College Algebra (6th Edition)

Published by Pearson
ISBN 10: 0-32178-228-3
ISBN 13: 978-0-32178-228-1

Chapter 6 - Matrices and Determinants - Exercise Set 6.5: 28

Answer

$-20$

Work Step by Step

To evaluate an nth-order determinant, where $n > 2$, 1. Select a row or column about which to expand. 2. Multiply each element $a_{ij}$ in the row or column by $(-1)^{i+j}$ times the determinant obtained by deleting the ith row and the $j\mathrm{t}\mathrm{h}$ column in the given array of numbers. 3. The value of the determinant is the sum of the products found in step 2. ----- We select column 2, (all 2s ). The first position, (i,j)=(1,2), 1+2=3, defines the sequence of alternating signs starting with a minus. $\left|\begin{array}{lll} 1 & 2 & 3\\ 2 & 2 & -3\\ 3 & 2 & 1 \end{array}\right|=$ $=-2\cdot\left|\begin{array}{ll} 2 & -3\\ 3 & 1 \end{array}\right|+2\cdot\left|\begin{array}{ll} 1 & 3\\ 3 & 1 \end{array}\right|-2\cdot\left|\begin{array}{ll} 1 & 3\\ 2 & -3 \end{array}\right|$ $=2[-(2+9)+(1-9)-(-3-6)]$ $=2(-11-8+9)$ $=2(-10)$ $=-20$
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