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1 QUEEN S UNIVERSITY FACULTY OF ENGINEERING AND APPLIED SCIENCE DEPARTMENT OF MATHEMATICS AND STATISTICS APSC 174 FINAL EXAMINATION - APRIL 213 INSTRUCTORS: ALAJAJI, MOLAHAJLOO, WEATHERBY INSTRUCTIONS This eamination is 3 hours in length and consists of 6 questions. READ THE QUESTIONS CAREFULLY! Answer all questions, writing clearly in the space provided. If you need more room, continue to answer on the back of the previous page or on pages 8, 9, and 1, if etra space is needed, providing clear directions to the marker. SHOW ALL YOUR WORK, clearly and in order, if you wish to receive full credit. No tetbook, lecture note, calculator, computer, or other aid of any sort is allowed. PLEASE NOTE: Proctors are unable to respond to queries about the interpretation of eam questions. Do your best to answer the eam questions as written. Good luck! FOR MARKER S USE ONLY Question Mark Available Received TOTAL 1 This material is copyrighted and is for the sole use of students registered in APSC174 and writing this eamination. This material shall not be distributed or disseminated. Failure to abide by these conditions is a breach of copyright and may also constitute a breach of academic integrity under the University Senate s Academic Integrity Policy Statement. 1

2 1. Consider the system of linear equations given by: = 3, = 1, = 3, = 2, where we wish to solve for the quadruple ( 1, 2, 3, 4 ) of real numbers. (a) Write the augmented matri for this system. (6 pts) (b) Transform the augmented matri to row-echelon form using a sequence of elementary row operations (clearly indicate which operation you perform at each step). (8 pts) (c) Solve the system resulting from (b) using back-substitution, and use it to determine the set of all solutions to the original system of linear equations. (6 pts) (a) The augmented matri for this system is (b) R1 R2 R2 : R1 R4 R4 : R2 R3 R3 :

3 R3 2R4 R4 : The augmented matri is now in row-echelon form. (c) The system of linear equations corresponding to the augmented matri in row-echelon form obtained from the last step of (b) is: = = = = 3. Solving for 4 in the last equation yields 4 = 1. Backsubstituting this value into the third and second equations yields 3 = 1 2 ( 4 + 1) = 1 2 ( 1 + 1) = and 2 = 2 4 = 2 ( 1) = 3, respectively. Finally, backsubstituting the values for 2 and 3 in the first equation yields 1 = = 3 3+ =. Hence, the original system of linear equations has a unique solution given by: ( 1, 2, 3, 4 ) = (, 3,, 1). 3

4 2. Let A,B M 4 (R) be the 4 4 real matrices defined by: A = 1, B = (a) Let C and D be the matri products Compute C and D. (6 pts) C = AB and D = BA. (b) Determine the invertibility of A by eamining the column vectors of A. (7 pts) (c) Determine the invertibility of B by computing the determinant of B. (7 pts). (a) We have C = AB = (1 + + ( 6) + 2) ( ) ( ) ( ( 2) + ) ( ) (5 + ( 4) + + ) (3 + ( 2) + + ) (2 + ( 2) ) = ( + + ( 3) + ) ( ) ( ) ( + + ( 1) + ) ( ) ( ) ( ) ( ) = We also have D = BA = ( ) (3 + ( 1) + + 2) (( 2) ( 3) + ) ( ) ( ) ( + ( 4) + + 1) ( ( 1) + ) ( ) = ( ) ( ) (( 6) ) ( ) ( ) (3 + ( 8) + + ) (( 2) ) ( ) =

