Solutions to Final Exam 2011 (Total: 100 pts)
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1 Page of 5 Introduction to Linear Algebra November 7, Solutions to Final Exam (Total: pts). Let T : R 3 R 3 be a linear transformation defined by: (5 pts) T (x, x, x 3 ) = (x + 3x + x 3, x x x 3, x + 3x + 5x 3 ). Let A = [v, v, v 3 ] be the standard matrix for the linear transformation T. (a) Find A. Show work! A satisfies the following: x x + 3x + x 3 A x x 3 = x x x 3 x + 3x + 5x 3 For e = v = [v, v, v 3 ] Thus = Ae = A = [v, v, v 3 ] =, v = Ae =, e = , e 3 =, v 3 = Ae 3 =, 5 (b) Find v v. Show work! e e e 3 v v = 3 3 = [ 3, 3 3, 3 T ] = 9 5 (c) Let u, u, u 3 R 3. Suppose the volume of the parallelepiped determined by u, u, u 3 is. What is the volume of the parallelepiped determined by T (u ), T (u ), T (u 3 ). Write a brief explanation. Since the volume of the parallelepiped determined by T (u ), T (u ), T (u 3 ) is det(a) times the volume of the parallelepiped determined by u, u, u 3, it is det(a) = = = 4 = 8.. Let A and B be the following 4 4 matrices, and b R 4 given below. (3 pts) a b b b b b b b A = b a b b b b a b, B = b b b b b b b b, b = b b b a b b b b
2 Page of 5 (a) Find the determinant of A. Show work! First apply [, ; ], [, 3; ], [, 4; ] to columns, eg. add one times the second column to the first, etc,. we have a b b b a + 3b b b b b b b b a b b b b a b = a + 3b a b b a + 3b b a b = (a + 3b) a b b b a b b b b a a + 3b b b a b b a b b b = (a + 3b) a b a b = (a + 3b)(a b) 3. a b (b) Find the condition that the set of columns of A is linearly dependent. Write a brief explanation. The set of columns of A is linearly dependent if and only if Ax = has a nontrivial solution. The latter condition is equivalent to A being singular (not invertible). Thus the condition is equivalent to det(a) =. Therefore a + 3b = or a = b. (c) Find the eigenvalues of A, and their multiplicities. Let f(x) be the characteristic polynomial of A. Then a x b b b f(x) = det(a xi) = b a x b b b b a x b = ((a x)+3b)(a x b) 3 b b b a x by replacing a by a x in the previous problem. Hence f(x) = (x (a + 3b))(x (a b)) 3, and a + 3b is an eigenvalue with multiplicity and a b with multiplicity 3 unless a+3b = a b, in which case A has only one eigenvalue a with multiplicity 4 and b =. (d) Suppose b. Find a linearly independent set of vectors {v, v, v 3 } such that Bv = Bv = Bv 3 =. Show that it is actually linearly independent. By solving the equation Bx = by augmented matrix, we have b b b b x b b b b b b b b, x x 3 = s +t +u b b b b x 4 Let v = [,,, ] T, v = [,,, ] T and v 3 = [,,, ] T. If sv + tv + uv =, then [ s t u, s, t, u] = [,,, ]. In this case we must have s = t = u =. Therefore the set of vectors {v, v, v 3 } is linearly independent. (e) Let T = [b, v, v, v 3 ]. (i) Show that T is invertible, and (ii) there is a diagonal matrix D such that T AT = D.
3 Page 3 of 5 T = In our case above, Hence T is invertible., and det(t ) = 4 = 4. AT = A[b, v, v, v 3 ] = [(a + 3b)b, (a b)v, (a b)v, (a b)v 3 ] = T D, where D is a diagonal matrix with a + 3b, a b, a b, a b on the diagonal. Since T is invertible, T AT = D. Remarks. First note that Bb = (4b) b. So BT = B[b, v, v, v 3 ] = [Bb, Bv, Bv, Bv 3 ] = [4bb,,, ] = [b, v, v, v 3 ]diag(4b,,, ) = T diag(4b,,, ), where diag(4b,,, ) is a diagonal matrix with 4b,,, on its diagonal. Thus B is diagonalized by T. Note that 4b, b cannot be written as a linear combination of {v, v, v 3 }. Thus the columns of T are linearly independent and T is invertible. Since A = (a b)i + B, we are done. Check the following. T AT = T ((a b)i+b)t = (a b)i+t BT = (a b)i+diag(4b,,, ) = D. 3. Let A, x and b be a matrix and vectors given below. Assume Ax = b. (35 pts) 3 x A = 5 3, x = x x 3, b = 3 3 x 4 4 (a) Evaluate det(a). Briefly explain each step. The first 4 steps are exactly the same as in (d). Then [3,, 5], [4, ; 7] and we have the determinant by expanding along the first and the second column. The rest are easy A = = (b) Applying the Cramer s rule and express x and x 3 as quotients of determinants. Do not evaluate determinants. x = det(a ) det(a), x 3 = det(a 3) det(a) with A = , A 3 =
4 Page 4 of 5 (c) Let B = adj(a), the adjugate of A. Determine the (, 3)-entry of B. Do not evaluate the determinant involved. B,3 = ( ) Let B be the augmented matrix of Ax = b. Let C be a matrix obtained from B after applying a series of elementary row operations. 3 5 B = 5 3 3, C = (d) Write a sequence of operations applied to B to obtain C using [i; c], [i, j], [i, j; c] notation. [, ] [, ; 3] [3, ; ] [4, ; ] [, 4] [; ]. (e) Find a 4 4 matrix P such that P B = C. P = E(; )E(, 4)E(4, ; )E(3, ; )E(, ; 3)E(, ) =. 3 (f) Express P as a product of elementary matrices using the notation E(i; c), E(i, j), E(i, j; c). P = (E(; )E(, 4)E(4, ; )E(3, ; )E(, ; 3)E(, )) = E(, )E(, ; 3)E(3, ; )E(4, ; )E(, 4)E(; ). (g) Explain that P in (e) is uniquely determined. Let C be the 4 4 matrix consisting of the first 4 columns of C. Since the first four columns of B forms A, we have P A = C. Since det(a) in (a), A is invertible. Thus P = C A and P is uniquely determined. 4. Let e, e, e 3, v, v and v 3 R 3 be as follows. ( pts) e =, e =, e 3 =, v =, v =, v 3 = (a) Find the reduced row echelon form of the following matrix. (Show work.) 3 4 8
5 Page 5 of (b) Using (a), explain that {v, v, v 3 } is linearly independent. Let B = [v, v, v 3 ]. Then {v, v, v 3 } is linearly independent if and only if B is invertible. B is invertible if and only if its reduced echelon form is I. This is the case as we have seen above. Hence {v, v, v 3 } is linearly independent. (c) Find the standard matrix A of a linear transformation T : R 3 R 3 such that T (v ) = e, T (v ) = e and T (v 3 ) = e 3. As in the previous problem let B = [v, v, v 3 ]. Then by out assumption, we have AB = A[v, v, v 3 ] = [Av, Av, Av 3 ] = [e, e, e 3 ] = I. Thus B is invertible and A = B and by (a) B = 4 6
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