Science:Math Exam Resources/Courses/MATH152/April 2011/Question B 06 (e)
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Question B 06 (e) 

Consider the linear systems described below. In cases (a)(d), write a single, possible reduced row echelon form of the augmented matrix of the system. Repeat part (d) (i.e. three equations in two unknowns with no solution) but in this case only, list all possible corresponding reduced row echelon forms. 
Make sure you understand the problem fully: What is the question asking you to do? Are there specific conditions or constraints that you should take note of? How will you know if your answer is correct from your work only? Can you rephrase the question in your own words in a way that makes sense to you? 
If you are stuck, check the hint below. Consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it! 
Hint 

Recall there is only no solution if the third equation is not a linear combination of the first two. I.e. how can you make sure that the third equation is not a linear combination of the first two? 
Checking a solution serves two purposes: helping you if, after having used the hint, you still are stuck on the problem; or if you have solved the problem and would like to check your work.

Solution 

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Please rate my easiness! It's quick and helps everyone guide their studies. We want to continue with part (d) but now find all matrices that produce no solution to 3 equations in 2 unknowns. We start with an augmented matrix in the form for some numbers . Notice we have put the matrix in reduced rowechelon form up to the first two rows. We did this because we know that if there were only two equations then we could find a unique solution depending the values of and . The problem now is for those same that would work in the case we want to find a condition on that creates no solution. Notice we can rowreduce that last row as follows where indicates an operation on row . Notice if then we have a row of zeros in the matrix. This implies for a vector x with components and that which would always hold true. Therefore, in this case there would be a solution. As discussed in part (d) this means that the third equation was secretly a linear combination of the first two equations and thus didn't really add another constraint to the problem. Since we are seeking there to be no solution we see then that we must require, 