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: 25

Answer

$-75$

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 the column 3 (because of the zeros). The first position, (i,j)=(1,3), 1+3=4, defines the sequence of alternating signs starting with a +. $\left|\begin{array}{lll} 3 & 1 & 0\\ -3 & 4 & 0\\ -1 & 3 & -5 \end{array}\right|=$ $=+0\cdot\left|\begin{array}{ll} -3 & 4\\ -1 & 3 \end{array}\right|-0\cdot\left|\begin{array}{ll} 3 & 1\\ -1 & 3 \end{array}\right|+(-5)\cdot\left|\begin{array}{ll} 3 & 1\\ -3 & 4 \end{array}\right|$ $=-5[3(4)-1(-3)]$ $= -5(15)=-75$
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