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Unformatted text preview: 2.23 Let f be a linear transformation from IR2 to IR2 such that f 1 1 = a 3 1 2 and f 2 1 = 1 2a Find all values of a, if any, for which f = 5 . 5 SOLUTION If we can express 1 2 as a linear combination of 1 1 2 =s +t 2 1 1 1 1 and 2 , say 1 Then we can use the linearity of f to compute f 1 2 . There are several ways to do this. The simplest in IR2 is to express one of the standard basis vectors as a linear combination of the given vectors. This is easy enough: 1 2 1 = − 0 1 1 Hence, the the linearity of f , f 1 0 =f = 2 1 −f 1 1 . 1 a 1−a − = 2a 3 2a − 3 Now it is easy: As you see, 1 2 1 1 1 =2 − , and so, by the linearity of f , 2 1 0 = 2f =2 1 1 −f 1 0 f a 1−a 3a − 1 − = 3 2a − 3 9 − 2a 3a − 1 5 = 9 − 2a 5 and 15 which is equivalent to the pair of Therefore, f equations 1 2 = 5 5 reduces to 3a − 1 = 5 9 − 2a = 5 . 21/september/2005; 22:09 ...
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 Fall '08
 Gladue
 Calculus

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