Science:Math Exam Resources/Courses/MATH307/December 2008/Question 01

{{#incat:MER QGQ flag|{{#incat:MER QGH flag|{{#incat:MER QGS flag|}}}}}}

MATH307 December 2008

  •  Q1  •  Q2  •  Q3  •  Q4  •  Q5  •  Q6 (a)  •  Q6 (b)  •  Q7  •  Q8  •  Q9  •  Q10  •  

   Other MATH307 Exams

  •  December 2012  •  April 2013  •  April 2012  •  April 2006  •  December 2008  •  December 2010  •  

Question 01
Let $A={\begin{bmatrix}2&4&-6&0&-4\\-1&-2&3&0&2\\3&6&-9&0&-6\end{bmatrix}}$ Determine the rank of A and find a basis of the left null space N(AT) of 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
Usually the quickest way to find the rank of a matrix A is to use row-reduction. But hold your horses: Since we need to work with AT anyway we can get the rank of A by considering AT only.

Hint 2
To find the left null space N(AT) row-reduce AT. Since the rank of A equals the rank of AT, row-reducing AT will also reveal the rank of A.

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.

Solution

The rank can be found from the rref of A or of $A^{T}$ however ${\text{rref}}(A^{T})$ is easier to find for this matrix, and necessary to calculate N(AT). So we start by calculating ${\text{rref}}(A^{T})$ . ${\begin{aligned}A={\begin{pmatrix}2&4&-6&0&-4\\-1&-2&3&0&2\\3&6&-9&0&-6\end{pmatrix}},\quad A^{T}={\begin{pmatrix}2&-1&3\\4&-2&6\\-6&3&-9\\0&0&0\\-4&2&-6\end{pmatrix}}\end{aligned}}$ Now calculating ${\text{rref}}(A^{T})$ is quite easy because all of the rows are multiples of the first row and hence linearly dependent. See that $R_{1}={\frac {R_{2}}{2}}={\frac {-R_{3}}{3}}={\frac {-R_{5}}{2}}$. Hence $\mathrm {rref} (A^{T})={\begin{pmatrix}1&{\frac {-1}{2}}&{\frac {3}{2}}\\0&0&0\\0&0&0\\0&0&0\\0&0&0\end{pmatrix}}$ The rank of A is equal to the number of leading zeroes in rref(A) so the rank of A is 1. Now to find the nullspace of $A^{T}$ we start by writing the matrix as an equation equal to zero. ${\begin{aligned}{\begin{bmatrix}1&{\frac {-1}{2}}&{\frac {3}{2}}\\0&0&0\\0&0&0\\0&0&0\\0&0&0\end{bmatrix}}{\begin{bmatrix}x_{1}\\x_{2}\\x_{3}\\\end{bmatrix}}={\begin{bmatrix}0\\0\\0\\0\\0\\\end{bmatrix}}\end{aligned}}$ So ${\begin{bmatrix}x_{1}-{\frac {x_{2}}{2}}+{\frac {3x_{3}}{2}}\\0\\0\\0\\0\end{bmatrix}}={\begin{bmatrix}0\\0\\0\\0\\0\end{bmatrix}}$ $x_{1}-{\frac {x_{2}}{2}}+{\frac {3x_{3}}{2}}=0\therefore x_{1}={\frac {x_{2}}{2}}-{\frac {3x_{3}}{2}}$ Now since there are two independent variables, N(A) will have two vectors. We can find them by setting up a vector with the equation we found above. ${\begin{aligned}{\begin{bmatrix}{\frac {x_{2}}{2}}-{\frac {3x_{3}}{2}}\\x_{2}\\x_{3}\end{bmatrix}}=x_{2}{\begin{bmatrix}{\frac {1}{2}}\\1\\0\end{bmatrix}}+x_{3}{\begin{bmatrix}-{\frac {3}{2}}\\0\\1\end{bmatrix}}\end{aligned}}$ And therefore we find that $N(A^{T})=\mathrm {span} \left(\left\{{\begin{bmatrix}{\frac {1}{2}}\\1\\0\end{bmatrix}}{\begin{bmatrix}-{\frac {3}{2}}\\0\\1\end{bmatrix}}\right\}\right)$

{{#incat:MER CT flag||

Click here for similar questions

MER QGH flag, MER QGQ flag, MER QGS flag, MER QGT flag, MER Tag Four fundamental subspaces, Pages using DynamicPageList3 parser function, Pages using DynamicPageList3 parser tag

}}