Science:Math Exam Resources/Courses/MATH152/April 2016/Question A 17

MATH152 April 2016

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Question A 17
Consider a linear system with 7 equations for 8 unknowns. Circle all possible types of solution sets that could result: (a) The system has no solutions. (b) The system has a unique solution. (c) The system has exactly 8 distinct solutions. (d) The system has a one-parameter family of solutions. (e) The system has a two-parameter family 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
Consider the matrix representation of the system.

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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. First write the linear system in the form of $Ax=b$ where $A$ is the coefficient matrix with $7$ rows and $8$ columns, $x$ is a $8\times 1$ vector representing the unknowns while $b$ is a $8\times 1$vector representing the right-hand side. Now we perform Gaussian Elimination to the augmented matrix $\left[A\mid b\right]$ to get its reduced row echelon form. We call the reduced row echelon form $\left[{\tilde {A}}\mid {\tilde {b}}\right]$. If there is a zero row in ${\tilde {A}}$ but the entry of ${\tilde {b}}$ on this row is non-zero, then there is no solution. So (a) is possible. The system has a unique solution if the corresponding homogeneous system $Ax=0$ has zero solution. Note that the matrix $A$ has $8$ columns and $7$ rows, its column spaces are linear dependent. This implies that there is a nonzero vector in the null space, so (b) is wrong. (c) is not possible. If we assume for the sake of contradiction that the linear system has exactly eight solutions ${\mathbf {x}}^{1},{\mathbf {x}}^{2},{\mathbf {x}}^{3},{\mathbf {x}}^{4},{\mathbf {x}}^{5},{\mathbf {x}}^{6},{\mathbf {x}}^{7},{\mathbf {x}}^{8}$ , then $y={\frac {1}{8}}({\mathbf {x}}^{1}+{\mathbf {x}}^{2}+{\mathbf {x}}^{3}+{\mathbf {x}}^{4}+{\mathbf {x}}^{5}+{\mathbf {x}}^{6}+{\mathbf {x}}^{7}+{\mathbf {x}}^{8})$ also solves the linear system, yielding a contradiction. The matrix $A$ has $8$ columns and $7$ rows, the maximal possible rank of matrix $A$ is $7.$ If $A$ has rank $7$, that is, no zero rows in ${\tilde {A}}$ , it is possible for the linear system to have one parameter family of solutions. If there is one zero row in the reduced row echelon form $[{\tilde {A}}\mid {\tilde {b}}]$ , then the matrix $A$ has rank $6$. In this case, it is possible for the system to have two parameter family of solutions. Thus, the possible results are $\color {blue}{(a)(d)(e)}$.

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