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Unformatted text preview: Assignment 5 Math 417 Winter 2009 Due February 20 3.1#2. A vector ~x = x y is in the kernel of the given A if and only if ~ 0 = A~x = 2 3 6 9 x y = 2 x + 3 y 6 x + 9 y , i.e, if and only if x and y satisfy the simultaneous linear equations 2 x +3 y = 0 and 6 x +9 y = 0. The second equation here is redundant, being 3 times the first. (If you didnt notice this right away, youd see it when you start rowreducing the augmented matrix of this system of equations.) So the kernel consists of vectors satisfying 2 x + 3 y = 0. Such vectors have the form 3 2 y y (obtained by solving the linear equation for x in terms of y ). In other words, they are obtained by multiplying the vector ~v = 3 2 1 by an arbitrary number. So this single vector ~v spans the kernel of A . The kernel is also spanned by any vector obtained by multiplying ~v by a nonzero number. So people who dislike fractions could use 3 2 or 3 2 just as well as ~v . 3.1#6. A calculation like that in the previous problem shows that ~x = x y z is in the kernel of A if and only if x + y + z = 0. (There are three equations, but the second and third are not only redundant as in the previous problem but actually identical to the first equation.) The general solution is ~x =  y z y z = y  1 1 + z  1 1 . So the kernel is spanned by the two vectors  1 1 and  1 1 . 1 3.1#10. By chance, the matrix A is already in echelon form. The system A~x = ~ 0 has only one free variable, namely x 3 . Setting x 3 = 1, we find the solution ~x = 1 2 1 and so ker( A ) is the span of this vector. 3.1#16. The image of A consists of the vectors of the form A~x = 1 2 3 1 2 3 1 2 3 x y z = x + 2 y + 3 z x + 2 y + 3 z x + 2 y + 3 z = ( x + 2 y + 3 z ) 1 1 1 ....
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This note was uploaded on 10/13/2009 for the course MATH 417 taught by Professor Eveet during the Spring '09 term at University of MichiganDearborn.
 Spring '09
 eveet
 Linear Algebra, Algebra, Linear Equations, Equations

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