Assignment 7
Math 417 Winter 2009 Due March 13
5 1 as a linear combination 3.4#10. The question is whether we can express x = 3 1 2 c1 v1 + c2 v2 of v1 = 0 and v2 = 1 . By inspection we see that, if there is to be 1 0 such an expression, the coecient c1
Assignment 6
Math 417 Winter 2009 Due March 6
3.2#46. The kernel of this matrix A consists of all the solutions to the homogeneous linear sytem Ax = 0. To nd a basis for the kernel, begin by solving that linear system by row-reduction. That needs no work,
Assignment 5
Math 417 Winter 2009 Due February 20
3.1#2. A vector x = x is in the kernel of the given A if and only if y 0 = Ax = 23 69 x 2x + 3y = , y 6x + 9y
i.e, if and only if x and y satisfy the simultaneous linear equations 2x+3y = 0 and 6x+9y = 0.
Assignment 4
Math 417 Winter 2009 Due February 6
2.3#6. ab cd ad bc ab + ba ad bc 0 d b = = . c a cd dc bc + ad 0 ad bc
2.3#14. As always, the product of two matrices is dened if the number of columns of the rst equals the number of rows of the second. So
Assignment 3
Math 417 Winter 2009 Due January 30
14 2.1#6. Yes, it is linear. Its matrix is 2 5. 36 2.1#8. The easiest way to nd the inverse of a linear transformation is to invert its matrix. So we compute, by row-reduction (saving space by doing two st
Assignment 2
Math 417 Winter 2009 Due January 23
1.2#42. First, lets assign variables for the trac volume along each one of the four street segments marked with question marks. Let x1 be the trac on JFK St. between Winthrop and Mt. Auburn. Let x2 be the t
Assignment 1
Math 417 Winter 2009 Due January 16
1.1#18. The augmented matrix of the 1 1 1 given system is 23a 3 8 b . 22c
We reduce it to echelon form. Subtracting the rst row from each of the others, we get a 12 3 0 1 5 b a . 0 0 1 c a Subtracting twice
Supplementary Assignment
Math 417 Winter 2009 Do Not Submit
8.1#8. We rst nd the eigenvalues of A by computing its characteristic polynomial det(A I2 ) = det 3 3 = (3 )(5 ) 9 = (6 )(4 ). 3 5
Then we compute the corresponding eigenspaces. = 6 : E6 = ker =
Assignment 11
Math 417 Winter 2009 Due April 20
7.1#2. We know that Av = v . (Note that = 0 since A is invertible and hence ker(A) = cfw_0.) Then A1 (v ) = v and so A1 v = 1 A1 (v ) = 1 v . Therefore, v is an eigenvector of A1 with eigenvalue 1 . 7.1#4. S
Assignment 10
Math 417 Winter 2009 Due April 10
6.1#10. Using the 11 1 2 det 13 method discussed in class. 1 111 3 = det 0 1 2 6 025 111 = det 0 1 2 = 1 001
R2 R2 R1 R3 R3 R1 R3 R3 2R2
6.1#20. Using the method discussed in class. 1 k 1 11 1 det 1 k + 1 k
Assignment 9
Math 417 Winter 2009 Due March 27
5.3#16. Since A and B are symmetric, we have A = AT and B = B T . Therefore, (A + B )T = AT + B T = A + B , which means that A + B is symmetic too. 5.3#26. Just like in the case of the product of two matrices
Assignment 8
Math 417 Winter 2009 Due March 20
5.1#40. Were given that (among other things) v2 v2 = a22 = 9. Since v2 v2 = v2 2 , we conclude that v2 = 3 (not 3, because lengths are never negative). 5.1#42. Using the fact that x v1 + v2
2 2
= x x for ever