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 - 3.4 Summary of Determinants - Problems - Page 241: 13

Answer

$A^{-1}=\begin{bmatrix} \frac{1}{4} &0 & \frac{-1}{4} & 0\\ \frac{-4}{19} & \frac{-7}{38} & \frac{31}{38} & \frac{3}{19} \\ \frac{-17}{76} &\frac{-11}{38}& \frac{111}{76} & \frac{2}{19} \\ \frac{-1}{76} & \frac{15}{38} & \frac{-65}{76} & \frac{-1}{19} \end{bmatrix}$

Work Step by Step

Given: $A=\begin{bmatrix} 5 & -1 & 2 & 1 \\ 3 & -1 & 4 & 5 \\ 1 & -1 & 2 &1 \\ 5 & 9 & -3 & 2 \end{bmatrix} $ We will gradually find: $C_{11}=-38$ $C_{12}=32$ $C_{13}=34$ $C_{14}=2$ $C_{21}=0$ $C_{22}=28$ $C_{23}=44$ $C_{24}=-60$ $C_{31}=38$ $C_{32}=-124$ $C_{33}=-222$ $C_{34}=130$ $C_{41}=0$ $C_{42}=-24$ $C_{43}=-16$ $C_{44}=8$ $\det(A)=5.(-38)-1.32+2(34)+1.2=$ Since $D=-152\ne0$, the matrix is invertible. $ adj(A)=\begin{bmatrix} -38 &0& 38 & 0\\ 32 & 28 & -124 & -24\\ 34& 44& -222 & -16\\ 2 & -60 & 130 & 8 \end{bmatrix}$ Use the adjoint method to find $A^{-1}:$ $A^{-1}=\frac{1}{\det(A)}adj(A)$ $A^{-1}=\frac{1}{-152}\begin{bmatrix} -38 &0& 38 & 0\\ 32 & 28 & -124 & -24\\ 34& 44& -222 & -16\\ 2 & -60 & 130 & 8 \end{bmatrix}$ $A^{-1}=\begin{bmatrix} \frac{1}{4} &0 & \frac{-1}{4} & 0\\ \frac{-4}{19} & \frac{-7}{38} & \frac{31}{38} & \frac{3}{19} \\ \frac{-17}{76} &\frac{-11}{38}& \frac{111}{76} & \frac{2}{19} \\ \frac{-1}{76} & \frac{15}{38} & \frac{-65}{76} & \frac{-1}{19} \end{bmatrix}$
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