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: 23

Answer

72

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 first row (because of the zeros). The first position, (i,j)=(1,1), 1+1=2, defines the sequence of alternating signs starting with a +. $\left|\begin{array}{lll} 3 & 0 & 0\\ 2 & 1 & -5\\ 2 & 5 & -1 \end{array}\right|=$ $=+3\cdot\left|\begin{array}{ll} 1 & -5\\ 5 & -1 \end{array}\right|-0\cdot\left|\begin{array}{ll} 2 & -5\\ 2 & -1 \end{array}\right|+0\cdot\left|\begin{array}{ll} 2 & 1\\ 2 & 5 \end{array}\right|$ $=3[1(-1)-5(5)]$ $=3(24)$ $= 72$
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