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

1 0 chapter test 5 3 find parametric equations of the

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1 = 0 . Chapter Test 5. 3. Find parametric equations of the line through the point (5 , - 2 , 1) that is parallel to the vector u = (3 , - 2 , 5). Solution. x = 5+3 t , y = - 2 - 2 t , z = 1+5 t , -∞ < t < (see p.266). 4. Find an equation of the plane passing through the points (1 , 2 , - 1), (3 , 4 , 5) , (0 , 1 , 1). Solution. x y z 1 1 2 - 1 1 3 4 5 1 0 1 1 1 = 0 (see the previous T. Exercise 5, p. 271). Thus x 2 - 1 1 4 5 1 1 1 1 - y 1 3 0 - 1 1 5 1 1 1 + z 1 2 3 4 0 1 1 1 1 - 1 2 - 1 3 4 5 0 1 1 = 0 or x - y + 1 = 0.
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44 CHAPTER 5. LINES AND PLANES 5. Answer each of the following as true or false. (b) If u × v = 0 and u × w = 0 then u × ( v + w ) = 0. (c) If v = - 3 u , then u × v = 0 . (d) The point (2 , 3 , 4) lies in the plane 2 x - 3 y + z = 5. (e) The planes 2 x - 3 y + 3 z = 2 and 2 x + y - z = 4 are perpendicular. Solution. (b) True, using Theorem 5.1 (p.260) properties (c) and (a): u × ( v + w ) = u × v + u × w = 0 + 0 = 0 . (c) True, using Theoretical Exercise T.5 (p.263), or directly: if u = ( u 1 , u 2 , u 3 ) then for v = ( - 3 u 1 , - 3 u 2 , - 3 u 3 ) the cross product i j k u 1 u 2 u 3 - 3 u 1 - 3 u 2 - 3 u 3 = 0 because (factoring out - 3) it has two equal rows. (d) False. Verification: 2 × 2 - 3 × 3 + 1 × 4 = - 1 6 = 0. (e) False. The planes are perpendicular if and only if the corresponding normal vectors are perpendicular, or, if and only if these vectors have the dot product zero. (2 , - 3 , 3) (2 , 1 , - 1) = 4 - 3 - 3 = - 2 6 = 0.
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Chapter 6 Vector Spaces Page 302. T2. Let S 1 and S 2 be finite subsets of R n and let S 1 be a subset of S 2 . Show that: (a) If S 1 is linearly dependent, so is S 2 . (b) If S 2 is linearly independent, so is S 1 . Solution. Since S 1 is a subset of S 2 , denote the vectors in the finite sets as follows: S 1 = { v 1 , v 2 , ..., v k } and S 2 = { v 1 , v 2 , ..., v k , v k +1 , ..., v m } . (a) If S 1 is linearly dependent, there are scalars c 1 , c 2 , ..., c k not all zero, such that c 1 v 1 + c 2 v 2 + ... + c k v k = 0 . Hence c 1 v 1 + c 2 v 2 + ... + c k v k + 0 v k +1 + ... +0 v m = 0 , where not all the scalars are zero, and so S 2 is linearly dependent too. (b) If S 1 is not linearly independent, it is (by Definition) linearly de- pendent. By (a), S 2 is also linearly dependent, a contradiction. Hence S 1 is linearly independent. T4. Suppose that S = { v 1 , v 2 , v 3 } is a linearly independent set of vectors in R n . Show that T = { w 1 , w 2 , w 3 } is also linearly independent, where w 1 = v 1 + v 2 + v 3 , w 2 = v 2 + v 3 and w 3 = v 3 . Solution. Take c 1 w 1 + c 2 w 2 + c 3 w 3 = 0 , that is, c 1 ( v 1 + v 2 + v 3 ) + 45
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46 CHAPTER 6. VECTOR SPACES c 2 ( v 2 + v 3 ) + c 3 v 3 = 0 . Hence c 1 v 1 + ( c 1 + c 2 ) v 2 +( c 1 + c 2 + c 3 ) v 3 = 0 and by linearly independence of S , c 1 = c 1 + c 2 = c 1 + c 2 + c 3 = 0. But this homogeneous system has obviously only the zero solution. Thus T is also linearly independent. T6. Suppose that S = { v 1 , v 2 , v 3 } is a linearly independent set of vectors in R n . Is T = { w 1 , w 2 , w 3 } , where w 1 = v 1 , w 1 = v 1 + v 2 and w 1 = v 1 + v 2 + v 3 , linearly dependent or linearly independent ? Justify your answer. Solution. T is linearly independent. We prove this in a similar way to the previous Exercise T4. Page 316. 28 (Bonus). Find all values of a for which { ( a 2 , 0 , 1) , (0 , a, 2) , (1 , 0 , 1) } is a basis for R 3 .
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