2008 Algebra Test 2

2008 Algebra Test 2 - . We have A T A = 4 12 12 46 A T b =...

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1. Math1151 Algebra S2 2008. Test 2. Version 1a. (1) Find all values of a for which the matrix A = 1 1 a 2 - 2 - 1 a 3 0 is invertible. Solution. We compute det A = a (6 + 2 a ) + (3 - a ) = (2 a + 3)( a + 1). Hence the matrix A is invertible for all a 6 = - 1 , - 3 / 2. (2) Given the vectors a = ( - 1 , 4 , 2), b = ( - 2 , - 1 , 3) and c = (8 , - 5 , - 7), ﬁnd (a) a vector which is perpendicular to both a and b Solution . The vector a × b = ± ± ± ± ± ± - 1 4 2 - 2 - 1 3 e 1 e 2 e 3 ± ± ± ± ± ± = (14 , - 1 , 9) is perpendicular to both a and b . (b) the projection of c onto a . Solution. proj a c = c · a | a | 2 a = - 8 - 20 - 14 1 + 16 + 4 ( - 1 , 4 , 2) = (2 , - 8 , - 4) . (3) Find the line y = α + βx which best ﬁts, in the least squares sense, the points (1 , 3), (2 , - 1), (4 , 2) and (5 , 0). Solution. The line of best ﬁt y = α + βx is given by the solution of the equation A T A ² α β ³ = A T b where A = 1 1 1 2 1 4 1 5 and b = 3 - 1 2 0
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Unformatted text preview: . We have A T A = 4 12 12 46 A T b = 4 9 and solving this gives ( , ) = (19 / 10 ,-3 / 10). (4) Let A = (2 , 7 ,-5). Find the point P on the plane : x-2 y + z = 4 such that AP is perpendicular to the plane. Solution. First note that n = (1 ,-2 , 1) T is normal to the plane , so the point P such that AP is perpendicular is given by the intersection of the line ` ( ) = a + n and , where A = (2 , 7 ,-5) T . Solving the equation (2 + )-2(7-2 ) + (-5 + ) = 4 gives = 7 / 2, so P is the point with coordinate vector (11 / 2 , ,-3 / 2) T . 1...
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This document was uploaded on 03/19/2012.

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