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Unformatted text preview: MAT 1332: Calculus for Life Sciences A course based on the book Modeling the dynamics of life by F.R. Adler Supplementary material University of Ottawa Frithjof Lutscher, with Jing Li and Robert Smith? February 24, 2010 MAT 1332: Additional Course Notes 1 Linear Algebra III  Inverses and Determinants We know that, for every nonzero real number, x, there exists an inverse, x 1 = 1 /x such that the product xx 1 = 1 . In the last section, we learned how to multiply matrices (and when this is possible). Here we ask whether inverses exist for matrices as well, and how we can compute them. Introductory example The multiplication of two square matrices can give the identity matrix: 2 5 1 3 3 5 1 2 = 1 1 . But of course, not every (square) matrix has an inverse. The zero matrix, for example, is not invertible. Definition: A square matrix A is invertible if there exists another square matrix B of the same dimension, such that AB = I = BA. We then write A 1 = B. Note: If a matrix is not square, then it cannot be invertible. How can we find out whether a given matrix has an inverse and what the inverse is? Lets go back to the example above with A = 2 5 1 3 . We want to find a matrix B = b 11 b 12 b 21 b 22 , such that AB = A b 11 b 21 A b 12 b 22 = 1 1 . Hence, we have to solve the two systems A b 11 b 21 = 1 , A b 12 b 22 = 1 . We can do this simultaneously, using the three allowed elementary row operations of multipli cation, addition and interchange. 2 5 1 1 3 1 R 1 2 R 2 2 5 1 1 1 2 R 1+5 R 2 2 6 10 1 1 2 1 MAT 1332: Additional Course Notes 2 . 5 R 1 , R 2 1 3 5 1 1 2 Once we have reduced the left hand side to the identity matrix, we have the inverse of the original matrix on the right hand side. Algorithm to compute an inverse matrix: Given a square matrix A, we take the identity matrix I and write the system [ A  I ] . Then we use the three row operations to reduce the left hand side to the identity matrix. If this is possible, we end up with [ I  B ] = [ I  A 1 ] . If we cannot reduce the left hand side to the identity matrix, then the original matrix A is not invertible. Example 1 Find the inverse of A = 3 7 1 3 . We write 3 7 1 1 3 1 R 1 R 2 1 3 1 3 7 1 R 2 3 R 1 1 3 1 2 1 3 . 5 R 2 1 3 1 1 1 / 2 3 / 2 R 1 3 R 2 1 3 / 2 7 / 2 1 1 / 2 3 / 2 Hence, A is invertible and its inverse is A 1 = 3 / 2 7 / 2 1 / 2 3 / 2 . Example 2 Find the inverse of A = 1 2 2 4 . We write 1 2 1 2 4 1 R 1 2 R 2 1 2 1 2 1 Because we have a row of zeros on the left hand side, we cannot transform the left hand side into the identity matrix by elementary row operations. Therefore, the matrix A is not invertible. 2 MAT 1332: Additional Course Notes 3 Example 3 Find the inverse of A = 1 2 3 1 3 4 2 4 . We write 1 2 3 1 1 3 4 1 2 4 1 R 2 R 1 1 2 3 1 1 1 1 1 2 4 1  R 2 R 3 2 R 2 1 2 3 1 1 1 1 1 2 2 2 1 . 5 R 3 1 2 3 1 1 1...
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 Spring '07
 MUNTEANU
 Calculus, Algebra, Addition

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