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Lecture

# Tutorial Problem Set #1 Solutions

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Mathematics
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MAT223H1
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Department of Mathematics, University of Toronto MAT223H1F - Linear Algebra I Fall 2013 Tutorial Problems 1 1. Consider the system of linear equations x1+ x 2 2 ax 1 bx =24 cx + dx = 8: 1 2 (a) Find non-zero values of a;b;c and d such that the system has no solutions. (b) Find non-zero values of a;b;c and d such that the system has a unique solution. (c) Find non-zero values of a;b;c and d such that the system has in▯nitely many solutions. Give reasons in each case. Solution #1: We will give two solutions to this problem. In this ▯rst solution, we will work with the equations directly; in the second, we will use matrices and some of the theory developed in the course so far. (a) If we let a = b = 1 (and take any non-zero values for c and d), then the ▯rst two equations become x1+ x 2 2 x + x = 4: 1 2 But then 2 = 4, which is nonsense. So the system has no solutions in this case. (b) Notice that if we let c = d = 4 then the third equation becomes 4x1+ 4x =28 or equivalently, after dividing through by 4, 8 x 1 x =2 4 = 2; which is exactly the ▯rst equation. So we may ignore the third equation in this case and consider the simpler system x1+ x 2 2 ax 1 bx =24: Now it’s easy to ▯nd examples where this system has a unique solution. Keeping in mind that we want to ▯nd non-zero values of a and b, let’s try a = ▯1 and b = 1. By solving the resulting simultaneous equations, we get the unique solution x1= ▯1 and x = 2. 1 of 4 (c) Just as in the previous part, we may take c = d = 4 and work with the simpler system x1+ x2= 2 ax1+ bx2= 4: If we now take a = b = 2, then we see that the second equation becomes identical with the ▯rst, and so our system reduces to just one equation: x1+ x2= 2: This equation clearly has in▯nitely many solutions, because for any2value 2f x , say x = t, we can simply take1x = 2 ▯ t and get a solution to the equation. Solution #2: Now let’s approach this problem with matrices. The system of equations gives us the augmented matrix 2 3 1 1 2 4 5 a b 4 : c d 8 (a) This system will have no solutions if, after reduction, some row is all zeros to the left of the bar and nonzero to the right of the bar. For example, if we take a = b = 1, then we have 2 3 2 3 1 1 2 R2▯R1 1 1 2 4 1 1 45 ▯▯▯▯▯! 40 0 2 ; 4 4 8 c d 8 as desired. (b) This system will have a unique solution if, after reduction, it has two leading 1s and a row of zeros. It’s very easy to arrange that we get a row of zeros: simply take c = d = 4, and apply one row operation to get 2 3 2 3 1 1 2 1 1 2 R 34R1 4a b 45 ▯▯▯▯▯! 4a b 45: 4 4 8 0 0 0 Now it’s easy to get a matrix with two leading 1s: let a = ▯1 and b = 1, and note that 2 3 2 3 2 3 2 3 1 1 2 1 1 2 R ▯ 1 1 1 2 1 0 ▯1 4▯1 1 45 ▯▯▯▯!1 40 2 65 ▯▯▯!2 40 1 35 ▯▯▯▯! 4 0 1 3 5 : 0 0 0 0 0 0 0 0 0 0 0 0 ▯ ▯ ▯ ▯ x ▯1 This gives us the unique solution = . x2 3 (c) This system will have in▯nitely many solutions if, after reduction, it has one leading 1 and two rows of zeros. Mimicking what we did in the previous part, we can take a = b = 2 and c = d = 4, and then row reduce: 2 3 2 3 1 1 2 R 22R1 1 1 2
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