Answer
$5,5$
Work Step by Step
Deleting the $3rd$ row and column gives us $\left[\begin{array}{c c} 1& 0\\ -3& 5\end{array} \right]$.
Thus, we have:
$M=1(5)-0(-3)=5$, $A=M(-1)^{3+3}=M=5$.
College Algebra 7th Edition
by Stewart, James; Redlin, Lothar; Watson, Saleem
Published by
Brooks Cole
ISBN 10:
1305115546
ISBN 13:
978-1-30511-554-5
Chapter 6, Matrices and Determinants - Section 6.4 - Determinants and Cramer's Rule - 6.4 Exercises - Page 533: 16
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