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

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 (b)
In this question we will work with polynomials of degree 3 written $\displaystyle p(x)=a_{1}x^{3}+a_{2}x^{2}+a_{3}x+a_{4}$ (b) Show that dim(N(A)) = 2 and bind a basis a1, a2 for N(A).

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.

Hint 1
How does the dimension of the nullspace relate to the number of pivot columns of the matrix?

Hint 2
Recall we can find the nullspace by row-reducing.

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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. In part (a) we found that $A={\begin{bmatrix}12&4&1&0\\0&0&1&0\end{bmatrix}}$ The matrix A has 2 pivot columns and so by rank-nullity theorem $\displaystyle {}\dim(N(A))=4-\dim(R(A))=4-2=2$ and so there are two basis vectors for the nullspace. To find them we could row reduce A or notice that, at x = 0, $\displaystyle {}3a_{1}(0)^{2}+2a_{2}(0)+a_{3}=0$ gives $a_{3}=0$. Using the value at x=2 $\displaystyle {}12a_{1}+4a_{2}+a_{3}=12a_{1}+4a_{2}=0,$ gives $a_{2}=-3a_{1}$. Since neither equation depends on $a_{4}$ then it is a free variable. Hence, the basis of N(A) is ${\vec {a}}={\begin{bmatrix}a_{1}\\a_{2}\\a_{3}\\a_{4}\end{bmatrix}}={\begin{bmatrix}a_{1}\\-3a_{1}\\0\\a_{4}\end{bmatrix}}={\begin{bmatrix}1\\-3\\0\\0\end{bmatrix}}a_{1}+{\begin{bmatrix}0\\0\\0\\1\end{bmatrix}}a_{4}$ Hence, ${\vec {a_{1}}}={\begin{bmatrix}1\\-3\\0\\0\end{bmatrix}},{\vec {a_{2}}}={\begin{bmatrix}0\\0\\0\\1\end{bmatrix}}$

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