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# tat01 - T HE U NIVERSITY OF S YDNEY P URE M ATHEMATICS...

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T HE U NIVERSITY OF S YDNEY P URE M ATHEMATICS 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 = 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 0 - 3 - 2 - 2 - 1 3 5 9 R 3 := R 3 + R 1 -------→ 1 1 2 11 0 - 3 - 2 - 2 0 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 . a ) Use Gaussian elimination to find all solutions of the equation A x = 0 , where x = x 1 x 2 x 3 and 0 = 0 0 0 . b ) Hence show that some non–zero linear combination of the columns of A is equal to 0 .

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