MM1_Chapter_10

# 21 1 new we now apply rule iii twice subtracting row

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Unformatted text preview: ract row 1 from row 3 (that is, r3 = r3 − r1 ): 1 3 0 −2 det(A) = −2 × 0 −3 0 −1 0 4 1 −3 . 1 0 2 3 Now, we have only one non-zero entry in the column 1, namely, the number 1 at the top. Expanding about column 1 gives that −2 det(A) = −2 × 1 × −3 −1 1 −3 1 0. 2 3 280 10. Matrices So we have reduced the problem to that of ﬁnding the determinant of a 3 × 3 matrix. We now apply the same procedure to ﬁnd this determinant. Indeed, rule (i) allows us to take out the common factors −1 and 3 from columns 1 and 3, respectively: 2 det(A) = −2 × (−1) × 3 × 3 1 1 −1 1 0 2 1 2 =6× 3 1 1 −1 1 0. 2 1 Next, we use rule (ii) to interchange columns 1 and 2 to make the entry in the top left corner 1: 1 2 −1 det(A) = 6 × (−1) × 1 3 0. 21 1 new We now apply rule (iii) twice. Subtracting row 1 from row 2 (that is, r2 = r2 − r1 ) new and subtracting twice row 1 from row 3 (that is, r3 = r3 − 2r1 ) gives that 1 2 −1 det(A) = −6 × 0 1 1. 0 −3 3 Now, we have only one non-zero entry in the column 1, namely, the number 1 at the top. Expanding about column 1 gives that det(A) = −6 × 11 . −3 3 Rule (i) allows us to take out the common factor 3 from row 2: det(A) = −18 × 11 . −1 1 Hence det(A) = −18×[1×1−1×(−1)] = −18×[1+1] = −18×2 = −36. 2 Example 10.37 (determinant of the identity matrix) The n × n identity matrix In has determinant det(In ) = 1. (10.7) Proof: Let us start with the 2 × 2 matrix I2 . We have det(I2 ) = 10 01 = 1 × 1 − 0 × 0 = 1. (10.8) 10. Matrices 281 For the 3 × 3 identity matrix I3 , we ﬁnd by expanding the determinant det(I3 ) about the ﬁrst row det(I3 ) = 100 010 001 =1× 10 01 = 1 × det(I2 ) = 1 × 1 = 1, where we have used (10.8) in the second last step. Now we use induction and assume that we have already veriﬁed that det(In−1 ) = 1. Then we want to show that det(In ) = 1. By expanding the determinant det(In ) of the n × n identity matrix In about the ﬁrst row, we ﬁnd (in analogy to above) det(In ) = 1 × det(In−1 ) = 1 × 1 = 1, where have used the assumption that det(In−1 ) = 1 in the second last step. 2 The next lemma gives information about the relation between the determinant det(A B ) of the product A B of two square matrices A and B and the determinants of the individual matrices det(A) and det(B ). Lemma 10.38 (determinant of the product of matrices) If A and B are n × n matrices, then the n × n matrix A B has the determinant det(A B ) = det(A) × det(B ) = det(A) det(B ). (10.9) Note that, in general, for two n × n matrices det(A + B ) = det(A) + det(B ). Example 10.39 (determinant of product of matrices) Determine the determinant det(A B ) where 111 A= 0 1 1 001 and 100 B = 2 2 0 . 333 Solution: We will determine the determinant in two ways: (a) by using (10.9); and (b) by computing A B ﬁrst and then computing the determinant of A B . 282 10. Matrices (a) We have, from expanding det(A) about the ﬁrst column and from expanding det(B ) about the ﬁrst row, det(A) = 111 011 001 = 1× 11 01 = 1 × (1 × 1 − 1 × 0) = 1, det(B ) = 100 220 333 = 1× 20 33 = 1 × (2 × 3 − 0 × 3) = 6. From (10.9), we ﬁnd...
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## This document was uploaded on 01/31/2014.

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