Sample Final Exam: Solutions Problem. A linear transformation T : R R 4 is given by () x x T = x 4. x + (a) Find the standard matrix A of this transformation; (b) Find a basis and the dimension for Range(T ) =Col(A); (c) Find a basis and the dimension for Ker(T ) =Null(A). (d) Does the vector 5 b = belong to Range(T )? Solution. (a) The standard matrix of T is A = [ T (e ) T (e ) ] = 4. (b) Row reducing A, we obtain 4 4 4 6. Thus every column of A has a pivot, and the pivot columns, and 4, form a basis for Range(T ) =Col(A). Then dim Range(T ) =. (c) Since every column of A has a pivot, the equation Ax = has no free variables, and therefore has only the trivial solution x =. In other words, Ker(T ) =Null(A) = {}, this subspace has no basis, and dim Ker(T ) =.
(d) Writing the augmented matrix A b and reducing it, we obtain A b = 5 4 4 4 5 6 4.6. Since the right-hand side column has a pivot, the system Ax = b is inconsistent, i.e., b does not belong to Range(T ) =Col(A). Problem. (a) Compute the determinant: 4 8 8 8 5 6 8 8 8 7 8 8 8 (b) Find the volume of the tetrahedron ABCD with vertices A(,, ), B(,, ), C(, 5, ), D(4,, ). (Hint: The volume of a parallelepiped whose set of vertices includes A, B, C, D is 6 times more.) (c) Solve the system of equations by Cramer s method: ; { 5x = 7 x + = 8 Solution. (a) 4 8 8 8 5 6 8 8 8 7 8 8 8 4 8 8 5 = ( ) + 6 8 8 7 8 = ( ) 4+ 4 8 5 6 8 7 = ( ) + 4 5 6 7 = 6(8 ) =.
(b) The volume of the tetrahedron ABCD is V = 6 det 4 5 = 6 det 4 = [ 4 4 6 det + det + det ] = 6 + ( 9) + 7 = =. 6 (c) We have 5 det A = det = 8, det A (b) = det Therefore, 7 = 5, det A 8 (b) = det x = det A (b) det A = 5 8, = det A (b) det A = 9 8. 5 7 = 9. 8 Problem. Let A =, B = 5. 5 (a) Find A ; (b) Find B T ; (c) Find a matrix X such that AX = B T. Solution. (a) We will use the formula A = Adj A det A. We calculate det A = det = det +5 det = 5+ = 5, 5 Adj A = C C C 5 C C C = C C C 5 5 5 5 5 = 7 4. 4
4 Then A = 5 5 / / / 7 4 = 7/5 4/5 /5. 5 4 /5 /5 4/5 (b) B T = 5. / / / (c) X = A B T = 7/5 4/5 /5 5 = 5/ / /5. /5 /5 4/5 / /5 Problem 4. (a) Find the eigenvalues and eigenvectors of the matrix A = 4 ; 5 (b) Diagonalize A, i.e., represent it as A = P DP where P is an invertible matrix, and D is a diagonal matrix; (c) Verify that AP = P D. Solution. (a) The characteristic polynomial of A, p(a) = det(ti A) = det t t 4 = (t 5)(t )(t ), t 5 has three zeros, λ = 5, λ =, λ =, which are the eigenvalues of A. The corresponding eigenvectors are the solutions to the equations (A λ i I)x =, i =,,. We have A 5I = 4 4 4 4 5, therefore, the corresponding equations can be written as x + x =, + 5x =. Taking x as a free variable, we obtain that = 5 x, x = x = 5 x x = x, and x = x, or x = x = x / 5/. Putting x =, we obtain the eigenvector x x () = 5
corresponding to the eigenvalue λ = 5. Next, A I = 4 6, therefore, the corresponding equations can be written as x x =, x =. Taking as a free variable, we obtain that x = x = and =, or x = x =. Putting =, we obtain the eigenvector x x () = corresponding to the eigenvalue λ =. Finally, A I = 4, 4 therefore, the corresponding equations can be written as x + + x =, x =. Taking as a free variable, we obtain that x =, =, and x =, or x = x =. Putting =, we obtain the eigenvector x () = x corresponding to the eigenvalue λ =. (b) D = λ λ = 5, P = [ x () x () x ()] = 5. λ 5 (c) AP = 4 5 = 5 5 = 5 5 = P D. 5
6 Problem 5. Given vectors u =, u =, u =, (a) Show that these vectors form a basis for R ; (b) Use the Gram Schmidt process to find an orthogonal basis for R. Solution. (a) The matrix u u u = has three pivots, therefore, the vectors u, u, and u form a basis for R. (b) Set v = u =, v = u u v v = =, v v v = u u v v u v v = = /. v v v v 6 / These vectors v =, v =, v = / / form an orthogonal basis for R. Problem 6. Let A = {a, a, a } and B = {b, b, b } be two bases for a vector space V, and suppose a = 4b b, a = b + b + b, and a = b b. (a) Find the change-of coordinate matrix P B A ; (b) Find [x] B for x = a + 4a + a. Solution. (a) P B A = [a ] B [a ] B [a ] B = 4. (b) [x] B = P B A [x] A = 4 4 = 8.