hmw5 - Solutions to Homework 5. Math 110, Fall 2006. Prob...

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Solutions to Homework 5. Math 110, Fall 2006. Prob 2.5.3. ( a ) a 2 b 2 c 2 a 1 b 1 c 1 a 0 b 0 c 0 , ( b ) a 0 b 0 c 0 a 1 b 1 c 1 a 2 b 2 c 2 , ( c ) 0 - 1 0 1 0 0 - 3 2 1 , ( d ) 2 1 1 3 - 2 1 - 1 3 1 , ( e ) 5 - 6 3 0 4 - 1 3 - 1 2 , ( f ) - 2 1 2 3 4 1 - 1 5 2 . Prob 2.5.6. The matrix Q is simply the change of basis matrix where the change is from the standard basis to the basis β . That is, Q is the matrix whose columns are vectors in β . (a) [ L A ] β = ± 6 11 - 2 - 4 ² , Q = ± 1 1 1 2 ² . (b) [ L A ] β = ± 3 0 0 - 1 ² , Q = ± 1 1 1 - 1 ² . (c) [ L A ] β = 2 2 2 - 2 - 3 - 4 1 1 2 , Q = 1 1 1 1 0 1 1 1 2 . (d) [ L A ] β = 6 0 0 0 12 0 0 0 18 , Q = 1 1 1 1 - 1 1 - 2 0 1 . Prob 2.5.7. (a) By the linearity of T , T ( x,y ) = ( ax + by,cx + dy ) for some a , b , c , d . To find these numbers, notice that every vector on the line remains unchanged under the action of T , i.e., ( ax + bmx,cx + dmx ) = T ( x,mx ) = ( x,mx ) for all x . This implies a + bm = 1, c + dm = m . Now, every vector perpendicular to the line gets reflected about the line, i.e., ( amx - bx,cmx - dx ) = T (
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hmw5 - Solutions to Homework 5. Math 110, Fall 2006. Prob...

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