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

T28 show that if a is an n n matrix then a can be

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T.28. Show that if A is an n × n matrix, then A can be written uniquely as A = S + K , where S is symmetric and K is skew-symmetric. Solution. Suppose such a decomposition exists. Then A T = S T + K T = S - K so that A + A T = 2 S and A - A T = 2 K . Now take S = 1 2 ( A + A T ) and K = 1 2 ( A - A T ). One verifies A = S + K , S = S T and K T = - K similarly to the previous Exercise 26. T.32. Show that if A x = b is a linear system that has more than one solution, then it has infinitely many solutions. Solution. Suppose u 1 6 = u 2 are two different solutions of the given linear system. For an arbitrary real number r , such that 0 < r < 1, consider w r = r u 1 + (1 - r ) u 2 . This is also a solution of the given system: A w r = A ( r u 1 + (1 - r ) u 2 ) = r ( A u 1 ) + (1 - r )( A u 2 ) = r b + (1 - r ) b = b . First observe that w r / ∈ { u 1 , u 2 } . Indeed, w r = u 1 implies u 1 = r u 1 + (1 - r ) u 2 , (1 - r )( u 1 - u 2 ) = 0 and hence u 1 = u 2 , a contradiction. Similarly, w r 6 = u 2 . Next, observe that for 0 < r, s < 1, r 6 = s the corresponding solutions are different: indeed, w r = w s implies r u 1 + (1 - r ) u 2 = s u 1 + (1 - s ) u 2 and so ( r - s )( u 1 - u 2 ) = 0 , a contradiction. Page 89. T.11. Let u and v be solutions to the homogeneous linear system A x = 0 .
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18 CHAPTER 1. MATRICES (a) Show that u + v is a solution. (b) For any scalar r , show that r u is a solution. (c) Show that u - v is a solution. (d) For any scalars r and s , show that r u + s v is a solution. Remark. We have interchanged ( b ) and ( c ) from the book, on purpose. Solution. We use the properties of matrix operations. (a) A ( u + v ) Th . 1 . 2(b) = A u + A v = 0 + 0 = 0 . (b) A ( r u ) Th . 1 . 3(d) = r ( A u ) = r 0 = 0 . (c) We can use our previous (b) and (a): by (b), for r = - 1 we have ( - 1) v = - v is a solution; by (a) u +( - v ) = u - v is a solution. (d) Using twice our (b), r u and s v are solutions and, by (a), r u + s v is a solution. T.12. Show that if u and v are solutions to the linear system A x = b , then u - v is a solution to the associated homogeneous system A x = 0 . Solution. Our hypothesis assures that A u = b and A v = b . Hence A ( u - v ) Th . 1 . 2(b) = A u - A v = b - b = 0 and so u - v is a solution to the associated homogeneous system A x = 0 . Page 86 (Bonus). Find all values of a for which the resulting linear system has (a) no solution, (b) a unique solution, and (c) infinitely many solutions. 23. x + y - z = 2 x + 2 y + z = 3 x + y + ( a 2 - 5) z = a . Solution. We use Gauss-Jordan method. The augmented matrix is 1 1 - 1 2 1 2 1 3 1 1 a 2 - 5 a . The first elementary operations give 1 1 - 1 2 1 2 1 3 1 1 a 2 - 5 a - R 1 + R 2 , 3 1 1 - 1 2 0 1 2 1 0 0 a 2 - 4 a - 2 .
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19 Case 1 . a 2 - 4 = 0, that is a ∈ {± 2 } . (i) a = 2. Then the last matrix gives (final Step after Step 8; in what follows we refer to the steps in the Procedure p. 65 - 68) the following reduced row echelon form 1 1 - 1 2 0 1 2 1 0 0 0 0 - R 2 + R 1 1 0 - 3 1 0 1 2 1 0 0 0 0 . The corresponding system is x = 1 + 3 z y = 1 - 2 z , which has ( z is an arbi- trary real number), infinitely many solutions. (ii) a = - 2. The last matrix is 1 1 - 1 2 0 1 2 1 0 0 0 - 4 and so our last equation is 0 × x + 0 × y + 0 × z = - 4. Hence this is an inconsistent system (it has no solutions).
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