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# tat04 - T HE U NIVERSITY OF S YDNEY P URE M ATHEMATICS...

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T HE U NIVERSITY OF S YDNEY P URE M ATHEMATICS Linear Mathematics 2010 Tutorial 4 — Solutions 1. You are given the following data points: x 1 0 1 2 y 4 1 2 1 Construct a Lagrange basis { p 0 , p 1 , p 2 , p 3 } of P 3 using the x values from the data set. Hence find the unique cubic polynomial p ( x ) that fits the data exactly. Estimate the value of y when x = 1 2 . Solution By definition, the vectors in the Lagrange basis of P 3 are as follows: p 0 ( x )= x ( x 1)( x 2) ( 1)( 1 1)( 1 2) = 1 6 x ( x 1)( x 2) , p 1 ( x )= ( x +1)( x 1)( x 2) (0+1)(0 1)(0 2) = 1 2 ( x +1)( x 1)( x 2) , p 2 ( x )= ( x +1)( x 0)( x 2) (1+1)(1 0)(1 2) = 1 2 x ( x +1)( x 2) , p 3 ( x )= ( x +1)( x 0)( x 1) (2+1)(2 0)(2 1) = 1 6 x ( x +1)( x 1) Hence the polynomial that fits the data is p ( x ) = 4 p 0 ( x ) + p 1 ( x ) 2 p 2 ( x ) + p 3 ( x ) . This simplifies to p ( x )= x 3 4 x +1 . When x = 1 2 we estimate that y p ( 1 2 )= 1 8 2+1= 7 8 = 0 . 875 . 2. The matrix J is a reduced row echelon form of the matrix A . A = parenleftBigg 1 2 0 4 5 2 5 3 2 1 4 10 7 8 3 8 19 9 14 13 7 17 19 50 18 parenrightBigg J = parenleftBigg 1 0 0 40 29 0 1 0 - 18 - 12 0 0 1 4 1 0 0 0 0 0 0 0 0 0 0 parenrightBigg a ) Find a basis for the column space of A . b ) Find a basis for the null space of A . c ) Verify that the sum of the rank of A and the nullity of A is the number of columns of A . Solution a ) The first three columns of J are a basis for the column space of J , so the first three columns of A are a basis for the column space of A . So a basis for the column space of A is braceleftBiggparenleftBigg 1 2 4 8 7 parenrightBigg , parenleftBigg 2 5 10 19 17 parenrightBigg , parenleftBigg 0 3 7 9 19 parenrightBiggbracerightBigg . Math 2061: Tutorial 4 — Solutions S.B and A.M 28/1/2010

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Linear Mathematics Tutorial 4 — Solutions Page 2 b ) Let x = parenleftBig x 1 . . . x 5 parenrightBig be in the null space of A . As Null( A )=Null( J ) , looking at J we see that x 4 and x 5 are free variables and that x 1 = 40 x 4 29 x 5 , x 2 =18 x 4 +12 x 5 , x 3 = 4 x 4 x 5 . Hence x = parenleftBigg x 1 x 2 x 3 x 4 x 5 parenrightBigg = parenleftBigg - 40 x 4 - 29 x 5 18 x 4 +12 x 5 - 4 x 4 - x 5 x 4 x 5 parenrightBigg = x 4 parenleftBigg - 40 18 - 4 1 0 parenrightBigg + x 5 parenleftBigg - 29 12 - 1 0 1 parenrightBigg , for all x 4 ,x 5 R . Consequently, a basis for the null space of A is braceleftBiggparenleftBigg - 40 18 - 4 1 0 parenrightBigg , parenleftBigg - 29 12 - 1 0 1 parenrightBiggbracerightBigg . c ) The rank of A is the dimension of the column space of A , so rank( A )=3 . The nullity of A is the dimension of the null space of A , so nullity( A )=2 .
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tat04 - T HE U NIVERSITY OF S YDNEY P URE M ATHEMATICS...

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