Science:Math Exam Resources/Courses/MATH221/December 2011/Question 06 (b)

MATH221 December 2011

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

   Other MATH221 Exams

  •  December 2008  •  December 2009  •  December 2007  •  April 2010  •  April 2009  •  December 2011  •  April 2013  •  

Question 06 (b)

Consider the vectors $v_{1}=\left({\begin{array}{c}1/2\\1/2\\1/2\\1/2\end{array}}\right),\quad v_{2}=\left({\begin{array}{c}1/2\\1/2\\-1/2\\-1/2\end{array}}\right),$ $v_{3}=\left({\begin{array}{c}1/2\\-1/2\\1/2\\-1/2\end{array}}\right),\quad v_{4}=\left({\begin{array}{c}1/2\\-1/2\\-1/2\\1/2\end{array}}\right)$ b) Let b = (1,2,3,4)T. Then write b as a linear combination of the vectors v1, v2, v3, v4.

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
We are trying to find scalars $x_{1},x_{2},x_{3},x_{4}$ such that the following holds: $v_{1}x_{1}+v_{2}x_{2}+v_{3}x_{3}+v_{4}x_{4}=b$ Write this in matrix notation and then follow the steps from part (a).

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 1

Found a typo? Is this solution unclear? Let us know here. Please rate my easiness! It's quick and helps everyone guide their studies. From Question 6 (a), we know that $\{v_{1},v_{2},v_{3},v_{4}\}$ is orthogonal basis. Since $|v_{i}|=1$ for $i=1,2,3,4$, it is actually orthonormal basis. Then, we have $b=\sum _{i=1}^{4}(b,v_{i})v_{i}.$ Computing the inner products, we have $(b,v_{1})=\left({\begin{bmatrix}1\\2\\3\\4\end{bmatrix}},{\begin{bmatrix}1/2\\1/2\\1/2\\1/2\end{bmatrix}}\right)=1\cdot {\frac {1}{2}}+2\cdot {\frac {1}{2}}+3\cdot {\frac {1}{2}}+4\cdot {\frac {1}{2}}=5,$ $(b,v_{2})=\left({\begin{bmatrix}1\\2\\3\\4\end{bmatrix}},{\begin{bmatrix}1/2\\1/2\\-1/2\\-1/2\end{bmatrix}}\right)=1\cdot {\frac {1}{2}}+2\cdot {\frac {1}{2}}-3\cdot {\frac {1}{2}}-4\cdot {\frac {1}{2}}=-2,$ $(b,v_{3})=\left({\begin{bmatrix}1\\2\\3\\4\end{bmatrix}},{\begin{bmatrix}1/2\\-1/2\\1/2\\-1/2\end{bmatrix}}\right)=1\cdot {\frac {1}{2}}-2\cdot {\frac {1}{2}}+3\cdot {\frac {1}{2}}-4\cdot {\frac {1}{2}}=-1,$ and $(b,v_{4})=\left({\begin{bmatrix}1\\2\\3\\4\end{bmatrix}},{\begin{bmatrix}1/2\\-1/2\\-1/2\\1/2\end{bmatrix}}\right)=1\cdot {\frac {1}{2}}-2\cdot {\frac {1}{2}}-3\cdot {\frac {1}{2}}+4\cdot {\frac {1}{2}}=0.$ Therefore, $b$ can be written as $b=5v_{1}-2v_{2}-v_{3}.$

Solution 2

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 create the augmented matrix $[v_{1}v_{2}v_{3}v_{4}b]$ and use Gaussian elimination. $\left[{\begin{array}{ccccc}1/2&1/2&1/2&1/2&1\\1/2&1/2&-1/2&-1/2&2\\1/2&-1/2&1/2&-1/2&3\\1/2&-1/2&-1/2&1/2&4\end{array}}\right]$ We multiply the whole matrix by 2 to get $\left[{\begin{array}{ccccc}1&1&1&1&2\\1&1&-1&-1&4\\1&-1&1&-1&6\\1&-1&-1&1&8\end{array}}\right]$ Subtracting the first row from the following three rows, we get $\left[{\begin{array}{ccccc}1&1&1&1&2\\0&0&-2&-2&2\\0&-2&0&-2&4\\0&-2&-2&0&6\end{array}}\right]$ We rearrange the rows so that the second row becomes the last row; each of the last three rows is also multiplied by -1/2. $\left[{\begin{array}{ccccc}1&1&1&1&2\\0&1&0&1&-2\\0&1&1&0&-3\\0&0&1&1&-1\end{array}}\right]$ We subtract row 2 from row 1 and row 2 from row 3 to get $\left[{\begin{array}{ccccc}1&0&1&0&4\\0&1&0&1&-2\\0&0&1&-1&-1\\0&0&1&1&-1\end{array}}\right]$ We subtract row 3 from row 1 and row 3 from row 4 to get $\left[{\begin{array}{ccccc}1&0&0&1&5\\0&1&0&1&-2\\0&0&1&-1&-1\\0&0&0&-2&0\end{array}}\right]$ Multiplying row 4 by -1/2, we get $\left[{\begin{array}{ccccc}1&0&0&1&5\\0&1&0&1&-2\\0&0&1&-1&-1\\0&0&0&1&0\end{array}}\right]$ Which finally simplifies to $\left[{\begin{array}{ccccc}1&0&0&0&5\\0&1&0&0&-2\\0&0&1&0&-1\\0&0&0&1&0\end{array}}\right]$ So our scalars will be $x_{1}=5,x_{2}=-2,x_{3}=-1$, and $x_{4}=0$. In other words $b=5v_{1}-2v_{2}-v_{3}$.