By Otto Bretscher, Kyle Burke
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This guide includes thoroughly worked-out suggestions for the entire odd-numbered workouts within the textual content.
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Additional info for Student Solutions Manual for Linear Algebra with Applications
For any m, the matrices L(m) and U (m) are triangular, with nonzero diagonal entries, so that they are invertible. 8, the matrix A(m) = L(m) U (m) is invertible as well. c. Using the hint, we write A = A(n−1) w v k = L x 0 t U 0 y . s We are looking for a column vector y, a row vector x, and scalars t and s satisfying these equations. The following equations need to be satisfied: v = L y, w = xU , and k = xy + ts. We find that y = (L )−1 v, x = w(U )−1 , and ts = k − w(U )−1 (L )−1 v. 57 Chapter 2 SSM: Linear Algebra We can choose, for example, s = 1 and t = k − w(U )−1 (L )−1 v, proving that A does indeed have an LU factorization.
So, . The matrix 1 2 1 2 1 2 1 2 represents an orthogonal 1 1 represents a projection combined with a scaling 1 1 b. This looks similar to a shear, with the one zero off the diagonal. Since the two diagonal 3 0 1 0 , showing that this matrix entries are identical, we can write =3 −1 3 − 31 1 represents a vertical shear combined with a scaling by a factor of 3. c. We are asked to write 3 4 4 −3 determined. This matrix, =k 3 k 4 k 4 k − k3 3 k 4 k 4 k − k3 , with our scaling factor k yet to be has the form of a reflection matrix a b b −a .
For B to be the inverse of A we must have AB = Ip+q . Let us partition B the same way as A: B= B11 B21 B12 , where B11 is p × p and B22 is q × q. B22 Then AB = A11 0 0 A22 B11 B21 B12 B22 = A11 B11 A22 B21 A11 B12 A22 B22 = Ip 0 0 Iq means that A11 B11 = Ip , A22 B22 = Iq , A11 B12 = 0, A22 B21 = 0. −1 This implies that A11 and A22 are invertible, and B11 = A−1 11 , B22 = A22 . This in turn implies that B12 = 0 and B21 = 0. We summarize: A is invertible if (and only if) both A11 and A22 are invertible; in this case A−1 = A−1 11 0 0 .
Student Solutions Manual for Linear Algebra with Applications by Otto Bretscher, Kyle Burke