(b) Find det A, giving your answer in terms of a. (1) Given that the area of the image of R is 18, (c) find the value of a. (3) (Total 7 marks)

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1 FP MATRICES PAST EXAM QUESTIONS Matrix questions are pretty much all standard and plain-vanilla. Q. is a bit more difficult just because it is a long question with a lot of working. Q. (d) and Q.3 onwards from the first printing have been deleted because they ask for knowledge now not included in FP.. The rectangle R has vertices at the points (, ), (, ), (, ) and (, ). 5 (a) Find the coordinates of the vertices of the image of R under the transformation given by the matrix A a, where a is a constant. (b) Find det A, giving your answer in terms of a. () Given that the area of the image of R is 8, (c) find the value of a. (Total 7 marks). The matrix R is given by R = (a) Find R. () (b) Describe the geometrical transformation represented by R. () City of London Academy

2 (c) Describe the geometrical transformation represented by R. () (Total 5 marks) 3. Use the method of mathematical induction to prove that, for n e +, (a) n n n n n (b) f(n) = 4 n + 6n is divisible by 3. (7) (7) (Total 4 marks) a 3 4. M =, where a is a real constant. 6 a (a) Given that a =, find M. (b) Find the values of a for which M is singular. () (Total 5 marks) 5. Write down the matrix that represents (a) an enlargement with centre (, ) and scale factor 8, () City of London Academy

3 (b) a reflection in the x-axis. () Hence, or otherwise, (c) find the matrix T that represents an enlargement with centre (, ) and scale factor 8, followed by a reflection in the x-axis. 6 k A = and B =, where k and c are constants. 4 6 c () (d) Find AB. Given that AB represents the same transformation as T, (e) find the value of k and the value of c. () (Total 9 marks) a 5 6. A =, where a is real. a 4 (a) Find det A in terms of a. () (b) Show that the matrix A is non-singular for all values of a. Given that a =, (c) find A. City of London Academy 3

4 (Total 8 marks) 7. M = (a) Describe fully the geometrical transformation represented by the matrix M. The transformation represented by M maps the point A with coordinates (p, q) onto the point B with coordinates (3, 4 ). (b) Find the value of p and the value of q. () (4) (c) Find, in its simplest surd form, the length OA, where O is the origin. () (d) Find M. () The point B is mapped onto the point C by the transformation represented by M. (e) Find the coordinates of C. () (Total marks) City of London Academy 4

5 a 8. R, where a and b are constants and a >. a b (a) Find R in terms of a and b. Given that R represents an enlargement with centre (, ) and scale factor 5, (b) find the value of a and the value of b. (5) (Total 8 marks) a 9. A, where a is a constant. 4 (a) Find the value of a for which the matrix A is singular. 3 B 4 () (b) Find B. The transformation represented by B maps the point P onto the point Q. Given that Q has coordinates (k 6, 3k + ), where k is a constant, (c) show that P lies on the line with equation y = x + 3. (Total 8 marks) City of London Academy 5

6 a. Given that X =, where a is a constant, and a, (a) find X in terms of a. Given that X + X = I, where I is the identity matrix, (b) find the value of a. (Total 6 marks) 3. A =, B = C = 3, (a) Describe fully the transformations described by each of the matrices A, B and C. (4) It is given that the matrix D = CA, and that the matrix E = DB. (b) Find D. () 3 (c) Show that E = () The triangle ORS has vertices at the points with coordinates (, ), ( 5, 5) and (4, ). This triangle is transformed onto the triangle OR S by the transformation described by E. (d) Find the coordinates of the vertices of triangle OR S. (4) City of London Academy 6

7 (e) Find the area of triangle OR S and deduce the area of triangle ORS. (Total 4 marks) k. A, where k is constant. k k A transformation T : is represented by the matrix A. (a) Find the value of k for which the line y = x is mapped onto itself under T. (b) Show that A is non-singular for all values of k. (c) Find A in terms of k. () MARK SCHEME a 4 8 a a 8. (a) M A A 3 (Total 8 marks) (b) det A = = a ( 4) = a + 4 B (c) Area of R = B City of London Academy 7

8 rea of R = 8 Area scale factor is 9 = a + 4 a 5 M A 3 [7]. (a) R = M A (b) Rotation of 9, clockwise (about (,)) B, B (c) Rotation of 45 clockwise Bft [5] n n 3. (a), for n =, B n n true for n = Assume true for n = k k k k k k MA// k k k k k ( k ) k MA ( k ) ( k ) true for n = k + if true for n = k A 7 + true for n by induction City of London Academy 8

