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**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...

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