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chapter7 naswer - Chapter 7 Diagonalization and Quadratic...

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170 Chapter 7 Diagonalization and Quadratic Forms Section 7.1 Exercise Set 7.1 1. (b) Since A is orthogonal, 9 41 2 52 5 2 5 1 3 4 55 31 6 12 5 2 5 0. T AA ⎡⎤ ⎢⎥ == −− ⎣⎦ 3. (a) For the given matrix A , 1010 10 0101 01 . T A is orthogonal with inverse . T A = (b) For the given matrix A , 111 1 222 2 1 2 . T = = A is orthogonal with inverse 11 22 . T A = Note that the given matrix is the standard matrix for a rotation of 45 ° . (c) 2 1 1 3 1 2 2 ⎛⎞ =+ + = ⎜⎟ ⎝⎠ r so the matrix is not orthogonal. (d) For the given matrix A , 1 26 3 12 1 2 1 66 6 6 3 1 33 326 3 0 0 100 010 001 T = = A is orthogonal with inverse 1 6 1 3 0 . T A = (e) For the given matrix A , 1111 2222 1 1 2666 1 1 1000 0100 0010 0001 T ⎤⎡ ⎥⎢ = ⎦⎣ = A is orthogonal with inverse 5 1 5 5 1 . T A = (f) 2 2 2 1 7 1 1 12 2 3 ⎛⎞⎛⎞ = r The matrix is not orthogonal. 7. (a) 1 32 cos , π = 3 sin = 3 1 3 1 21 3 3 6 x y + ⎡ ⎤ ⎢ ⎥ ⎣ ⎦ + Thus, () 1333 3 (, ) . , xy ′′ = −+ +
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SSM: Elementary Linear Algebra Section 7.1 171 (b) Since a rotation matrix is orthogonal, Equation (2) gives 5 3 1 2 22 53 3 1 2 3 5 2 1 . x y ⎡⎤ ⎢⎥ == ⎣⎦ + Thus, 55 33 1 (, ) . , xy ⎛⎞ = −+ ⎜⎟ ⎝⎠ 9. If 123 {, , } B = uu u is the standard basis for 3 , R and B ′′ = uuu is the rotated basis, then the transition matrix from B to B is 3 1 3 1 0 0 01 0 0 1 0 0 0 cos sin sin cos P ππ (a) 1 3 1 3 1 5 1 5 1 0 1 01 0 2 5 0 3 2 3 xx P yy zz = = −− = Thus 15 1 5 32 3 2 2 (, , ) . ,, xyz ′′′ =−− + (b) 3 1 3 1 3 1 3 1 0 1 0 6 3 0 3 6 3 P ⎡ ⎤ ⎢ ⎥ ⎣ ⎦ = Thus 13 1 3 36 3 2 2 . = 11. (a) If B = is the standard basis for 3 R and , B = then 1 0 cos [] , sin B θ = u 2 0 1 0 , B = u and 3 0 sin . cos B = u Thus the transition matrix from B to B is 0 0 0 cos sin , sin cos P θθ = i.e., . P ⎤⎡ ⎥⎢ = ⎦⎣ Then 1 0 0 0 cos sin . sin cos AP (b) With the same notation, 1 1 0 0 , B ⎡ ⎤ ⎢ ⎥ = ⎢ ⎥ ⎣ ⎦ u 2 0 , cos sin B = u and 3 0 , sin cos B = u so the transition matrix from B to B is 10 0 0 0 cos sin sin cos P = and 0 0 0 . cos sin sin cos A = 13. Let . abba A +− = Then 20 02 () , T ab AA + = + so a and b must satisfy 1 2 . += 17. The row vectors of an orthogonal matrix are an orthonormal set. 2 1 11 1 aa =+ + r Thus a = 0. 2 2 2 6 66 bb + r Thus 2 6 . b 2 3 2 3 cc ⎛⎞⎛⎞ + ⎜⎟⎜⎟ ⎝⎠⎝⎠ r
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Chapter 7: Diagonalization and Quadratic Forms SSM: Elementary Linear Algebra 172 Thus 1 3 . c The column vectors of an orthogonal matrix are an orthonormal set, from which it is clear that b and c must have opposite signs. Thus the only possibilities are a = 0, 2 6 , b = 1 3 c =− or a = 0, 2 2
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This note was uploaded on 01/11/2012 for the course MATH 100 taught by Professor Loveys during the Fall '11 term at McGill.

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chapter7 naswer - Chapter 7 Diagonalization and Quadratic...

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