Science:Math Exam Resources/Courses/MATH152/April 2013/Question A 05
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Question A 05 

Consider Choose p and q such that M have rank 2. Give an example of a vector b for which the linear system has no solution. 
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 hints below. Read the first one and consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it! If after a while you are still stuck, go for the next hint. 
Hint 1 

Since has rank 2, the columns in must be linear dependent. Which column in is redundant, which two columns are linearly independent? 
Hint 2 

is only solvable if is a linear combination of the columns of . How do you find a vector, which is not a linear combination of the two linear independent columns? 
Checking a solution serves two purposes: helping you if, after having used all the hints, 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. As calculated in A04, we take ,
Since has rank 2, the columns in must be linear dependent. We see that the second column is 7 times the first column. Hence, the second column is redundant. The first and the third column of are linearly independent since the third one has a 5 as second entry, where the first one has a 0. Therefore, there is no possible scalar such that columns 1 and 3 are multiples.
We want to find a vector, which is not a linear combination of the two linear independent columns. Once possible choice is to find a vector which orthogonal (and hence linear independent) to the first and third column by performing the cross product,
Note that there are several choices that could be made and this method is just a quick and effective way for finding a suitable vector. 