9 (b) f() = = 9 = 3 3 B true for n = Assume true for n = k, f(k) = 4 k + 6k is divisible by 3 f(k + ) = 4 k + + 6(k + ) MA = 4 4 k + 6(k + ) A f(k + ) f(k) = 3 4 k + 6 M f(k + ) = 3(4k+ ) f(k) which is divisible by 3 A true for n = k + if true for n = k A 7 true for n + by induction [4] (a) M Determinant: ( 88) B 6 M B: must be M: for correct attempt at changing elements in major diagonal and changing signs in minor diagonal. Three or four of the numbers in the matrix should be correct eg allow one slip A: for any form of the correct answer, with correct determinant then isw. Special case: a not replaced is BMA M A 3 (b) Setting and using a 8 to obtain a =. M City of London Academy 9

10 City of London Academy A cso Two correct answers, a = ±3, with no working is MA Just a = 3 is MA, and also one of these answers rejected is A. Need 3 to be simplified (not ). [5] 5. (a) B (b) B (c) M A M: Accept multiplication of their matrices either way round (this can be implied by correct answer) A: cao (d) M A A 3 M: Correct matrix multiplication method implied by one or two correct terms in correct positions. A: for three correct terms in correct positions a T AB c k c k c k

11 nd A: for all four terms correct and simplified (e) 6k c 8 and 4k c Form equations and solve simultaneously M k = and c = 4 A Alternative method M: AB = T B = A T = and compare elements to find k and c. Then A as before. M: follows their previous work but must give two equations from which k and c can be found and there must be attempt at solution getting to k = or c =. A: is cao (but not cso may follow error in position of 4k + c earlier). [9] 6. (a) det A = a(a + 4) ( 5 ) = a + 4a + MA s Correct use of ad bc for M (b) a + 4a + = (a + ) + 6 MAft Positive for all values of a, so A is non-singular Acso 3 s Attempt to complete square for M Alt Attempt to establish turning point (e.g. calculus, graph) M Minimum value 6 for Aft Positive for all values of a A is non-singular for A cso Alt Attempt at b 4ac for M. Can be part of quadratic formula City of London Academy

12 Their correct 4 for first A No real roots or equivalent, so A is non-singular for final Acso (c) A 4 5 = B for BMA 3 s Swap leading diagonal, and change sign of other diagonal, with numbers or a for M Correct matrix independent of their award final A [8] 7. (a) 45 or 4 rotation (anticlockwise), about the origin B, B More than one transformation / (b) p 3 q 4 M p q = 6 and p + q = 8 or equivalent MA p = 7 and q = both correct A 4 Second M for correct matrix multiplication to give two equations Alternative City of London Academy

13 City of London Academy 3 M = First MA Second M A (c) Length of OA (= length of OB) = M, A Correct use of their p and their q award M (d) M = MA (e) so coordinates are ( 4, 3 ) MA Accept column vector for final A. Order of matrix multiplication needs to be correct to award Ms [] 8. (a) R = M A A , ab a a a b a b a

14 term correct: M A A or 3 terms correct: M A A (b) Puts their a + a = 5 or their a + b = 5 M, or their (a + a)(a + b ) (a + ab)(a + b) = 5 (or to 5), Puts their a +ab = or their a + b = Solve to find either a or b a = 3, b = 3 A, A 5 Alternative for (b) Uses R column vector = 5 column vector, and equates rows to M, M give two equations in a and b only Solves to find either a or b as above method M A A M M as described in scheme (In the alternative scheme column vector can be general or specific for first M but must be specific for nd M) M requires solving equations to find a and/or b (though checking that correct answer satisfies the equations will earn this mark) This mark can be given independently of the first two method marks. So solving M = 5Mfor example gives MMMAA in part (b) Also putting leading diagonal = and other diagonal = 5 is MMMAA (No possible solutions as a >) A A for correct answers only Any Extra answers given, e.g. a = 5 and b = 5 or wrong answers deduct last A awarded So the two sets of answers would be A A Just the answer. a = 5 and b = 5 is A A Stopping at two values for a or for b no attempt at other is AA Answer with no working at all is marks M M [8] City of London Academy 4

15 9. (a) Use 4a ( ) = a, M, A Allow sign slips for first M (b) Determinant: (3 4) ( ) = (Δ) M B 4 = M Acso 3 3 Allow sign slip for determinant for first M (This mark may be awarded for / appearing in inverse matrix.) Second M is for correctly treating the by matrix, ie for Watch out for determinant (3 + 4) ( + ) = M then final answer is A k 6 4( k 6) (3k ) (c), = M, Aft 33k ( k 6) 3(3k ) k Lies on y = x + 3 A 3 k 3 Alternative: 3 x 3x ( x 3), =, M, A, 4 x 3 x 4( x 3) City of London Academy 5

