Science:Math Exam Resources/Courses/MATH307/December 2012/Question 01 (d)

MATH307 December 2012

  •  Q1 (a)  •  Q1 (b)  •  Q1 (c)  •  Q1 (d)  •  Q1 (e)  •  Q2 (a)  •  Q2 (b)  •  Q3 (a)  •  Q3 (b)  •  Q3 (c)  •  Q3 (d)  •  Q3 (e)  •  Q4 (a)  •  Q4 (b)  •  Q4 (c)  •  Q4 (d)  •  Q4 (e)  •  Q5 (a)  •  Q5 (b)  •  Q5 (c)  •  Q5 (d)  •  Q6 (a)  •  Q6 (b)  •  Q6 (c)  •  Q6 (d)  •  Q6 (e)  •  Q6 (f)  •  

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Question 01 (d)

In this question we will work with polynomials of degree 3 written $\displaystyle p(x)=a_{1}x^{3}+a_{2}x^{2}+a_{3}x+a_{4}$ (d) Using the equation in (c) find the equation Cs=c satisfied by $\mathbf {s} =[s_{1},s_{2}]^{T}$ if p(x) (i) has coefficient vector $\mathbf {a} =s_{1}\mathbf {a} _{1}+s_{2}\mathbf {a} _{2}$ (and therefore has zero slope at x = 0 and x = 2). (ii) passes through the points (0,1), (1,2) and (2,2). Does this equation have a solution? Give a reason.

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
To check if a linear system has a solution, we can look at the relationships between two of the fundamental subspaces of a matrix. Which ones should we look at? How are they related?

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. Since ${\vec {a}}=s_{1}{\vec {a_{1}}}+s_{2}{\vec {a_{2}}}=s_{1}{\begin{bmatrix}1\\-3\\0\\0\end{bmatrix}}+s_{2}{\begin{bmatrix}0\\0\\0\\1\end{bmatrix}}$$={\begin{bmatrix}1&0\\-3&0\\0&0\\0&1\end{bmatrix}}{\begin{bmatrix}s_{1}\\s_{2}\end{bmatrix}}$ Using the equation from part (c), we have $B{\vec {a}}={\vec {b}}$ ${\begin{bmatrix}8&4&2&1\\1&1&1&1\\0&0&0&1\end{bmatrix}}{\begin{bmatrix}1&0\\-3&0\\0&0\\0&1\end{bmatrix}}{\vec {s}}={\begin{bmatrix}2\\2\\1\end{bmatrix}}$ Therefore, $C={\begin{bmatrix}-4&1\\-2&1\\0&1\end{bmatrix}}$ and ${\vec {c}}={\vec {b}}={\begin{bmatrix}2\\2\\1\end{bmatrix}}$. Notice that for there to be a solution we need ${\vec {c}}$ in the range of C, R(C). We know that R(C) is the orthogonal complement to the nullspace of $C^{T}$, $N(C^{T})$. Therefore, onsider the basis vector of $N(C^{T})$ ${\begin{bmatrix}-4&-2&0\\1&1&1\end{bmatrix}}{\vec {x}}=0\rightarrow {\begin{bmatrix}2&1&0\\0&1&2\end{bmatrix}}{\vec {x}}=0$ ${\vec {x}}={\begin{bmatrix}x_{1}\\x_{2}\\x_{3}\end{bmatrix}}={\begin{bmatrix}1\\-2\\1\end{bmatrix}}x_{3}$ Since ${\begin{bmatrix}1\\-2\\1\end{bmatrix}}\cdot {\begin{bmatrix}2\\2\\1\end{bmatrix}}=2-4+1\neq 0$ In other words, $N(C^{T})$ is not orthogonal to ${\vec {c}}$. This means that ${\vec {c}}$ cannot possibly be in the range of C and so there will be no solution to the problem.

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