Problems. Universidad San Pablo - CEU. Mathematical Fundaments of Biomedical Engineering 1. Author: First Year Biomedical Engineering
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1 Universidad San Pablo - CEU Mathematical Fundaments of Biomedical Engineering 1 Problems Author: First Year Biomedical Engineering Supervisor: Carlos Oscar S. Sorzano September 15, 013
2 1 Chapter 3 Lay, ( a Let u = and v = b ( c 0, where a, b, c are positive (for simplicity. Compute the area of the parallelogram determined by u, v, u+v and 0, and compute the determinants of the matrices ( u v and ( v u. Draw a picture and explain what you nd. The area of the parallelogram is base times height. In this case: u u+v y v x A = cb The determinant of ( u v is a c b 0 = a 0 bc = bc The determinant of ( v u is c a 0 b = cb a 0 = cb We see that A = abs ( ( u v = abs ( ( v u Lay, Combine the methods of row reduction and cofactor expansion to compute the determinant
3 3 1 4 D = r 3 r 3 + r 1 D = = ( D = = ( 1+1 ( 9 D = 9(3 4 ( 3 ( 4 = 0 Lay, Assume 5g 5h 5i Lay, Assume Lay, Assume = 7. Calculate = 5 = 7. Calculate = = 7. Calculate 5g 5h 5i =. = 5 7 = = 7 d + a e + b f + c. d + a e + b f + c = d e f = = 7 = 14 Lay, Use determinants to decide if the set of vectors 6, 0, and is linearly independent.
4 = 11 The three vectors are linearly independent because their determinant is dierent from 0. Lay, Show that if A is invertible, then det{a 1 } = 1 det{a} If A is invertible, then Taking determinants on both sides AA 1 = I det{aa 1 } = det{i} det{a} det{a 1 } = 1 det{a 1 } = 1 det{a} Lay, 3..3 Find a formula for det{ra} when A is an n n matrix. Consider the column decomposition of A Then A = ( a 1 a... a n ra = ( ra 1 ra... ra n det{ra} = ra 1 ra... ra n = r a 1 ra... ra n = r a 1 a... ra n = r n a 1 a... a n = r n det{a} Lay, Let A and B square matrices. Show that even though AB and BA may not be equal, it is always true that det{ab} = det{ba} By applying properties of the determinants det{ab} = det{ba} det{a} det{b} = det{b} det{a} Lay, Use Cramer's rule to solve the following equation system 5x 1 + 7x = 3 x 1 + 4x = 1 3
5 x 1 = x = = = 5 6 = = 1 6 Lay, Determine the values of the parameter s for which the system below has a unique solution. Applying Cramer's rule x 1 = 5 4 s 6s 4 9 s x = 6s 5 9 6s 4 9 s 6sx 1 + 4x = 5 9x 1 + sx = = 5 s 4 ( 6s s 4 9 = 10s+8 1s 36 = 10(s (s+ 3(s 3 = 5(s (s+ 3(s 3 = 6s ( 5 9 6s s 4 9 = 1s+45 1s 36 s (s+ 3(s 3 = 1(s (s+ 3(s 3 = This equation system has a unique solution if the denominator of the fractions above do not vanish, that is, s ± 3. Lay, Calculate the adjugate of the matrix A = Then, use it to calculate A 1. For calculating the adjugate of the matrix A we need to calculate all its cofactors 4
6 The adjoint is = 0 C 1 = ( = 3 C 13 = ( = 3 C 1 = ( = 1 C = ( = 1 C 3 = ( = C 31 = ( = 0 C 3 = ( = 3 = 6 C 11 = ( C 33 = ( A = For calculating A 1 we need the determinant of A. We use the cofactor expansion along the second row A = a 1 C 1 + a C + a 3 C 3 = 3 1 = 3 Now A 1 = 1 A (A T = = Lay, Find the area of the parallelogram whose vertices are (-1,0, (0,5, (1,-4, (,1. Let us draw the parallelogram: 5
7 6 4 x x 1 Calling x A = ( 1, 0, x B = (0, 5, x C = (1, 4, the sought area is the absolute value of the determinant of the vectors x B x A and x C x A. x B x A = ( 0 5 ( 1 0 = ( 1 5 x C x A = ( 1 4 ( 1 0 = ( 4 abs ( x B x A x C x A = abs ( = 14 Lay, Use the concept of volume to explain why the determinant of a 3 3 matrix is zero i A is not invertible. From the invertible matrix theorem, we know that a matrix is invertible i its columns are linearly independent. So the statement of this problem can be restated as the determinant of a 3 3 matrix is zero i the three columns of A are linearly dependent. On the other side interpreting the determinant of A as the volume of the parallelepiped formed by the three columns, the problem is the volume of the parallelepiped formed by three vectors is zero i the three columns of A are linearly dependent. If the three vectors are linearly dependent, they span a subspace of dimension or 1. In both cases, there is no real parallelepiped but a parallelogram or a segment and the volume of the parallelepiped is 0. Let us show that if the volume of the parallepiped is zero, then three columns are linearly dependent. Let's assume they are linearly independent. Then, they would actually span a three-dimensiional space, and the volume of the parallelepiped formed by the three would not be zero. But this is a contradiction with our hypothesis. So the three vectors have to be linearly dependent. Lay, Let T : R n R m be a linear transformation, and let p be a vector and S a set in R n. Show that the image of p + S under T is the translated set T (p + T (S in R m. Any vector of the set p + S is of the form x = p + s 6
8 where s S. If we apply T to x and exploiting the fact that T is a linear transformation, we get T (x = T (p + s = T (p + T (s The set of all vectors of the form T (s is actually T (S, so we have that, as stated by the problem, T (x T (p + T (S Lay, Let S be the tetrahedron in R 3 with vertices at the vectors 0, e 1, e and e 3 and let S be the tetrahedron with vertices at vectors 0, v 1, v and v 3. See the gure. a. Describe a linear transformation that maps S into S. b. Find a formula for the volume of the tetrahedron S using the fact Volume of S= 1 3 Area of the base Height. a. Consider the matrix A = ( v 1 v v 3 The tetrahedron S is formed by all those points that can be written in the form x = λ λ 1 e 1 + λ e + λ 3 e 3 with λ 0 + λ 1 + λ + λ 3 1 If we consider now Ax, we have 7
9 Ax = A(λ λ 1 e 1 + λ e + λ 3 e 3 = λ 0 A0 + λ 1 Ae 1 + λ Ae + λ 3 Ae 3 = λ λ 1 v 1 + λ v + λ 3 v 3 So this is a point in the tetrahedron S as required by the problem. b. The base of the tetrahedron S is a triangle with vertices 0, e 1 and e, whose area is Area triangular base = 1 Base Height = = 1. The height of the tetrahedron is the length of e 3, that is, 1. Finally Volume of S= 1 3 Area of the base Height = = 1 6 According to Theorem 5. in Chapter 4, the volume of S is Volume of S = det{a} Volume of S= 1 6 det{a} 8
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