Science:Math Exam Resources/Courses/MATH221/April 2010/Question 06

MATH221 April 2010

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[hide] Question 06
True or false (explain your answer): Suppose A is a 5 × 4 matrix and that $b,c$ are vectors in $\mathbb {R} ^{5}$. Given that $Ax=b$ has a unique solution, then $Ax=c$ 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.

[show] Hint 1
What must be true about $A$ if $A\mathbf {x} =\mathbf {b}$ has a unique solution?

[show] Hint 2
Think about what properties of $A$ are necessary in order for a solution to exist for $Ax=c$, for every vector $c$ in $\mathbb {R} ^{5}$.

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. 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

The answer to this statement is false: No solution to $A\mathbf {x} =\mathbf {c}$ must exist (however, if a solution exists, it must be unique). If $A\mathbf {x} =\mathbf {c}$ then $\mathbf {c}$ is a linear combination of the columns of $A$. However, the vector $\mathbf {c}$ is in $\mathbb {R} ^{5}$ 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 $A$ could properly represent any vector $\mathbf {c}$ and therefore we are not guaranteed that a solution exists to $A\mathbf {x} =\mathbf {c}$. We can illustrate this with an example. Let $A={\begin{bmatrix}1&0&0&0\\0&1&0&0\\0&0&1&0\\0&0&0&1\\0&0&0&0\end{bmatrix}}$, $\mathbf {b} ={\begin{bmatrix}1\\0\\0\\0\\0\end{bmatrix}}$ and $\mathbf {c} ={\begin{bmatrix}0\\0\\0\\0\\1\end{bmatrix}}$. Then $A\mathbf {x} =\mathbf {b}$ has the unique solution $\mathbf {x} ={\begin{bmatrix}1\\0\\0\\0\end{bmatrix}}$, but $A\mathbf {x} =\mathbf {c}$ has no solution. However, if a solution to $A\mathbf {x} =\mathbf {c}$ exists, then it must be unique: If $A\mathbf {x} =\mathbf {b}$ has a unique solution then it must have a trivial nullspace, i.e. there must not be a vector $\mathbf {y} \not =\mathbf {0}$ such that $A\mathbf {y} =\mathbf {0}$ as otherwise $A(\mathbf {x} +\mathbf {y} )=A\mathbf {x} +A\mathbf {y} =\mathbf {b} +\mathbf {0} =\mathbf {b}$ and $\mathbf {x} +\mathbf {y}$ is a solution violating the uniqueness assumption of $A\mathbf {x} =\mathbf {b}$. So assume that $\mathbf {r}$ and $\mathbf {s}$ are both solutions of $A\mathbf {x} =\mathbf {c}$. We need to show that the solution is unique, that is, $\mathbf {r} =\mathbf {s}$. Subtracting the two equations we obtain $A\mathbf {r} -A\mathbf {s} =\mathbf {c} -\mathbf {c} =\mathbf {0}$ and thus $A(\mathbf {r} -\mathbf {s} )=\mathbf {0}$, that is, $\mathbf {r} -\mathbf {s}$ is in the nullspace of $A$. However, this nullspace only consists of the null vector, which implies that indeed $\mathbf {r} =\mathbf {s}$.