Stitz-Zeager_College_Algebra_e-book

When traveling downstream the river is helping carl

This preview shows page 1. Sign up to view the full content.

This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: trix A below. (Others will be discussed in the Exercises.) 3 1 A = 0 −1 2 1 2 5 4 We found det(A) = −13 by expanding along the ﬁrst row. Theorem 8.7 guarantees that we will get the same result if we expand along the second row. (Doing so would take advantage of the 0 there.) 3 1 det 0 −1 2 1 2 5 = 0C21 + (−1)C22 + 5C23 4 = (−1)(−1)2+2 det (A22 ) + 5(−1)2+3 det (A23 ) 32 24 = − det 31 21 − 5 det = −((3)(4) − (2)(2)) − 5((3)(1) − (2)(1)) = −8 − 5 = −13 In general, the sign of (−1)i+j in front of the minor in the expansion of the determinant follows an alternating pattern. Below is the pattern for 2 × 2, 3 × 3 and 4 × 4 matrices, and it extends naturally to higher dimensions. +− −+ +−+ − + − +−+ + − + − − + − + + − + − − + − + The reader is cautioned, however, against reading too much into these sign patterns. In the example above, we expanded the 3 × 3 matrix A by its second row and the term which corresponds to the second entry ended up being negative even though the sign attached to the minor is (+). These signs represent only the signs of the (−1)i+j in the formula; the sign of the corresponding entry as well as the minor itself determine the u...
View Full Document

{[ snackBarMessage ]}

Ask a homework question - tutors are online