Lab3Sol - NATIONAL UNIVERSITY OF SINGAPORE Department of...

Info icon This preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon
NATIONAL UNIVERSITY OF SINGAPORE Department of Mathematics MA1101R Laboratory 3 Semester I 2013/14 Linaer Combinations, Linear Span, Linear Independence, Transformation Matrices In this laboratory session, we will consolidate what we have learnt in lectures in- volving concepts of linear combinations/span/independence. We will also use the specially written matlab functions trans2x2 and trans3x3 to visualize the geomet- rical effect of certain matrices, when these matrices are multiplied to plane figures in R 2 . Activity 1 Enter the following matrix (carefully!) in the matlab command window. A = 0 2 1 - 3 2 - 2 2 1 - 1 0 5 - 2 4 0 1 3 4 5 2 9 0 2 1 2 7 - 1 0 2 . (i) Write down the reduced row-echelon form of A . 1 0 0 2 - 1 0 2 0 1 0 - 3 1 0 2 0 0 1 3 0 0 - 2 0 0 0 0 0 1 0 (ii) Let a 1 = 0 1 1 2 , a 2 = 2 - 1 3 1 , a 3 = 1 0 4 2 , a 4 = - 3 5 5 7 , a 5 = 2 - 2 2 - 1 , a 6 = - 2 4 9 0 , a 7 = 2 0 0 2 . 1
Image of page 1

Info icon This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
(So a i is the i -th column of A .) By just using the answer from (i), without performing any further computation in matlab , you should be able to answer the following questions. (a) What will be the reduced row-echelon form of the matrix ( a 1 a 2 a 3 | 0 )? 1 0 0 0 0 1 0 0 0 0 1 0 0 0 0 0 (b) Is { a 1 , a 2 , a 3 } a linearly independent set? Why? Yes, since c 1 a 1 + c 2 a 2 + c 3 a 3 = 0 has only the trivial solution, as seen from part (b). (c) Is span { a 1 , a 2 , a 3 , a 4 } = R 4 ? Why? No, since the reduced row echelon form of ( a 1 a 2 a 3 a 4 ) has a row of zeros.
Image of page 2
Image of page 3
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

What students are saying

  • Left Quote Icon

    As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

    Student Picture

    Kiran Temple University Fox School of Business ‘17, Course Hero Intern

  • Left Quote Icon

    I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

    Student Picture

    Dana University of Pennsylvania ‘17, Course Hero Intern

  • Left Quote Icon

    The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

    Student Picture

    Jill Tulane University ‘16, Course Hero Intern