Science:Math Exam Resources/Courses/MATH307/April 2006/Question 01 (a)

MATH307 April 2006

Work in progress: this question page is incomplete, there might be mistakes in the material you are seeing here.

  •  Q1 (a)  •  Q1 (b)  •  Q2 (a)  •  Q2 (b)  •  Q3 (a)  •  Q3 (b)  •  Q3 (c)  •  Q4 (a)  •  Q4 (b)  •  Q4 (c)  •  Q4 (d)  •  Q5 (a)  •  Q5 (b)  •  Q5 (c)  •  Q5 (d)  •  Q6 (a)  •  Q6 (b)  •  Q6 (c)  •  Q6 (d)  •  Q7 (a)  •  Q7 (b)  •  Q7 (c)  •  Q7 (d)  •  

   Other MATH307 Exams

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

Question 01 (a)
Let PA = LU be ${\begin{bmatrix}1&0&0\\0&0&1\\0&1&0\end{bmatrix}}{\begin{bmatrix}1&0&2&3\\2&1&8&8\\0&1&4&2\end{bmatrix}}={\begin{bmatrix}1&0&0\\0&1&0\\2&1&1\end{bmatrix}}{\begin{bmatrix}1&0&2&3\\0&1&4&2\\0&0&0&0\end{bmatrix}}$ and $b={\begin{bmatrix}b_{1}\\b_{2}\\b_{3}\end{bmatrix}}$ (a) Find a basis for each of the four fundamental subspaces.

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

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. This matrix A in this question is given in the form of an LU decomposition with partial pivoting where P is just the matrix which swaps the rows of A. The matrix P also would not affect R(A), N(A), R(AT) where we need to look at the matrix U (echelon form) to find the bases. To find a basis for the R(A), we must identify the pivot columns in matrix U, which turns out to be the first and second column. The pivots are highlighted in red. $U={\begin{bmatrix}{\color {Red}{1}}&0&2&3\\0&{\color {Red}{1}}&4&2\\0&0&0&0\end{bmatrix}}$ The basis for R(A) is then the corresponding columns in A to the pivot columns in U. The pivot columns of A are highlighted in red. $A={\begin{bmatrix}{\color {Red}{1}}&{\color {Red}{0}}&2&3\\{\color {Red}{2}}&{\color {Red}{1}}&8&8\\{\color {Red}{0}}&{\color {Red}{1}}&4&2\end{bmatrix}}$ $R(A)={\text{span}}\left({\begin{bmatrix}1\\2\\0\end{bmatrix}},{\begin{bmatrix}0\\1\\1\end{bmatrix}}\right)$ To find N(A), we can set $Ux=0$ and solve for x. This would give us 2 equations and 2 free variables. ${\begin{aligned}x_{1}&=-2x_{3}-3x_{4}\\x_{2}&=-4x_{3}-2x_{4}\\x_{3}&=x_{3}\\x_{4}&=x_{4}\end{aligned}}$ $x=x_{3}{\begin{bmatrix}-2\\-4\\1\\0\end{bmatrix}}+x_{4}{\begin{bmatrix}-3\\-2\\0\\1\end{bmatrix}}$ And therefore $N(A)={\text{span}}\left({\begin{bmatrix}-2\\-4\\1\\0\end{bmatrix}},{\begin{bmatrix}-3\\-2\\0\\1\end{bmatrix}}\right)$ The basis of R(AT) can be taken from the rows of U that contain pivots. $R(A^{T})={\text{span}}\left({\begin{bmatrix}1\\0\\2\\3\end{bmatrix}},{\begin{bmatrix}0\\1\\4\\2\end{bmatrix}}\right)$ To find the basis for N(AT), we must row reduce AT and solve for x in $A^{T}x=0$ From row reducing AT, you should acquire the matrix ${\begin{bmatrix}1&0&-2\\0&1&1\\0&0&0\\0&0&0\end{bmatrix}}$ Solving for x, you end up with 2 equations and 1 free variable. ${\begin{aligned}x_{1}&=2x_{3}\\x_{2}&=-x_{3}\\x_{3}&=x_{3}\end{aligned}}$ $x=x_{3}{\begin{bmatrix}2\\-1\\1\end{bmatrix}}$ And hence $N(A^{T})={\text{span}}\left({\begin{bmatrix}2\\-1\\1\end{bmatrix}}\right)$