Solution the first step is to get coordinates for the

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Solution: The first step is to get coordinates for the points vectorx = ( x 1 , x 2 , x 3 ) in this plane. Solve its equation for, say, x 3 , so x 3 = x 1 + 2 x 2 . Since we want Avectorx = 0, we write vectorx as a column vector and then use x 1 and x 2 as coordinates: vectorx = x 1 x 2 x 1 + 2 x 2 = x 1 1 0 1 + x 2 0 1 2 = x 1 vectorv + x 2 vectorw, where vectorv 1 and vectorv 2 are the two obvious column vectors. Since we want Avectorx = 0 for the vectors in the plane, this means 0 = A ( x 1 vectorv + x 2 vectorw ) = x 1 Avectorv + x 2 Avectorw. Because x 1 and x 2 can be anything, we want Avectorv = 0 and Avectorw = 0, that is, 0 = Avectorv = a 11 a 12 a 13 a 21 a 22 a 23 a 31 a 32 a 33 1 0 1 = a 11 + a 13 a 21 + a 23 a 31 + a 33 = A 1 + A 3 , 3
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where A 1 , A 2 , and A 3 are the three columns of A . This gives A 1 = A 3 . Similarly, since we want Avectorw = 0, 0 = Avectorw = a 11 a 12 a 13 a 21 a 22 a 23 a 31 a 32 a 33 0 1 2 = a 12 + 2 a 13 a 22 + 2 a 23 a 32 + 2 a 33 = A 2 + 2 A 3 , so A 2 + 2 A 3 = 0, that is, A 2 = 2 A 3 . To summarize, we find that A = A 3 2 A 3 A 3 , where A 3 can be any column vector, say A 3 = ( a, b, c ), then A = a 2 a a b 2 b b c 2 c c 4
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10. Linear maps F ( X ) = AX , where A is a matrix, have the property that F (0) = A 0 = 0, so they necessarily leave the origin fixed. It is simple to extend this to include a translation, F ( X ) = V + AX, where V is a vector. Note that F (0) = V . Find the vector V and the matrix A that describe each of the following mappings [here the light blue F is mapped to the dark red F ]. a). b). c). d). Solution: a). V = parenleftbigg 4 2 parenrightbigg , A = I b). V = parenleftbigg 4 2 parenrightbigg , A = parenleftbigg 1 0 0 2 parenrightbigg c). V = parenleftbigg 1 2 parenrightbigg , A = parenleftbigg 1 0 0 1 parenrightbigg d). V = parenleftbigg 1 2 parenrightbigg , A = parenleftbigg 1 1 0 1 parenrightbigg . 5
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11. [Bretscher, Sec. 2.4 #35] An n × n matrix A is called upper triangular if all the elements below the main diagonal , a 11 , a 22 , . . . a nn are zero, that is, if i > j then a ij = 0. a) Let A be the upper triangular matrix A = a b c 0 d e 0 0 f . For which values of a, b, c, d, e, f is A invertible? Solution: As always, in thinking about the invertability I think of solving the equations Avectorx = vector y . In this case, the equations are ax 1 + bx 2 + cx 3 = y 1 + dx 2 + ex 3 = y 2 fx 3 = y 3 Clearly, to always be able to solve the last equation for x 3 we need f negationslash = 0. This gives us x 3 , which we use in the second equation. It then can always be solved for x 2
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