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

MATH152 April 2017

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[hide] 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!

[show] 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.

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[show] 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 $\color {blue}{(a),(c),(e)}$ .