Linear Algebra and Its Applications (5th Edition)

Published by Pearson
ISBN 10: 032198238X
ISBN 13: 978-0-32198-238-4

Chapter 2 - Matrix Algebra - 2.2 Exercises - Page 111: 7

Answer

$A^{-1}=\begin{bmatrix}6&-1\\-5/2&1/2\\\end{bmatrix}$ $Ax = b_{1}$ => x= $\begin{bmatrix}-9\\4\\\end{bmatrix}$ $Ax = b_{2}$ => x= $\begin{bmatrix}11\\-5\\\end{bmatrix}$ $Ax = b_{3}$ => x= $\begin{bmatrix}6\\-2\\\end{bmatrix}$ $Ax = b_{4}$ => x= $\begin{bmatrix}13\\-5\\\end{bmatrix}$

Work Step by Step

a) $A=\begin{bmatrix}a&b\\c&d\\\end{bmatrix}=\begin{bmatrix}1&2\\5&12\\\end{bmatrix}$ $A^{-1}=\frac{1}{ad-bc}\begin{bmatrix}d&-b\\-c&a\\\end{bmatrix}$ $=\frac{1}{1(12)-2(5)}\begin{bmatrix}12&-2\\-5&1\\\end{bmatrix}$ $=\frac{1}{2}\begin{bmatrix}12&-2\\-5&1\\\end{bmatrix}$ $=\begin{bmatrix}6&-1\\-5/2&1/2\\\end{bmatrix}$ $x=A^{-1}b_{1}$=$\begin{bmatrix}6&-1\\-5/2&1/2\\\end{bmatrix}$$\begin{bmatrix}-1\\3\\\end{bmatrix}$=$\begin{bmatrix}-9\\4\\\end{bmatrix}$ $x=A^{-1}b_{2}$=$\begin{bmatrix}6&-1\\-5/2&1/2\\\end{bmatrix}$$\begin{bmatrix}1\\-5\\\end{bmatrix}$=$\begin{bmatrix}11\\-5\\\end{bmatrix}$ $x=A^{-1}b_{3}$=$\begin{bmatrix}6&-1\\-5/2&1/2\\\end{bmatrix}$$\begin{bmatrix}2\\6\\\end{bmatrix}$=$\begin{bmatrix}6\\-2\\\end{bmatrix}$ $x=A^{-1}b_{4}$=$\begin{bmatrix}6&-1\\-5/2&1/2\\\end{bmatrix}$$\begin{bmatrix}3\\5\\\end{bmatrix}$=$\begin{bmatrix}13\\-5\\\end{bmatrix}$ b) $\begin{bmatrix}A&b_{1}&b_{2}&b_{3}&b_{4}\\\end{bmatrix}$ ~$\begin{bmatrix}1&2&-1&1&2&3\\5&12&3&-5&6&5\\\end{bmatrix}$ ~$\begin{bmatrix}1&2&-1&1&2&3\\0&2&8&-10&-4&-10\\\end{bmatrix}$ ~$\begin{bmatrix}1&2&-1&1&2&3\\0&1&4&-5&-2&-5\\\end{bmatrix}$ ~$\begin{bmatrix}1&0&-9&11&6&13\\0&1&4&-5&-2&-5\\\end{bmatrix}$
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