1)^{2+4}.\begin{vmatrix} 1 & 0 &3\\ 3 & 9 & 0\\ 2 & 0 &3 \end{vmatrix}=1.(-27)=-27$ $C_{34}=(-1)^{3+4}.\begin{vmatrix} 1 & 0 &3\\ -2 & 1 & 1\\ 2 & 0 &3 \end{vmatrix}=(-1).(-3)=3$ $C_{44}=(-1)^{4+4}.\begin{vmatrix} 1 & 0 &3\\ -2 & 1 & 1\\ 3 & 9 &0 \end{vmatrix}=1.(-72)=-72$ a) We have $det(A)=a_{11}C_{11}+a_{21}C_{21}+a_{31}C_{31}+a_{41}C_{41}=-282$ b)$M_C=\begin{bmatrix} 84 &-46 & -29 &81\\ -162& 66&99&-27\\ 18 &38 & -11 &3\\ -30 &26 &130 &- 72 \end{bmatrix}$ c) $adj(A)=\begin{bmatrix} 84 &-162 & -18 &-30\\ -46& 60&38&26\\ -29&99 & -11 &130\\ 81 &-27 &3 &- 72 \end{bmatrix}$ d) Because $\det(A)=6 \ne 0$, we are able to find $A^{-1}$ by using the theorem: $A^{-1}=\frac{1}{\det(A)}adj(A)=\frac{-1}{282}.\begin{bmatrix} 84 &-162 & -18 &-30\\ -46& 60&38&26\\ -29&99 & -11 &130\\ 81 &-27 &3 &- 72 \end{bmatrix}$
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