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

# Solution gauss jordan method is used 1 1 1 2 3 2 1 3

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Solution. Gauss-Jordan method is used: 1 1 1 - 2 3 2 1 3 2 5 0 - 1 1 6 3 - 2 R 1 + R 2 1 1 1 - 2 3 0 - 1 1 6 - 1 0 - 1 1 6 3 - R 2 1 1 1 - 2 3 0 1 - 1 - 6 1 0 - 1 1 6 3 R 2 + R 3 1 1 1 - 2 3 0 1 - 1 - 6 1 0 0 0 0 4 . The corresponding equivalent system has the third equation 0 × x 1 + 0 × x 2 + 0 × x 3 + 0 × x 4 = 4, which has no solution. 2. Find all values of a for which the resulting linear system has (a) no solution, (b) a unique solution, and (c) infinitely many solutions. x + z = 4 2 x + y + 3 z = 5 - 3 x - 3 y + ( a 2 - 5 a ) z = a - 8 . Solution. Gauss-Jordan method is used: 1 0 1 4 2 1 3 5 - 3 - 3 a 2 - 5 a a - 8 ( - 2) R 1 + R 2 , 3 R 1 + R 3 1 0 1 4 0 1 1 - 3 0 - 3 a 2 - 5 a + 3 a + 4 3 R 2 + R 3

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24 CHAPTER 1. MATRICES 1 0 1 4 0 1 1 - 3 0 0 a 2 - 5 a + 6 a - 5 . As usually we distinguish two cases (notice that a 2 - 5 a + 6 = ( a - 2)( a - 3) = 0 a ∈ { 2 , 3 } ): Case 1 . a = 2 or a = 3. In both cases a + 1 6 = 0 and so the third equation of the corresponding system is 0 × x + 0 × y + 0 × z = a + 1, with no solution. Case 2. If a / ∈ { 2 , 3 } then a 2 - 5 a + 6 6 = 0 and the procedure continues with Step 4: 1 a 2 - 5 a +6 R 3 1 0 1 4 0 1 1 - 3 0 0 1 a - 5 a 2 - 5 a +6 - R 3 + R 2 , 1 1 0 0 4 - a - 5 a 2 - 5 a +6 0 1 0 - 3 - a - 5 a 2 - 5 a +6 0 0 1 a - 5 a 2 - 5 a +6 , with the corresponding equivalent system (unique solution) x = 4 - a - 5 a 2 - 5 a +6 , y = - 3 - a - 5 a 2 - 5 a +6 , z = a - 5 a 2 - 5 a +6 . 3. If possible, find the inverse of the following matrix 1 2 - 1 0 1 1 1 0 - 1 . Solution. We use the Practical Procedure (see p. 95, textbook): 1 2 - 1 1 0 0 0 1 1 0 1 0 1 0 - 1 0 0 1 - R 1 + R 3 1 2 - 1 1 0 0 0 1 1 0 1 0 0 - 2 0 - 1 0 1 2 R 2 + R 3 1 2 - 1 1 0 0 0 1 1 0 1 0 0 0 2 - 1 2 1 1 2 R 3 1 2 - 1 1 0 0 0 1 1 0 1 0 0 0 1 - 1 2 1 1 2 - R 3 + R 2 ,R 3 + R 1
25 1 2 0 1 2 1 1 2 0 1 0 1 2 0 - 1 2 0 0 1 - 1 2 1 1 2 ( - 2) R 2 + R 1 1 0 0 - 1 2 1 3 2 0 1 0 1 2 0 - 1 2 0 0 1 - 1 2 1 1 2 = = C . . . D . Since C has no zero rows, A - 1 exists (that is, A is nonsingular) and A - 1 = D = - 1 2 1 3 2 1 2 0 - 1 2 - 1 2 1 1 2 . 4. If A = - 1 - 2 - 2 2 , find all values of λ for which the homogeneous system ( λI 2 - A ) x = 0 has a nontrivial solution. Solution. The homogeneous system has a nontrivial solution if and only if λI 2 - A = λ + 1 2 2 λ - 2 is singular (see Theorem 1.13, p. 99). Use the practical procedure for finding the inverse: Case 1 . λ + 1 6 = 0 then this is the first pivot in λ + 1 2 1 0 2 λ - 2 0 1 1 λ +1 R 1 1 2 λ +1 1 λ +1 0 2 λ - 2 0 1 - 2 R 1 + R 2 1 2 λ +1 1 λ +1 0 2 λ - 2 - 4 λ +1 - 2 λ +1 1 = 1 2 λ +1 1 λ +1 0 2 λ 2 - λ - 6 λ +1 - 2 λ +1 1 = C . . . D . Now, if λ 2 - λ - 6 = ( λ + 2)( λ - 3) = 0 λ ∈ {- 2 , 3 } then C has a zero row and λI 2 - A is singular (as required). Case 2 . If λ + 1 = 0 then the initial matrix is 0 2 1 0 2 - 3 0 1 so that using Step 3 we obtain 2 - 3 0 1 0 2 1 0 1 - 3 2 0 1 2 0 2 1 0 1 - 3 2 0 1 2 0 1 1 2 0

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26 CHAPTER 1. MATRICES 1 0 3 4 1 2 0 1 1 2 0 Hence the coefficient matrix of the system is a nonsingular matrix with in- verse 3 4 1 2 1 2 0 and therefore the homogeneous system has a unique (trivial) solution.
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