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plugin-hw05 - Assignment 5 Math 417 Winter 2009 Due...

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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 = Ax = 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 didn’t notice this right away, you’d 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 0 + z - 1 0 1 . So the kernel is spanned by the two vectors - 1 1 0 and - 1 0 1 . 1

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§ 3.1#10. By chance, the matrix A is already in echelon form. The system Ax = 0 has only one free variable, namely x 3 . Setting x 3 = 1, we find the solution x = 1 - 2 1 0 and so ker( A ) is the span of this vector. § 3.1#16. The image of A consists of the vectors of the form Ax = 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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