29_hwc1SolnsODDA

29_hwc1SolnsODDA - dened. SOLUTION | v | = 2 | x | = 2 | y...

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As you can check, this is the same result that one gets from the formula for inverting 2 × 2 matrices. Also in exercise 3.30 , we found that with q ( x ) = x 3 - 5 x 2 + 7 x - 2, q ( B ) = 0. In this case, q (0) = - 2, and q (0) - q ( x ) xq (0) = 1 2 ( x 2 - 5 x + 7) . Therefore. by part (a) , B is invertible, and B - 1 = 1 2 ( B 2 - 5 B + 7 I ) = 1 2 1 2 - 2 0 2 - 2 0 - 2 4 . You can easily verify this result. A look ahead: It turns out that, given any n × n matrix A , there is a simple recipe for writing down a polynomial p ( x ) of degree n such that p ( A ) = 0. This polynomial is called the characteristic polynomial of A , and A is invertible if and only if p (0) 6 = 0. Section 4 4.1 Consider the following vectors: v = h 1 1 i x = h 1 - 1 i y = ± 2 - 1 2 ² z = - 1 - 1 3 2 Compute the length of each of these vectors, and the dot product of each pair of vectors for which it is
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Unformatted text preview: dened. SOLUTION | v | = 2 | x | = 2 | y | = 3 | z | = 15 v x = 0. 4.3 Consider the vectors v 1 = h 1 2 i v 2 = h 2 1 i and v 3 = h-2 1 i . (a) Compute v i v j for each i, j = 1 , 2 , 3. (b) What are the lengths of each of the three vectors? (c) What are the angles between each of the three pairs of vectors? Is any pair orthogonal? (d) Draw a diagram showing each of the three vectors as arrows. Do the lengths and angles that you computed look right? SOLUTION (a) v 1 v 2 = v 2 v 1 = 4 21 /september/ 2005; 22:09 30...
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This note was uploaded on 10/03/2010 for the course MATH 380 taught by Professor Gladue during the Fall '08 term at Roger Williams.

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