MM1_Chapter_10

We learn that a matrix a has an inverse matrix if and

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Unformatted text preview: det(A B ) = det(A) × det(B ) = 1 × 6 = 6. (b) Computing the matrix A B yields 111 100 653 AB = 0 1 1 2 2 0 = 5 5 3 . 001 333 333 To compute the determinant det(A B ), we use that, from Lemma 10.35 (iii), we may subtract the second row of the matrix from the first row without changing the new value of the determinant (that is, r1 = r1 − r2 ). Thus det(A B ) = 653 553 333 = 6−5 5 3 5−5 5 3 3−3 3 3 = 100 553. 333 Now we expand about the first row and get det(A B ) = 100 553 333 =1× 53 33 We get indeed the same result. 10.4 = 1 × (5 × 3 − 3 × 3) = 15 − 9 = 6. 2 Invertible Matrices In this section, we introduce the inverse matrix of an (invertible) square n × n matrix A. The inverse matrix A−1 of A satisfies A A−1 = A−1 A = In . If we compare 10. Matrices 283 invertible matrices A to real numbers x = 0, then the inverse matrix A−1 corresponds to the number 1/x. We learn that a matrix A has an inverse matrix if and only if det(A) = 0. We introduce the adjoint matrix, and we learn a formula for the inverse matrix which involves the adjoint matrix and the determinant. Definition 10.40 (invertible and inverse matrix) Let A be an n × n matrix. If there exists an n × n matrix B such that A B = In and B A = In , (10.10) then A is invertible. The matrix B satisfying (10.10) is called the inverse (matrix) of A, and usually the inverse matrix of A is denoted by A−1 . In Definition 10.40 we would prefer that, if there exists a matrix B satisfying (10.10), then this matrix is the only matrix satisfying (10.10). In other words, we would like the inverse matrix to be unique. To show that this is indeed true, assume that there exits another n × n matrix C satisfying A C = In and C A = In . (10.11) Multiplying the first formula in (10.11) from the left by the matrix B yields B (A C ) = B In ⇒ (B A) C = B ⇒ In C = B ⇒ C = B, where we have used the associative law of matrix multiplication (see Theorem 10.20) and (10.1). Since C = B , we see that the inverse matrix is indeed uniquely determined. Example 10.41 (matrix and its inverse) Consider the matrices A and B , given by A= 23 34 and B= −4 3 3 −2 . Show that the matrix B is the inverse matrix to the matrix A. Solution: We compute A B and B A, and find AB = = 23 34 −4 3 3 −2 2 × (−4) + 3 × 3 3 × (−4) + 4 × 3 2 × 3 + 3 × (−2) 3 × 3 + 4 × (−2) 284 10. Matrices = BA = 10 01 , −4 3 3 −2 23 34 = (−4) × 2 + 3 × 3 3 × 2 + (−2) × 3 = 10 01 (−4) × 3 + 3 × 4 3 × 3 + (−2) × 4 . Since A B = B A = I2 , we know that B is the inverse matrix to A. We also see that A is the inverse matrix of B . 2 Remark 10.42 (not every matrix is invertible) It is easy to see that the zero matrix O= 00 00 does not have an inverse and is thus not invertible. Indeed, if B is any other 2 × 2 matrix, then OB = 00 00 B= 00 00 and BO =B 00 00 = 00 00 . We see that not every square matrix is invertible. If an n × n matrix A has an inverse A−1 , then taking the determinant on both sides of the equation In = A−1 A and using (10.9) yields det(In ) = det(A−1 A) = det(A−1 ) det(A). From (10.7), we know that det(In ) = 1, and thus 1 = det(A−1 ) det(A). (10.12) From (10.12), we know that neither det(A) nor det(A−1 ) can be zero, s...
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This document was uploaded on 01/31/2014.

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