Science:Math Exam Resources/Courses/MATH307/April 2005/Question 01 (a)

MATH307 April 2005

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)  •  Q2 (c)  •  Q2 (d)  •  Q3 (a)  •  Q3 (b)  •  Q3 (c)  •  Q4 (a)  •  Q4 (b)  •  Q4 (c)  •  Q4 (d)  •  Q5  •  Q6 (a)  •  Q6 (b)  •  Q7  •  Q8 (a)  •  Q8 (b)  •  Q8 (c)  •  Q8 (d)  •  Q8 (e)  •  Q8 (f)  •  Q9 (a)  •  Q9 (b)  •  Q9 (c)  •  Q9 (d)  •  Q9 (e)  •  

   Other MATH307 Exams

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

Question 01 (a)
Consider the symmetric matrix $B={\begin{pmatrix}2&-2&0&4\\-2&3&2&1\\0&2&3&3\\4&1&3&2\end{pmatrix}}$ Find the $\displaystyle LDU$ decomposition of B.

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. We want to find a lower triangular matrix, a diagonal matrix, and an upper triangular matrix, which we call $L,D,U$ respectively such that when multiplied together, we get $B=LDU$. First we perform row operations on $B$ until $B$ becomes an upper triangular matrix, keeping track of the the row operations performed: ${\begin{aligned}B={\begin{pmatrix}2&-2&0&4\\-2&3&2&1\\0&2&3&3\\4&1&3&2\end{pmatrix}}\rightarrow {\begin{pmatrix}2&-2&0&4\\0&1&2&5\\0&2&3&3\\4&1&3&2\end{pmatrix}}\rightarrow {\begin{pmatrix}2&-2&0&4\\0&1&2&5\\0&2&3&3\\0&5&3&-6\end{pmatrix}}\\\rightarrow {\begin{pmatrix}2&-2&0&4\\0&1&2&5\\0&0&-1&-7\\0&5&3&-6\end{pmatrix}}\rightarrow {\begin{pmatrix}2&-2&0&4\\0&1&2&5\\0&0&-1&-7\\0&0&-7&-31\end{pmatrix}}\rightarrow {\begin{pmatrix}2&-2&0&4\\0&1&2&5\\0&0&-1&-7\\0&0&0&18\end{pmatrix}}.\end{aligned}}$ The row operations performed are: $E_{1}$: row1 + row2 $E_{2}$: -2*row1 + row4 $E_{3}$: -2*row2 + row3 $E_{4}$: -5*row2 + row4 $E_{5}$: -7*row3 + row4 These row operations translate into the following 5 elementary matrices $E_{i}$ as described above: ${\begin{aligned}E_{1}={\begin{pmatrix}1&0&0&0\\1&1&0&0\\0&0&1&0\\0&0&0&1\end{pmatrix}},E_{2}={\begin{pmatrix}1&0&0&0\\0&1&0&0\\0&0&1&0\\-2&0&0&1\end{pmatrix}},E_{3}={\begin{pmatrix}1&0&0&0\\0&1&0&0\\0&-2&1&0\\0&0&0&1\end{pmatrix}}\\E_{4}={\begin{pmatrix}1&0&0&0\\0&1&0&0\\0&0&1&0\\0&-5&0&1\end{pmatrix}},E_{5}={\begin{pmatrix}1&0&0&0\\0&1&0&0\\0&0&1&0\\0&0&-7&1\end{pmatrix}}.\end{aligned}}$ Hence, we have ${\begin{aligned}{\begin{pmatrix}2&-2&0&4\\0&1&2&5\\0&0&-1&-7\\0&0&0&18\end{pmatrix}}=E_{5}E_{4}E_{3}E_{2}E_{1}B={\begin{pmatrix}1&0&0&0\\1&1&0&0\\-2&-2&1&0\\7&9&-7&1\end{pmatrix}}B.\end{aligned}}$ Let $E_{L}=E_{5}E_{4}E_{3}E_{2}E_{1}$ describe the matrix product of elementary matrices required to take $B$ to an upper triangular matrix. Similarly, we perform 3 more row operations to make the diagonal entries of $B$ be all ones: $E_{6}$: 1/2*row1 $E_{7}$: -1*row3 $E_{8}$: 1/18*row4 and we let $E_{D}=E_{8}E_{7}E_{6}$ to describe the matrix product of elementary matrices required to make the diagonal entries of $B$ be all ones. That is, we have our desired upper triangular matrix $U$ after multiplying $E_{D},E_{L},B$: $E_{D}E_{L}B=E_{D}{\begin{pmatrix}2&-2&0&4\\0&1&2&5\\0&0&-1&-7\\0&0&0&18\end{pmatrix}}={\begin{pmatrix}1&-1&0&2\\0&1&2&5\\0&0&1&7\\0&0&0&1\end{pmatrix}}=U$ In summary, ${\begin{aligned}E_{L}={\begin{pmatrix}1&0&0&0\\1&1&0&0\\-2&-2&1&0\\7&9&-7&1\end{pmatrix}},E_{D}={\begin{pmatrix}1/2&0&0&0\\0&1&0&0\\0&0&-1&0\\0&0&0&1/18\end{pmatrix}},\\U={\begin{pmatrix}1&-1&0&2\\0&1&2&5\\0&0&1&7\\0&0&0&1\end{pmatrix}}\end{aligned}}$ Since $E_{D}E_{L}B=U$, we can invert $E_{D}$ and $E_{L}$ to attain $D$ and $L$ respectively: ${\begin{aligned}E_{D}E_{L}B=U\Rightarrow B=E_{L}^{-1}E_{D}^{-1}U\\\Rightarrow D=E_{D}^{-1}={\begin{pmatrix}2&0&0&0\\0&1&0&0\\0&0&-1&0\\0&0&0&18\end{pmatrix}}\\\Rightarrow L=E_{L}^{-1}={\begin{pmatrix}1&0&0&0\\-1&1&0&0\\0&2&1&0\\2&5&7&1\end{pmatrix}}\end{aligned}}$. Hence, we have ${\begin{aligned}B=LDU={\begin{pmatrix}1&0&0&0\\-1&1&0&0\\0&2&1&0\\2&5&7&1\end{pmatrix}}{\begin{pmatrix}2&0&0&0\\0&1&0&0\\0&0&-1&0\\0&0&0&18\end{pmatrix}}{\begin{pmatrix}1&-1&0&2\\0&1&2&5\\0&0&1&7\\0&0&0&1\end{pmatrix}}.\end{aligned}}$