This preview shows pages 1–4. Sign up to view the full content.
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: MATH 223, Linear Algebra Fall, 2007 Assignment 2 Solutions 1. Find the rank of each of the matrices below: (a) 1 2 3 4 5 6 7 8 9 10 11 12 (over the rationals) (b) 1 1 1 1 1 1 (over Z 2 ) (c) 1 2 i 3 + 2 i 4 3 i 2 + i 1 4 + 3 i 3 2 i (over C ) Solution: Recall that the rank of a matrix A is defined to be the number of leading ones in the reduced row echelon form of A (this is the same thing as the row canonical form of A ). Thus, to find the ranks of these matrices, we need to put them into reduced row echelon form. Using row reduction, we find that for each matrix above, the reduced row echelon form is: (a) 1 1 1 1 2 3 (b) 1 1 1 1 (c) 1 2 i 1 i 2 Notice that in each case the number of leading ones (and hence the rank) is equal to 2. 1 2. Consider the following 3 3 real matrices: A =  2 1 8 1 1 7 3 4 B = 5 7 6 3 9 2 2 C = 6 3 1 2 4 5 1 1 8 . Find (by hand!) the following expressions: (a) AB (b) BA (c) AB BA (d) ABC Solution: Using the definition of matrix multiplication, we easily find: (a) AB =  20 13 5 25 17 16 7 8 21 (b) BA =  31 5 12 42 3 33 6 30 . (c) Subtracting the second answer from the first gives AB BA = 11 18 7 17 20 17 1 8 9 (d) Finally, multiplying the first answer on the right by C , we find ABC =  151 117 5 200 159 68 47 10 215 3. Show that x =  1 1 1 is a solution to the equation x 3 + 1 = 0 . Can you think of another 3 3 real matrix that is a solution to this equation? Bonus: find all diagonal complex matrices that are solutions to x 3 + 1 = 0. 2 Solution: Note that the 1 and 0 in x 3 +1 = 0 stand for the 3 3 identity matrix and the 3 3 zero matrix (i.e. the 3 3 matrix all of whose entries are zero) respectively. We calculate: x 2 =  1 1 1  1 1 1 =  1 1 1 , so we have x 3 = x x 2 =  1 1 1  1 1 1 =  1 1 1 , and therefore x 3 + 1 =  1 1 1 + 1 1 1 = = 0 , as required....
View
Full
Document
This note was uploaded on 04/29/2008 for the course MATH 223 taught by Professor Loveys during the Fall '07 term at McGill.
 Fall '07
 Loveys
 Linear Algebra, Algebra, Matrices

Click to edit the document details