Precalculus: Mathematics for Calculus, 7th Edition

Published by Brooks Cole
ISBN 10: 1305071751
ISBN 13: 978-1-30507-175-9

Chapter 10 - Section 10.6 - Determinants and Cramer's Rule - 10.6 Exercises - Page 742: 17

Answer

$M_{12}=-12$ and $A_{12}=12$

Work Step by Step

The minor of an element is defined $a_{ij}$of a determinant is defined as the determinant obtained by deleting its ith row and jth columns and the minor $M$ of an element $a_{ij}$ is abbreviated as $M_{ij}$. Further, the cofactor of an element $a_{ij}$ is abbreviated as $A_{ij}$. where $A_{ij}=(-1)^{i+j} M$ Here, $M_{12}=\begin{vmatrix}-3&2\\0&4\end{vmatrix}=(-3)(4)-0=-12$ and $A_{12}=(-1)^{1+2} M_{12}=(-1)^3(-12)=12$
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