16 x 6, which was of the form (k 6, 3k + ) A 3x 3 x 3x y Or, =, and solves M 4 y k 6 x 4y 3k simultaneous equations Both equations correct and eliminate one letter to get x = k or y = k + 3 or x y = 3 or equivalent. Completely correct work (to x = k and y = k + 3), and conclusion lies on y = x + 3 M for multiplying matrix by appropriate column vector A correct work (ft wrong determinant) A for conclusion A A [8]. (a) The determinant is a M X a = M A 3 a Attempt ad bc for first M a for second M det (b) I = B a Attempt to solve, or a, or, a a a City of London Academy 6

17 or a M To obtain a = 3 only A cso 3 Alternatives for (b) If they use X + I = X they need to identify I for B, then attempt to solve suitable equation for M and obtain a = 3 for A If they use X + X = O, they can score the Bthen marks for solving If they use X 3 + I = O they need to identify I for B, then attempt to solve suitable equation for M and obtain a = 3 for A Final A for correct solution only [6]. (a) A represents an enlargement scale factor 3 (centre O) M A B represents reflection in the line y = x B C represents a rotation of, i.e.45 (anticlockwise) (about O) 4 B 4 Enlargement for M 3 for A (b) M A Answer incorrect, require CD for M City of London Academy 7

18 (c) B Answer given so require DB as shown for B (d) = so (, ), (9, ) and (5, 75) MAAA Coordinates as shown or written as,, 75 for each A (e) Area of Δ OR S is B Determinant of E is 8 or use area scale factor of enlargement So area of ΔORS is = 87.5 MA B Divide by theirs for M [4]. (a) k t t( k 4) k k t t( k) M t( + k) = t(k 4) dm k = 9 A 3 City of London Academy 8

19 (b) det A = k + ( k) (Must be seen in part (b)) M = (k ) +, which is always positive M A is non-singular Acso 3 nd M: Alternative is to use quadratic formula on the quadratic equation, or to use the discriminant, with a comment about no real roots, or can t equal zero, or a comment about the condition for singularity. 4 8 x A Conclusion. (c) A k = MA k k k k M: Need, k s unchanged and attempt to change their det A sign for either (leaving as top right) or k (leaving as bottom left). [4]. No Report available for this question.. No Report available for this question. 3. No Report available for this question. 4. Generally this was a very accessible question where the vast majority of candidates gained full marks. They had a clear understanding of the process to find the inverse matrix and were able to apply it successfully in most cases. There were some arithmetical errors in finding the determinant and some candidates could not deal with the structure of the inverse matrix. A few did not substitute the given value for a. A much longer method involving simultaneous equations was much more prone to errors and fortunately not seen very often. In part (b) most understood the definition of a singular matrix and were able to solve their quadratic equation to give two accurate values for a. Some rejected the negative solution however, losing the last mark. City of London Academy 9

