Tat01 - THE UNIVERSITY OF SYDNEY PURE MATHEMATICS Linear Mathematics 2010 Tutorial 1 — Solutions 1 a Calculate the matrix product 3 0 4 1 1 2 1 3

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Unformatted text preview: THE UNIVERSITY OF SYDNEY PURE MATHEMATICS Linear Mathematics 2010 Tutorial 1 — Solutions 1. a ) Calculate the matrix product 3 0 4 1 1 2- 1 3 5 5- 2 4 . b ) Find the unique solution of the linear system of equations which corresponds to the aug- mented matrix 3 0 4 31 1 1 2 11- 1 3 5 9 . Solution a ) 3 0 4 1 1 2- 1 3 5 5- 2 4 = 31 11 9 . b ) Let A = 3 0 4 1 1 2- 1 3 5 . Then we want to find x = x 1 x 2 x 3 such that A x = 31 11 9 . Notice that the matrix A must be invertible because the question says that there is a unique solution. (It is easy to see that A is invertible because det A =- 7 6 = 0 —check!) The shortest way to solve this question is to observe that ( 5- 2 4 is a solution by part (a). As det A =- 7 the matrix A is invertible, so x = 5- 2 4 is the unique solution. For completeness, we also find the solution using Gaussian elimination: 3 0 4 31 1 1 2 11- 1 3 5 9 R 1 ↔ R 2----→ 1 1 2 11 3 0 4 31- 1 3 5 9 R 2 := R 2- 3 R 1-------→ 1 1 2 11- 3- 2- 2- 1 3 5 9 R 3 := R 3 + R 1-------→ 1 1 2 11- 3- 2- 2 4 7 20 R 2 := R 2 + R 1-------→ 1 1 2 11 0 1 5 18 0 4 7 20 R 3 := R 3- 4 R 2-------→ 1 1 2 11 0 1 5 18 0 0- 13- 52 R 3 :=- 1 13 R 3-------→ 1 1 2 11 0 1 5 18 0 0 1 4 Therefore x 3 = 4 , x 2 = 18- 5 x 3 =- 2 , x 1 = 11- x 2- 2 x 3 = 5 . Using back substitution the unique solution is 5- 2 4 , as before. 2. Let A = 1- 3 2 2- 5 6- 1 0- 8 ....
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This note was uploaded on 02/06/2012 for the course MATH 2061 taught by Professor Notsure during the Three '09 term at University of Sydney.

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Tat01 - THE UNIVERSITY OF SYDNEY PURE MATHEMATICS Linear Mathematics 2010 Tutorial 1 — Solutions 1 a Calculate the matrix product 3 0 4 1 1 2 1 3

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