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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 row-reducing 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 non-zero 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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