Science:Math Exam Resources/Courses/MATH152/April 2015/Question A 12

MATH152 April 2015

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[hide] Question A 12
Circle all possible solution sets for linear systems of five equations in three unknowns. (a) a unique solution. (b) no solutions. (c) an infinite number of solutions. (d) exactly two 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
Recall the role of Gaussian Elimination in solving linear systems.

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.

[show] Solution

In order to solve a linear system, we may perform Gaussian Elimination to get the row echelon form of the system's augmented matrix. If there is a zero row in the matrix representing the linear system but the right-hand side is non-zero, then there is no solution. If there are more than two zero rows in the linear system matrix (i.e., the left-hand side of the augmented matrix), then it is possible for the system to have infinitely many solutions. If there are only two zero rows, and the remaining three rows of the left-hand side form an invertible $3\times 3$ matrix, then there exists a unique solution. (d) is not possible. If we assume for the sake of contradiction that the linear system has exactly two solutions $\mathbf {v} ={\begin{bmatrix}x_{1}&y_{1}&z_{1}\end{bmatrix}}^{\top }$ and $\mathbf {w} ={\begin{bmatrix}x_{2}&y_{2}&z_{2}\end{bmatrix}}^{\top }$, then ${\frac {1}{2}}(\mathbf {v} +\mathbf {w} )$ also solves the linear system, yielding a contradiction. Therefore, the answer is (a), (b), (c).