College of Engineering and Computer ScienceMechanical Engineering DepartmentMechanical Engineering 501ASeminar in Engineering AnalysisFall 2004 Number: 17472 Instructor: Larry CarettoSeptember 7 Homework SolutionsPage 364, problem 3 – Are all ordered polynomials in x, of degree not exceeding three, a vector space or not? If they are find the dimension and basis.Any possible polynomial of degree three or less can be expressed by the equation p =a0 + a1x + a2x2 + a3x3. If we define a basis for these polynomials as the four functions f0 = 1, f1 = x, f2 = x2, and f3 = x3, we see that there four functions allow us to express any polynomial as the weighted sum of these four functions. (I.e., we can write p =a0f0 + a1f1 + a2f2 + a3f3.) Thus, these four functions form the basis for a vector space and consequently, the dimension of the space, which is the number of basis functions, is four.This vector space satisfies all the requirements of commutativity and associativity for vector addition and distributivity and associativity for scalar multiplication. The scalar zero, 0, is the null element and one is the unit element.Page 365, problem 19 – Show that the vectors [2 0 3 0 8]T and [3 2 –2 4 0]T are orthogonal.Two vectors are orthogonal if their inner product is zero. (The inner product is the sum of the products of each component.) For the two vectors shown, the inner product is (2)(3) + (0)(2) +(3)(-2) + (0)(4) + (8)(0) = 6 + 0 – 6 + 0 + 0 = 0. Thus, the two vectors are orthogonal.Page 365, problem 21 – Find all vectors, v, orthogonal to a = [1 2 0]T. Do they form a vector space?A vector, v, with components [v1 v2 v3]T will have an inner product with a that is equal to 1v1 + 2v2 + 0v3 = v1 + 2v2. All vectors for which this inner product is zero will be orthogonal to a. Since v3 isnot part of this formula, v3 can have any value, say x. If v2 = y, then v1 = -2y will give a zero inner product. Thus any vector of the form [-2y y x] will be orthogonal to a.These vectors satisfy all the requirements of commutativity and associativity for vector addition and distributivity and associativity for scalar multiplication. The scalar zero, 0, is the null element and one is the unit element. They are not a basis for an arbitrary three-dimensional vector space, but they do form a vector space for this limited case.Page 330, problem 15 – Solve the following set of equations using Gauss elimination or indicate if no solution exists. 10x + 4y - 2z = -4-3w - 17x + y + 2z = 2 w + x + y = 6 8w - 34x + 16y - 10z = 4We first rearrange the equations as shown below to simplify the elimination process. w + x + y = 6 [I] 10x + 4y - 2z = -4 [II]-3w - 17x + y + 2z = 2 [III] 8w - 34x + 16y - 10z = 4 [IV]Engineering Building Room 1333 Mail Code Phone: 818.677.6448E-mail: [email protected] 8348 Fax: 818.677.7062Next we replace equation [III] by [III] + 3[1] and equation [IV] by [IV] – 8[I] to eliminate the w coefficients in all equations below the first.w + x + y = 6 [I] 10x + 4y - 2z = -4 [II] [-17x +3]x + [ 1 + 3]y + 2z = [2 + 3(6)] [III] [-34 – 8]x + [16 – 8]y - 10z = [4 – 8(6)] [IV]Carrying out the indicated multiplications and subtractions gives. W + x + y = 6 [I] 10x + 4y - 2z = -4 [II] -14x + 4y + 2z = 20 [III] -42x + 8y – 10z = -44 [IV]Now we add (14/10) times equation [II] to equation three and (42/10) times equation [II] to equation [IV] to eliminate the x coefficient in the equations [III] and [IV].][10)6)(42(4410)2)(42(1010)4)(42(810)10)(42(42][10)6)(14(2010)2)(14(210)4)(14(410)10)(14(14I VzyxI IIzyxCarrying out the indicated arithmetic in these two equations, the four-equation set becomes W + x + y = 6 [I] 10x + 4y - 2z = -4 [II] + 9.6y – 0.8z = 14.4 [III] + 24.8y – 18.4z = -60.8 [IV]Finally we subtract 24.8/9.6 times equation [III] from equation [IV] to complete the upper triangularform.][6.9)4.14)(8.24(8.606.9)8.0)(8.24(4.186.9)6.9)(8.24(8.24 IVzyThe upper triangular form of the equations obtained after completing the indicated arithmetic is shown below. W + x + y = 6 [I] 10x + 4y - 2z = -4 [II] + 9.6y – 0.8z = 14.4 [III] – (49/3)z = -98 [IV]We now complete the solution by back substitution. First we solve equation [IV] for z = -98/(-49/3) or z = 6. Next, we use this value of z in equation [III] to solve for y.26.9)6(8.04.146.98.04.14zyNow we use y = 2 and z = 6 to solve equation [II] for x.010)2(4)6(2410424yzxFinally, with x, y, and z known, we can solve equation [1] for w.4102616xywEngineering Building Room 1333 Mail Code Phone: 818.677.6448E-mail: [email protected] 8348 Fax: 818.677.7062Page 336, problem 5 – Are the vectors [2 –4], [1 9], and [3 5] linearly dependent or independent?Since these are two-dimensional vectors, they must be linearly dependent. (The maximum number of linearly independent vectors equals the dimension of the space.) In this case, it is easy to see that [3 5] = [2 –4] + [1 9].Page 336, problem 9 – Find the rank of the matrix at the right.We see that this matrix has two columns, which is less than the number of rows (three). Thus, its rank can be at most two, the number of columns. Are the columns linearly independent?361248By inspection we can see that the first column is –2 times the second column. Thus, this matrix has only one linearly independent column so its rank is one. Notice that the first row is –4 times the middle row and the last row is –3 times the middle row. Thus, the matrix has only one linearly independent row. We expect this since either the number of linearly independent rows or the number of linearly independent columns determines the rank of a matrix. Both are the same.Page 337, problem 15 – Find the rank of the matrix shown at the right.We see that this matrix has three columns, which is less than the number of rows