5 (b) Let α 1,α 2,α 3,α 4 R such that α 1 + 3α 2 2α 3 + 2α 4 = α α α α 4 = α 1 2α 2 + 2α 3 α 3 1 α 2 Hence, by comparing the last two coordinates, we see that α 2 = α 3 =. Thus, = α 1 2α 2 + 2α 3 = α 1 by comparing the second coordinate, and finally, by comparing the first coordinate, we get = α 1 + 3α 2 2α 3 + 2α 4 = α 4 and so α 4 =. Therefore, α 1,α 2,α 3,α 4 are all. Thus, the column vectors of A are linearly independent and hence A is an invertible matri. (c) By a class theorem, we can epand the determinant of a matri along any row of the matri. So we will epand along the fourth row of B. We could epand along any row, but the fourth row (or the third row) are the easiest to epand along. So, we have det(b) = 4 ( 1) 4+j b 4j det([b] 4,j ) j= = ( 1) 4+1 (1) det ( 1) 4+2 (4) det ( ) ( ( ) ( )) = ( 1)( 1) 3+3 det + 4 ( 1) 2+2 (1) det + ( 1) 2+3 (1) det ( ) ( ) = ( 1)((5)(1) (3)(2)) + 4( ) (1)(1) (2)(3) (1)() (3)(3) = 1 + 4( 5 + 9) = = 17. Thus, by a class theorem, B is invertible. 5

6 3. Let A M 4 (R) be the 4 4 real matri defined by: 3 3 A = (a) Determine the set of all eigenvalues of A. (1 pts) (b) For each eigenvalue of A, determine a basis for the eigenspace of A corresponding to that eigenvalue. (1 pts) (a) From class, we know that λ R is an eigenvalue if and only if det(λi A) = where I is the identity matri. We have λ 3 λ 3 λ 3 det(λi A) = det = (λ 3) det λ λ λ + 1 λ + 1 ( ) λ = (λ 3) 2 det = (λ 3) 2 (λ + 1) 2. λ + 1 So λ R is an eigenvalue if and only if (λ 3) 2 (λ + 1) 2 = if and only if λ = 3 or λ = 1. So the eigenvalues for A are 3 and 1. (b) Case 1: Let λ = 3. Then, the eigenspace of 3 is defined as ker(3i A). We have ker(3i A) = ker = ker = 2 3 R 4 : = = 2 3 R 4 : =

7 By comparing coordinates in this last equation we have 4 4 = and =. This implies 3 = 4 = after some algebraic manipulation. So ker(3i A) = 2 3 R 4 : 3 = 4 = = 2 R 4 = (( ) )) 1 1 So a basis for the eigenspace of 3 is,(. Case 2: Let λ = 1. Then, the eigenspace of 1 is defined as ker( I A). We have ker(3i A) = ker = ker = 2 3 R 4 4 : = = 2 3 R 4 4 : 2 4 = 4 By comparing coordinates in this last equation we have 4 1 =, 4 2 = and 4 =. This implies 1 = 2 = 4 =. So 1 ker( I A) = 2 3 R 4 : 1 = 2 = 4 = = 3 R 4 = 3 1 So a basis for the eigenspace of 1 is (( 4 1 )). 7

8 4. Let V and W be real vector spaces, and let L : V W be a linear transformation. Let V denote the zero vector of V and let W denote the zero vector of W. (a) Show that L( V ) = W. (5 pts) (b) Show that the kernel Ker(L) of L is a vector subspace of V. (5 pts) (c) Show that the image (i.e. range) Im(L) of L is a vector subspace of W. (5 pts) (a) We have: L( V ) = L( V ) = L( V ) = W. (b) We check the 3 aioms of a vector subspace one by one: Since L( V ) = W it follows that V ker(l). v 1,v 2 ker(l): L(v 1 + v 2 ) = L(v 1 ) + L(v 2 ) = W + W = W, and hence v 1 + v 2 ker(l). α R, v ker(l): L(α v) = α L(v) = α W = W, and hence α v ker(l). Hence ker(l) is a vector subspace of V. (c) Again we check the 3 aioms of a vector subspace one by one: Since W = L( V ) it follows that W Im(L). w 1,w 2 Im(L): v 1,v 2 V such that w 1 = L(v 1 ) and w 2 = L(v 2 ); hence, w 1 +w 2 = L(v 1 ) + L(v 2 ) = L(v 1 + v 2 ), which shows that w 1 + w 2 Im(L). α R, w Im(L): v V such that w = L(v); hence α w = α L(v) = L(α v), which shows that α w Im(L). Hence Im(L) is a vector subspace of W. 8

