This preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: Name: ID#: Solutions to Midterm II Math 20 Introduction to Linear Algebra and Multivariable Calculus 22 November 2004 Show all of your work. Full credit may not be given for an answer alone. You may use the backs of the pages or the extra pages for scratch work. Do not unstaple or remove pages. This is a noncalculator exam. Students who, for whatever reason, submit work not their own will ordi narily be required to withdraw from the College. —Handbook for Students 1 1 1. (12 Points) Suppose that det a b c d e f g h k = 7 . Find the following, with justification. (i) det a b c 3 d 3 e 3 f g h k Solution. The determinant is multilinear in each of its rows, so the answer is 3 · 7 = 21 . (ii) det 3 a b c d e f g h k Solution. In this case the effect is that of multiplying each row by 3, so the deter minant gets multiplied by 3 three times. Thus the determinant of the given matrix is 3 3 · 7 = 27 · 7 = 189 . 1 1 1 (iii) det d e f g h k a b c Solution. Each trnasposition of the rows introduces a factor of 1 . We have det d e f g h k a b c = det d e f a b c g h k = det a b c d e f g h k = 7 . (iv) det a b c d + g e + h f + k g h k Solution. This matrix is our original matrix with the third row added to the sec ond. The determinant does not change under this kind of row operation, and so remains 7. / 12 2 2 2 2. (8 Points) Suppose B is a square matrix with B 2 = B . What can you say about det( B ) ? Solution. Since the determinant of a product is the product of the determinants, we have det( B ) 2 = det( B ) , so det( B ) is either 1 or ....
View
Full Document
 Spring '09
 MihaiBeligan
 Linear Algebra, Algebra, Multivariable Calculus, Matrices, Det, ab det

Click to edit the document details