20 5. Part (a) was generally well done. Some however did have the columns of their matrix the wrong way round. Part (b) was similar to (a), but slightly less well done, with both sign and column errors. Candidates who have been taught to sketch a graph and transform a unit triangle/square performed well throughout this question. There were a noticeable minority who appeared to have no idea about using matrices for transformations, which meant a loss of access to some fairly straightforward marks. Many candidates multiplied their matrices in the wrong order for part (c). In this particular case it made no difference, and was not penalised, but it suggests that many candidates were not aware of the correct order. The majority of candidates knew how to multiply matrices in part (d) and were able to achieve a correct answer. In part (e), most candidates were successful in obtaining the correct equations. Of those who didn t, the majority were able to follow through from their (c) and (d). As in many other questions, candidates should be encouraged to check that their final solutions are consistent with their matrices. 6. In part (a) some candidates gave the reciprocal of the determinant as their answer rather than the determinant itself. The successful attempts at part (b) employed methods involving the discriminant of the equation det A = ; completing the square on the determinant; and a calculus/graphical approach. Some candidates lost the final mark in this part through not being able to fully justify their answer. The discriminant approach was most common with calculus rarely used. A number of candidates seemed confused about the precise meaning of the terms in use - singular, non-singular, real, complex. Whether the roots of the quadratic were real, complex, non-zero or positive was not clear to some. Part (c) was well answered by the vast majority of candidates, but some candidates did not spot the given value of a. 7. In general this question was answered very well, and a high proportion of candidates gained full marks. There was, in some cases, uncertainty about the order needed to perform the matrix multiplication. A common error in part (a) was to omit the centre of rotation, but the correct angle and direction were almost always present. There were popular approaches to part (b), either involving the formation of a pair of linear simultaneous equations, or finding the inverse of matrix M. Slips with signs produced the most common errors. The marks in (c) were almost always gained, with the simplified version of the surd being easily obtained. If slips were seen in part (d), these were mainly due to errors with signs. Provided that M had been found correctly, the coordinates in part (e) were normally obtained correctly. Some candidates did not use the coordinates of point B, thereby losing the marks. In part (d) and part (e), other candidates referred back to the geometry of the situation, obtaining their correct answers with some ease. 8. In part (a) the product of the two matrices was usually executed correctly with few errors. Part (b) caused difficulties for some and there were a number of attempts where pages were covered in matrix work which led nowhere. The common errors included solving R = 5R instead of solving R = 5I. A sizeable minority used the determinant of the matrix, putting it equal to 5 or to 5. They usually did not give a second equation to enable them to find the two unknowns. The successful majority approached the solution by equating the elements of their matrix solution to part (a) to 5 and to as appropriate. Usually they obtained two sets of solutions a = 3 with b = 3 and a = 5 with b = 5. They then discarded the second set of solutions as they had been given the condition a >, but some candidates failed to discard the second set and lost the final A mark. 9. Those who understood the word singular put their determinant equal to zero and solved the subsequent equation. There were frequently sign errors leading to the solution a = ½ and other algebraic errors leading to a = instead of a = ½. In part (b) most understood the method for finding the inverse matrix, but there were a number of errors and the determinant was often given as 4 instead of. Part (c) could be approached in various ways. The most popular method was to multiply the inverse matrix by the column vector with elements k 6, and 3k +. The answer obtained was the column vector with elements k and k + 3. Candidates then needed to complete their solution by concluding that the point lies on the line y = x + 3. x k 6 Another approach involved using the original matrix and multiplying it by the column vector and equating to which leads to x = k and y = k + 3. Again a conclusion x 3, 3 k was needed. x Others used the original matrix and multiplied it by the column vector, k 6 again equating to This leads straight to the equation y = x + 3. It was clear, however, that some y. 3 k candidates were unfamiliar with transformation work using matrices and did not set the transformation matrix first and follow it by a column matrix as required.. In part (a) the method of finding the inverse of a x matrix was well known but sign errors caused some candidates to lose accuracy. Only a few candidates actually showed how they found the determinant, so that if they went wrong they lost a method mark. A common error seen was ( + a). In part (b) the correct Identity matrix was almost always seen. There was a City of London Academy

21 significant number of candidates who did not add the matrices correctly.. Part (a) was the most challenging part of this final question for many. Those who had a method of looking at images of base vectors were usually successful but in general no method was seen leaving just a description. Part (b) and part (c) were rarely incorrect. There were some numerical errors in part (d) but most were successful although they did not always write the answer as coordinates. A few tried to post multiply by E and gained M. In part (e) the common error here was to use 5 rather than 75 leading to an area of 95 for ΔOR S and a final answer of 7.5 for ΔORS.. Candidates often found parts (a) and (b) of this question more difficult than parts (c) and (d), so it was helpful that the four parts could be tackled independently of one another. Part (a) of the question was poorly answered. Candidates applied the matrix to a variety of different vectors including i + j, i + j, xi + yj, xi + yj, etc. Many correctly performed the multiplication on xi + xj but then failed to use the fact that the line was mapped onto itself. By far the most common mistake was to suppose that xi + xj was mapped onto itself, an assumption that gave contradictory values k = 5 and k =. Most candidates seemed unconcerned by this contradiction. In part (b), to show that the matrix A was non-singular for all values of k, most candidates knew that the determinant was involved and found it correctly. Although some simply established non-singularity for the specific value of k they had found in part (a), most were able to produce a convincing proof, either by completing the square for the quadratic function or by using the quadratic formula. Conclusions relating to non-singularity were sometimes omitted. The inverse matrix in part (c) was often correct, although it was common to see k k k, or other variations, instead of. k k k k k k k In part (d), many candidates opted to use the original matrix and solve simultaneous equations rather than to use the inverse matrix. Although there were many good solutions, a common mistake was to map Q onto P instead of P onto Q. In part (c), where a hence method was required, candidates clearly struggled to find the link between the given equation and the result from part (b). Those candidates who substituted x = cos into the given equation were more successful in recognising this link and many who used this technique thus managed to obtain full marks, providing their conclusion was clear. A significant minority of candidates made use of the calculator and clearly thought that simple substitution was enough, while others made very little real progress or completely omitted this part. City of London Academy

Time: 1 hour 30 minutes

Time: 1 hour 30 minutes Paper Reference(s) 6667/0 Edexcel GCE Further Pure Mathematics FP Bronze Level B Time: hour 30 minutes Materials required for examination papers Mathematical Formulae (Green) Items included with question