201f02mt2[1]

# 201f02mt2[1] - B U Department of Mathematics Math 201...

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Unformatted text preview: B U Department of Mathematics Math 201 Matrix Theory Fall 2002 Second Midterm This archive is a property of Boazii University Mathematics Department. The purpose of this archive is to organise and centralise the distribution of the exam questions and their solutions. g c This archive is a non-profit service and it must remain so. Do not let anyone sell and do not buy this archive, or any portion of it. Reproduction or distribution of this archive, or any portion of it, without non-profit purpose may result in severe civil and criminal penalties. 1. (a) Find the equation (do not solve) for the coefficients C, D, E in b = C + Dt + Et2 , the parabola which best fits the four points:(t, b) = (0, 0), (1, 1), (1, 3) and (2, 2). Solution: We have C = 0, C + D + E = 1, C + D + E = 3, C + 2D + 4E = 2. 1 0 0 0 C 1 1 1 1 D , b = . Let A = 1 1 1 , x = 3 E 1 2 4 2 We have the system Ax = b which is inconsistent. We need to look for its least square solution and solve the system: AT A = AT b, x i.e. 4 4 6 4 6 10 6 10 18 C 6 D = 8 E 12 b ) onto the subspace (b) (Fill in the blanks) In solving this problem you are projecting the vector ( spanned by ( the columns of A ) 1 1 2. Let A = 2 -1 -2 4 (a) Find orthonormal vectors e1 , e2 and e3 so that e1 and e2 form a basis for the column space of A. (b) Find the projection matrix P which projects onto the left nullspace of A. Solution: (a) Let 1 a = 2 , -2 1 b = -1 , 4 1 1 a = 2 e1 = a 3 -2 2 b 1 e2 = = 1 . b 3 2 b = b - (eT b)e1 , 1 Then e3 must be in the left nullspace N (AT ): AT y = 1 2 -2 1 -1 4 y1 y2 = y3 0 0 y1 = -2y3 , y2 = 2y3 , and y3 is free. dimN (AT ) = 1 -2 A basis for N (AT ) is : c = 2 .Then 1 -2 1 c 2 . = e3 = c 3 1 Now {e1 , e2 , e3 } is now an ON basis for R3 . (b) We have -2 1 P = e3 (eT e3 )-1 eT = e3 eT = 2 3 3 3 9 1 -2 2 1 4 -4 -2 1 4 2 = -4 9 -2 2 1 3. (a) Let P2 be the vector space of polynomials of degree less than or equal to 2. Suppose L : P2 R3 is the linear transformation defined by p(1) L[p(t)] = p(0) , p(t) P2 . p(-1) Find the matrix representation of L relative to the standard bases of P2 and R3 . Solution: We know that p1 = 1, p2 = t, p3 = t2 is the standard basis of P2 and eT = 1 1 0 0 , eT = 2 0 1 0 , eT = 3 0 0 1 is the standard basis of R3 . Then L[p1 ] = 1 1 1 3 i=1 = e1 +e2 +e3 , L[p2 ] = 1 0 -1 = e1 -e2 , L[p3 ] = 1 0 1 = e1 +e3 . Since L[pj ] = aij ei , where (aij ) is the matrix representation, we have 1 1 1 0 0 . A = (aij ) = 1 1 -1 1 (b) Let P be an n n matrix satisfying P 2 = P and let = 1 be real. Prove that the matrix: I - P is invertible and P (I - P )-1 = I + 1- Solution: Consider the nullspace of I - P : (I - P )x = 0 P x = x. Apply P : P 2 x = P x = P x ( - 1)P x = 0 P x = 0 x = 0. This means that the nullspace is trivial: dimN (I - P ) = 0; I - P is of rank n (I - P )-1 exists. Then (I - P )(I - P )-1 = (I - P )[I + = I + P ( Therefore, (I - P )-1 = I + P. 1- 2 P P ] = I - P - P2 + 1- 1- 1- - 1 -1+ )=I 1- 1- 1 0 1 0 0 1 0 1 . How many of the 24 terms in detA are nonzero? Justify your answer and 4. (a) Let A = 1 0 -1 0 0 -1 0 1 find detA. Solution: There are 4 nonzero terms in detA. These are: + - - + detA = -4. (b) Prove that if B is an n n matrix of rank n, then the adjugate matrix Bcof must also have rank n. Solution: We have BBcof = (detB)I, detB = 0. det(Bcof ) = (detB)n-1 = 0 Bcof has rank n. a11 a11 a13 a13 a22 a24 a22 a24 a33 a33 a31 a31 a44 a42 a44 a42 = = = = -1 -1 -1 -1 ...
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