Science:Math Exam Resources/Courses/MATH152/April 2017/Question A 14

MATH152 April 2017

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Question A 14
Circle all possible types of solution sets that a system of 3 equations in six unknowns can have: (a) No solutions (b) A unique solution (c) An infinite number of solutions (d) A two-dimensional set of solutions (e) A four-dimensional set of solutions

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
We can write the system of equations in the form of matrix vector multiplication. That is, $Ax=b$ where $x$ is the $6\times 1$ vector representing the $6$ unknowns while $A$ is the corresponding $3\times 6$ matrix.

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.

If you are stuck on a problem: Read the solution slowly and as soon as you feel you could finish the problem on your own, hide it and work on the problem. Come back later to the solution if you are stuck or if you want to check your work.
If you want to check your work: Don't only focus on the answer, problems are mostly marked for the work you do, make sure you understand all the steps that were required to complete the problem and see if you made mistakes or forgot some aspects. Your goal is to check that your mental process was correct, not only the result.

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. Following the hint, we can simplify the question to finding the solution of ${\textstyle Ax=b}$ where ${\textstyle A}$ is a ${\textstyle 3\times 6}$ matrix, ${\textstyle b}$ is a ${\textstyle 3\times 1}$ vector, and ${\textstyle x}$ is the ${\textstyle 6\times 1}$ vector representing the ${\textstyle 6}$ unknowns. It is not possible for ${\textstyle x}$ to be uniquely solved since ${\textstyle A}$ has more columns than rows. If ${\textstyle b}$ is not in the range of ${\textstyle A}$ , there exists no solution. If ${\textstyle b}$ is in the range of ${\textstyle A}$ , there are infinitely many solutions. Since we have ${\textstyle 6}$ unknowns but only ${\textstyle 3}$ equations, we have freedom to assign values to at least three unknowns. Thus, the answer is $\color {blue}{(a),(c),(e)}$ .