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Question 06 

True or false (explain your answer): Suppose A is a 5 × 4 matrix and that are vectors in . Given that has a unique solution, then has a unique solution as well. 
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 

What must be true about if has a unique solution? 
Hint 2 

Think about what properties of are necessary in order for a solution to exist for , for every vector in . 
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 

Found a typo? Is this solution unclear? Let us know here.
Please rate my easiness! It's quick and helps everyone guide their studies. The answer to this statement is false: No solution to must exist (however, if a solution exists, it must be unique). If then is a linear combination of the columns of . However, the vector is in and this space requires 5 vectors to form a basis. We can therefore not guarantee that the 4 vectors we have from the columns of could properly represent any vector and therefore we are not guaranteed that a solution exists to . We can illustrate this with an example. Let , and . Then has the unique solution , but has no solution. However, if a solution to exists, then it must be unique: If has a unique solution then it must have a trivial nullspace, i.e. there must not be a vector such that as otherwise and is a solution violating the uniqueness assumption of . So assume that and are both solutions of . We need to show that the solution is unique, that is, . Subtracting the two equations we obtain and thus , that is, is in the nullspace of . However, this nullspace only consists of the null vector, which implies that indeed . 