Download Investigations in Geometry (Mathematics Motivators) by Alfred S. Posamentier, Gordon Sheridan PDF

By Alfred S. Posamentier, Gordon Sheridan

Math Motivators! a chain of enrichment investigations in secondary arithmetic, is made of 4 books. They contain investigations in pre-algebra, algebra, geometry, and enterprise and customer arithmetic. every one ebook includes nearly 32 black-line masters that may be reproduced for pupil use. The investigations probe a large spectrum of mathematical recommendations and purposes at various degrees.

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Aii 0 . . 0] ·  b  , which is indeed 0 if i < j.  jj   ..   .  bnj  [ai1 59. a. Write the system Ly = b in components: y1 −3y1 y1 −y1 + y2 + 2y2 + 8y2 + y3 − 5y3 + y4 = −3 = 14 , so that y1 = −3, y2 = 14 + 3y1 = 5, = 9 = 33 y3 = 9 − y1 − 2y2 = 2, and y4 = 33 + y1 − 8y2 + 5y3 = 0:   −3  5 y= . 2 0   1  −1  b. Proceeding as in part (a) we find that x =  . 2 0 L(m) 0 U (m) and U = L3 L4 0 so that A(m) = L(m) U (m) , as claimed. U2 L(m) U (m) . Then A = LU = U4 L3 U (m) 61.

C. We are asked to write 3 4 4 −3 determined. This matrix, =k 3 k 4 k 4 k − k3 3 k 4 k 4 k − k3 , with our scaling factor k yet to be has the form of a reflection matrix a b b −a . This form further requires that 1 = a2 + b2 = ( k3 )2 + ( k4 )2 , or k = 5. Thus, the matrix represents a reflection combined with a scaling by a factor of 5. 41. refQ x = −refP x since refQ x, refP x, and x all have the same length, and refQ x and refP x enclose an angle of 2α + 2β = 2(α + β) = π. ) 43. Since y = Ax is obtained from x by a rotation through θ in the counterclockwise direction, x is obtained from y by a rotation through θ in the clockwise direction, that is, a rotation through −θ.

33 . To show the existence and uniqueness of this representation algebraically, choose a nonzero vector w1 in L1 and a nonzero w2 in L2 . Then the system x1 w1 + x2 w2 = 0 or x [w1 w2 ] 1 = 0 has only the solution x1 = x2 = 0 (if x1 w1 + x2 w2 = 0 then x2 x1 w1 = −x2 w2 is both in L1 and in L2 , so that it must be the zero vector). 4). Now set v1 = x1 w1 and v2 = x2 w2 to obtain the desired representation v = v1 + v2 . 3. Let v = v1 + v2 , w = w1 + w2 , so that v + w = (v1 + w1 ) + (v2 + w2 ) and kv = kv1 + kv2 .

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