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Unformatted text preview: Solutions to Assignment 8 Math 217, Fall 2002 4.6.26 If A is 6 4, what is the smallest possible dimension of Nul( A )? Well, we know that rank( A )+dim(Nul( A )) = 4. It is possible that rank( A ) = 4, that is, that dim(Col( A )) = 4 because A is 6 4. We conclude the smallest possible dimension of Nul( A ) is zero. 4.6.20 Suppose a nonhomogeneous system of nine linear equations in ten unknowns has a solution for all possible constants on the right sides of the equations. Is it possible to find two nonzero solutions of the associated homogeneous system that are not multiples of each other? Discuss. Again, we know that rank( A ) + dim(Nul( A )) = 10. If the system is consistent for all possible constants on the right side of the equation, then the matrix of coefficients must have a pivot in every row. As the matrix is 9 10, this means that the matrix must have 9 pivots. Thus there are 9 pivot columns, and so the rank( A ) = 9. It must be the case that dim(Nul( A )) = 1, and we conclude that a basis for the vector space of solutions to the homogeneous system of equations contains only one element. This means that every nonzero solution is a multiple of the one basis element. So it is not possible to find two nonzero vectors whichof the one basis element....
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This note was uploaded on 01/30/2011 for the course MATH 217 taught by Professor Staff during the Spring '10 term at Waterloo.
 Spring '10
 Staff
 Math

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