9 5. Let A and B be two real-valued 2 2 matrices in M 2 (R) given by A = ( a b c d where a,b,c,d,a,b,c and d are real numbers. (a) Show that det(a B) = det(a) det(b). (6 pts) ) ( ) a b and B = c d (b) Under what condition on the entries of A, is A invertible? (4 pts) (c) Use (a) to show that when A is invertible, then det(a 1 ) = 1 det(a). (5 pts) (a) We have Thus A B = = ( ) ( ) a b a b c d c d ( ) aa + bc ab + bd ca + dc cb + dd. det(a B) = (aa + bc )(cb + dd ) (ab + bd )(ca + dc ) = aca b + ada d + bcb c + bdc d aca b adb c bca d bdc d = ada d adb c + bcb c bca d = ad(a d b c ) bc(a d b c ) = (ad bc)(a d b c ). On the other hand, det(a) det(b) = (ad bc)(a d b c ). Thus det(a B) = det(a) det(b). (b) A is invertible iff its determinant is non-zero. Thus A is inbertible iff ad bc. (c) If A is invertible (which holds when ad bc ), then setting B = A 1 in (a) yields that det(a A 1 ) = det(a) det(a 1 ). 9

10 But det(a A 1 ) = det(i 2 ) = 1 where I 2 is the 2 2 identity matri. Hence 1 = det(a) det(a 1 ) and det(a 1 ) = 1 det(a). 1

11 6. Let V and W be real vector spaces, and let L : V W be a linear transformation such that Ker(L) = { V }. Assume that V has finite dimension n and let (v 1,v 2,,v n ) be a basis for V. Determine whether or not (L(v 1 ),L(v 2 ),,L(v n )) is a basis for Im(L). (1 pts) We will show that (L(v 1 ),L(v 2 ),,L(v n )) is a basis for Im(L) by proving that the set {L(v 1 ),L(v 2 ),,L(v n )} is both a generating set for Im(L) and linearly independent. Let w IM(L), then (by definition of Im(L)) there eits v V such that w = L(v). Since (v 1,v 2,,v n ) is a basis for V, then there eist reals α 1, α 2,, α n such that Thus w = L(v) v = α 1 v 1 + α 2 v α n v n. = L(α 1 v 1 + α 2 v α n v n ) = α 1 L(v 1 ) + α 2 L(v 2 ) + + α n L(v n ) where the third equality follows by linearity of L. Thus {L(v 1 ),L(v 2 ),,L(v n )} is a generating set for Im(L). Now assume that β 1 L(v 1 ) + β 1 L(v 1 ) + + β 1 L(v 1 ) = W for some reals β 1, β 2,, β n. But by the linearity of L, Thus we have that and hence β 1 L(v 1 ) + β 1 L(v 2 ) + + β 1 L(v n ) = L(β 1 v 1 + β 2 v β n v n ). L(β 1 v 1 + β 2 v β n v n ) = W β 1 v 1 + β 2 v β n v n Ker(L). But we know by assumption that Ker(L) = { V }. Therefore β 1 v 1 + β 2 v β n v n = V. Finally, since {v 1,v 2,,v n } is linearly independent (since (v 1,v 2,,v n ) is a basis for V), then the above equation ditrectly implies that β 1 = β 2 = = β n = and hence {L(v 1 ),L(v 2 ),,L(v n )} is linearly independent. We then conclude that (L(v 1 ),L(v 2 ),,L(v n )) is a basis for Im(L). 11