Assignment 5 Solutions

Assignment 5 Solutions - MAT1332 Assignment 5 Solutions...

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MAT1332 Assignment 5 Solutions Total =20 points 1. (6 points) For the system of linear equations x + 3 y + 9 z = 3 2 x + 7 y + 23 z = 2 x + ay + a 2 z = a (a) determine the values of a for which the system has (i) no solution, (ii) infinitely many solutions, (iii) a unique solution. (b) In case (ii) above describe all solutions. (c) If a = 1 find the inverse of the matrix A = 1 3 3 2 7 2 1 a a 2 The augmented matrix of the system is A = 1 3 9 3 2 7 23 2 1 a a 2 a We perform the following operations, where R i is row i : R 2 R 2 - 2 R 1 , R 3 R 3 - R 1 , R 3 R 3 - ( a - 3) R 2 , and obtain A 1 3 9 3 0 1 5 - 4 0 a - 3 a 2 - 9 a - 3 1 2 3 3 0 1 5 - 4 0 0 a 2 - 9 - 5 a + 15 5( a - 3) . Since a 2 - 9 - 5 a + 15 = ( a - 3)( a - 2) we get : 1 3 9 3 0 1 5 - 4 0 0 ( a - 3)( a - 2) 5( a - 3) If a = 2, then the last row of the matrix is is 0 0 0 - 5 . Hence the system is inconsistent. If a = 3 then M = 1 2 4 2 0 1 5 - 4 0 0 0 0 1 0 - 6 10 0 1 5 - 4 0 0 0 0 Hence the system has infinitely many solutions. 1

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If a / ∈ { 3 , 2 } , then ( a - 3)( a - 2) 6 = 0 and so the system is uniquely solvable The answer to question (a) is therefore : (i) The system in inconsistent if a = 2. 1 point (ii) The system has infinitely many solutions if a = 3. 1 point (iii) The system is uniquely solvable if a / ∈ { 2 , 3 } . 1 point (b) The RREF of the matrix is 1 0 - 6 15 0 1 5 - 4 0 0 0 0 The corresponding linear system is x - 6 z = 15 y + 5 z = - 4 Thus z is a free variable. Putting z = t , the general solution is (15 + 6 t, - 4 - 5 t, t ) ( t a free parameter) 1 point (c) We set up the 3 by 6 matrix for the inverse.
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