Linear Algebra with Applications (3rd Edition)

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ma253 , Fall 2007 — Problem Set 4 Solutions 1. Problems from the textbook: a. Section 2 . 3 1 The inverse is 8 - 3 - 5 2 . *2 The matrix is not invertible. *4 The inverse is 3 2 - 1 1 2 1 2 0 - 1 2 - 3 2 1 1 2 . *12 Hard to do by hand; doing it with the calculator is fine. You should have gotten 5 - 20 - 2 7 0 - 1 0 0 - 2 6 1 2 0 3 0 1 . 29 Row-reducing the matrix gives 1 0 2 - k 0 1 k - 1 0 0 k 2 - 3k + 2 , so that the matrix is invertible if and only if k 2 - 3k + 2 = 0 , i.e., if k = 1 and k = 2 . *30 Row-reducing the matrix gives 1 0 - c 0 1 b 0 0 0 , so this is not invertible for any values of b and c . 31 You have to consider separately the case when a = 0 and the case when a = 0 , because otherwise you can’t row-reduce. If a = 0 , the rref form of the matrix is 1 0 - c a 0 1 b a 0 0 0 , He may look like an idiot and talk like an idiot but don’t let that fool you. He really is an idiot. – Groucho Marx
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so that the matrix is not invertible in this case. On the other hand, when a = 0 , after one row-reduction step we get - b - c 0 0 0 c 0 0 b , so the rref can’t have a pivot in the second column, and again we see that the matrix is not invertible. So the matrix in question is never invertible. *34 The matrix A is invertible if and only if none of a , b , c is zero, and then the inverse is 1 a 0 0 0 1 b 0 0 0 1 c . The general case is the same: an n × n diagonal matrix is invertible if and only if none of the diagonal entries a 11 , a 22 , . . . , a nn is zero, and then the inverse will be a diagonal matrix whose entries are the reciprocals 1/a ii . *43 We basically did this in class: the transformation that sends x to A ( Bx ) is the one that corresponds to the matrix AB . So the problem is asking us to find ( AB ) - 1 (under the assumption that both A and B are invertible). We know that ( AB ) - 1 = B - 1 A - 1 , and that is the answer to the problem, i.e., x = B - 1 ( A - 1 y ) . b. Section 2 . 4 31 The problem has infinitely many solutions, as we can see by row-reducing the appropriate matrix. The solutions are: 2 + t - 1 + s - 1 - 2t 1 - s t s , where t and s can be arbitrary. *32 The equation Bx = 0 unpacks into a homogeneous linear system with three unknowns and two equations. We showed, early in the game, that such systems always have infinitely many solutions, so in particular there must be a non-zero solution x 0 . Now, if AB = I 3 , we would have x 0 = I 3 x 0 = ABx 0 = A ( Bx 0 ) = A0 = 0, which contradicts the fact that x 0 is not zero. Hence, it is impossible for AB to be equal to I 3 . When a true genius appears in the world, you may know him by this sign, that the dunces are all in confederacy against him. – Jonathan Swift
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33 Since an invertible matrix must yield a one-to-one transformation, it can’t send any non-zero vectors to zero. Now repeat the same proof as in 32.
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