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tut04s - THE UNIVERSITY OF SYDNEY PURE MATHEMATICS Linear...

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Unformatted text preview: THE UNIVERSITY OF SYDNEY PURE MATHEMATICS Linear Mathematics 2012 Tutorial 4 — Solutions 1. Let X = braceleftBigparenleftBig 6 3 2 parenrightBig , parenleftBig 2 3 1 parenrightBig , parenleftBig 4- 6 parenrightBigbracerightBig . Show that X is a linearly dependent subset of R 3 and find the dimension of Span( X ) . Solution Suppose that a parenleftBig 6 3 2 parenrightBig + b parenleftBig 2 3 1 parenrightBig + c parenleftBig 4- 6 parenrightBig = parenleftBig parenrightBig , for some a, b, c ∈ R . To find all possible a, b, c we have to solve the system of linear equations corresponding to the following augmented matrix: parenleftBigg 6 2 4 3 3- 6 2 1 parenrightBigg Row reduce-----→ parenleftBigg 1 2 1- 4 parenrightBigg . (Note that the row echelon form of a matrix is not unique, as it depends upon the sequence of row operations that you choose—the reduced row echelon form is, however, unique. Consequently, you might obtain a different row echelon form for the augmented matrix above . However, your row echelon form should still give the same set of solutions as the one above. This comment applies to all of the questions below which compute row echelon forms.) Looking at the row reduced form of the augmented matrix above, we can take c to be a free variable. The general solutions is c = r , for any r ∈ R , b = 4 c = 4 r , and a =- 2 c =- 2 r . In particular, taking r = 1 we have that- 2 parenleftBig 6 3 2 parenrightBig + 4 parenleftBig 2 3 1 parenrightBig + parenleftBig 4- 6 parenrightBig = parenleftBig parenrightBig . Hence, these three vectors are linearly dependent. Now, Span( X ) = Span parenleftBigbraceleftBigparenleftBig 6 3 2 parenrightBig , parenleftBig 2 3 1 parenrightBig , parenleftBig 4- 6 parenrightBigbracerightBigparenrightBig = Span parenleftBigbraceleftBigparenleftBig 6 3 2 parenrightBig , parenleftBig 2 3 1 parenrightBigbracerightBigparenrightBig , since parenleftBig 4- 6 parenrightBig is a linear combination of parenleftBig 6 3 2 parenrightBig and parenleftBig 2 3 1 parenrightBig by the last paragraph. Note that parenleftBig 6 3 2 parenrightBig is not a scalar multiple of parenleftBig 2 3 1 parenrightBig , so parenleftBig 6 3 2 parenrightBig and parenleftBig 2 3 1 parenrightBig are independent, and hence dim Span( X ) = 2 ....
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