College Algebra (10th Edition)

Published by Pearson
ISBN 10: 0321979478
ISBN 13: 978-0-32197-947-6

Chapter 8 - Section 8.4 - Matrix Algebra - 8.4 Assess Your Understanding - Page 598: 36

Answer

$A^{-1}=\left[\begin{array}{rr} {-2/b}&{3/b}\\ {1}&{-1}\end{array}\right]$

Work Step by Step

To find the inverse of an $n$ by $n$ nonsingular matrix $A,$ proceed as follows: STEP 1: Form the matrix $\left[A|I_{n}\right]$ STEP 2: Transform the matrix $\left[A|I_{n}\right]$ into reduced row echelon form. STEP 3: The reduced row echelon form of $\left[A|I_{n}\right]$ will contain the identity matrix $I_{n}$ on the left of the vertical bar; the $n$ by $n$ matrix on the right of the vertical bar is the inverse of $A.$ If the identity matrix can not be obtained on the left, A has no inverse. ---- STEP 1 $\left[A|I_{n}\right] = \left[\begin{array}{ll|ll} {b}&{3}&{1}&{0}\\ {b}&{2}&{0}&{1}\end{array}\right] \qquad \left(\begin{array}{l} .\\ R_{2}=r_{2}-r_{1}. \end{array}\right)$ STEP 2 $\rightarrow\left[\begin{array}{rr|rr} {b}&{3}&{1}&{0}\\ {0}&{-1}&{-1}&{1}\end{array}\right] \qquad \left(\begin{array}{l} R_{1}=r_{1}+3r_{2}.\\ \times(-1). \end{array}\right)$ $\rightarrow\left[\begin{array}{rr|rr} {b}& {0}&{-2}&{3}\\ {0}&{1}&{1}&{-1}\end{array}\right] \qquad \left(\begin{array}{l} \div b.\\ . \end{array}\right)$ $\rightarrow\left[\begin{array}{rr|rr} {1}& {0}&{-2/b}&{3/b}\\ {0}&{1}&{1}&{-1}\end{array}\right] $ STEP 3 $A^{-1}=\left[\begin{array}{rr} {-2/b}&{3/b}\\ {1}&{-1}\end{array}\right]$
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