Solucionario-de-algebra-lineal-Kolman-octava-edicion.pdf

B false for u v as above take w then e 1 e 2 e 1 but

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(b) False: for u , v as above: take w = 0 ; then e 1 e 2 = e 1 0 but e 2 6 = 0 . (c) True: indeed, if c u = 0 and c 6 = 0 then ( cu 1 , cu 2 , ..., cu n ) = (0 , 0 , ..., 0) or cu 1 = cu 2 = ... = cu n = 0, together with c 6 = 0, imply u 1 = u 2 = ... = u n = 0, that is, u = 0 . (d) False: compare with the correct formula on p. 246, Theoretical Exercise T9; for instance, if c = - 1, and u 6 = 0 then k- u k 6 = - k u k .
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40 CHAPTER 4. N-VECTORS (e) False: compare with the correct Theorem 4.5 (p.238); if u 6 = 0 and v = - u then k u + v k = k 0 k = 0 6 = 2 k u k = k u k + k- u k = k u k + k v k . (g) False: the dot product u v = - 1 + 0 + 0 6 = 0, so that the vectors are not orthogonal (see Definition p.238). (h) True: indeed, k u k = p u 2 1 + u 2 2 + ... + u 2 n = 0 implies u 2 1 + u 2 2 + ... + u 2 n = 0 and (for real numbers) u 1 = u 2 = ... = u n = 0, that is, u = 0 . (i) True: indeed, compute (once again by the definition p. 238) u (2 v + 3 w ) T7 , p . 246 = 2( u v ) + 3( u w ) = 2 × 0 + 3 × 0 = 0.
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Chapter 5 Lines and Planes Page 263. T2. Show that ( u × v ) w = u ( v × w ). Solution. Indeed, ( u × v ) w = ( u 2 v 3 - u 3 v 2 , u 3 v 1 - u 1 v 3 , u 1 v 2 - u 2 v 1 ) ( w 1 , w 2 , w 3 ) = = ( u 2 v 3 - u 3 v 2 ) w 1 + ( u 3 v 1 - u 1 v 3 ) w 2 + ( u 1 v 2 - u 2 v 1 ) w 3 = = u 1 ( v 2 w 3 - v 3 w 2 ) + u 2 ( v 3 w 1 - v 1 w 3 ) + u 3 ( v 1 w 2 - v 2 w 1 ) = = ( u 1 , u 2 , u 3 ) ( v 2 w 3 - v 3 w 2 , v 3 w 1 - v 1 w 3 , v 1 w 2 - v 2 w 1 ) = = u ( v × w ). T4. Show that ( u × v ) w = u 1 u 2 u 3 v 1 v 2 v 3 w 1 w 2 w 3 . Solution. In the previous Exercise we already obtained ( u × v ) w = u 1 ( v 2 w 3 - v 3 w 2 ) + u 2 ( v 3 w 1 - v 1 w 3 ) + u 3 ( v 1 w 2 - v 2 w 1 ). But this computation can be continued as follows: = u 1 ( v 2 w 3 - v 3 w 2 ) + u 2 ( v 3 w 1 - v 1 w 3 ) + u 3 ( v 1 w 2 - v 2 w 1 ) = = u 1 v 2 v 3 w 2 w 3 - u 2 v 1 v 3 w 1 w 3 + u 3 v 1 v 2 w 1 w 2 = u 1 u 2 u 3 v 1 v 2 v 3 w 1 w 2 w 3 , 41
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42 CHAPTER 5. LINES AND PLANES using the expansion of the determinant along the first row. T5. Show that u and v are parallel if and only if u × v = 0 . Solution. In Section 4.2 (p. 238), the following definition is given: two nonzero vectors u and v are parallel if | u v | = k u k k v k (that is, cos θ = ± 1, or, equivalently sin θ = 0, θ denoting the angle of u and v ). Notice that k u k 6 = 0 6 = k v k for nonzero vectors and u × v = 0 k u × v k = 0. Using the length formula k u × v k = k u k k v k sin θ we obtain sin θ = 0 if and only if u × v = 0 , the required result. Page 271. T5. Show that an equation of the plane through the noncollinear points P 1 ( x 1 , y 1 , z 1 ) , P 1 ( x 2 , y 2 , z 2 ) and P 1 ( x 3 , y 3 , z 3 ) is x y z 1 x 1 y 1 z 1 1 x 2 y 2 z 2 1 x 3 y 3 z 3 1 = 0 . Solution. Any three noncollinear points P 1 ( x 1 , y 1 , z 1 ) , P 1 ( x 2 , y 2 , z 2 ) and P 1 ( x 3 , y 3 , z 3 ) determine a plane whose equation has the form ax + by + cz + d = 0 , where a , b , c and d are real numbers, and a , b , c are not all zero. Since P 1 ( x 1 , y 1 , z 1 ), P 1 ( x 2 , y 2 , z 2 ) and P 1 ( x 3 , y 3 , z 3 ) lie on the plane, their coordi- nates satisfy the previous Equation: ax 1 + by 1 + cz 1 + d = 0 ax 2 + by 2 + cz 2 + d = 0 ax 3 + by 3 + cz 3 + d = 0 .
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43 We can write these relations as a homogeneous linear system in the unknowns a , b , c and d ax + by + cz + d = 0 ax 1 + by 1 + cz 1 + d = 0 ax 2 + by 2 + cz 2 + d = 0 ax 3 + by 3 + cz 3 + d = 0 which must have a nontrivial solution. This happens if and only if (see Corollary 3.4 p. 203) the determinant of the coefficient matrix is zero, that is, if and only if x y z 1 x 1 y 1 z 1 1 x 2 y 2 z 2 1 x 3 y 3 z 3
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