MATH1231/1241 Algebra S Test 2 v3b

Transcription

1 MATH/4 Algebra S 9 Test vb August, 7 These answers were written by Matthew David and Daniel Le, typed up by Brendan Trinh and edited by Henderson Koh and Aaron Hassan Please be ethical with this resource It is for the use of MathSoc members, so do not repost it on other forums or groups without asking for permission If you appreciate this resource, please consider supporting us by coming to our events and buying our T-shirts Also, happy studying We cannot guarantee that our answers are error-free or would receive full marks please notify us of any errors or typos at unswmathsoc@gmailcom, or on our Facebook page There are sometimes multiple methods of solving the same question Remember that in the real class test, you will be expected to explain your steps and working out Please note quiz papers that are NOT in your course pack will not necessarily reflect the style or difficulty of questions in your quiz Hints Simply show that T preserves addition and scalar multiplication Worked Sample Solutions Let p, q P Then p + q is a polynomial in P and def T ((p + q)(x)) (p + q)() (p + q)() p() + q() p() + q() p() q() + p() q()

2 T (p(x)) + T (q(x)) And so, T satisfies the addition property Next, let p P and α R Then αp is a polynomial in P with def T (αp(x)) α αp() αp() p() p() αt (p(x)) Thus, T also satisfies the scalar multiplication property Therefore, T is linear (i) Hints As usual, row-reduce the matrix A, and then the rank will be the number of leading columns in the row-echelon form of A, whilst the nullity is the number of nonleading columns in the row-echelon form You should find that rank (A) and nullity (A) Here is a sample solution Worked Sample Solution We row-reduce A But anticipating part (ii) of this question, we do so with b augmented in the right-hand part (the augmented part of the matrix will be irrelevant to the answer for (i), but will be handy for (ii) and putting it in now will save us from having to redo row-reductions for part (ii)) The appropriate augmented matrix is given by 4 R R +R R R +R 4 8 R R 4R Only the left-hand part of the matrix is relevant to part (i) As we can see, there are two leading columns in the row-echelon form, with three non-leading columns (remember, only looking to the left of the vertical line) This means that rank (A) and nullity (A) (ii) Hints Look at the row-echelon form of the augmented matrix from our working in part (i) If there is a solution to the matrix equation (ie if the right-hand column is non-leading), then b is indeed in the image of A Otherwise, it is not Here is a worked sample solution

3 Worked Sample Solution From part (i), we see that there exists a solution to the matrix equation Ax b since the right-hand column is non-leading in the row-echelon form Therefore, b is in the image of A Hints Recall that the general solution to the matrix differential equation dy dt Ay for some given constant matrix A is y c eλ t v + c eλ t v, where λk and vk are eigenvalueeigenvector pairs of A Worked Sample Solution The equivalent matrix form of the system of differential equations is dy dt 8 y, y y y Using the given eigenvalues and eigenvectors of the matrix A, we thus see that the general solution to the system of differential equations is y y y c e5t + c 4 e7t, c, c arbitrary constants Writing it out component-wise, we equivalently have that the general solution to the given system of differential equations is y c e5t + c e7t y c e5t + 4c e7t 4 (i) Hints Make use of the fact that pk ck is a probability distribution Worked Sample Solution Since pk is a (discrete) probability distribution function, then we know that 5 X pk k 5 X ck k Solving for c, we find that c( ) 55c, This is the general solution provided that the eigenvectors v and v are linearly independent (ie the matrix A is not what is called defective), which will probably always be the case for the purposes of MATH/4

4 and hence c 55 (ii) Hints Since, pk is a (discrete) probabilty (mass) function, to find the probability that the random variable X takes a certain value, we simply substitute this into pk This should be easy now that we have c from part (i) Worked Sample Solution We have P (X ) (iii) Hints Since X is a discrete random variable, we can simply add up the probabilities of the outcomes that satisfy the event {X 4} Alternatively, we can note that this event is the complement of the event {X 5} and use our knowledge of probability of complementary events to answer the question Worked Sample Solution Using a similar method to part (ii) for each case, we know that P (X 4) P (X 4) + P (X ) + P (X ) + P (X ) c 4 + c + c + c 55 6 Simpler calculation Note that since X only takes integer values from to 5, the event {X 4} is the complement of the event {X 5} Therefore, P (X 4) P (X 5) () () () (4)

5 MATH/4 Algebra S 9 Test v7a August, 7 These answers were written by Treves Li, typed up by Brendan Trinh and edited by Henderson Koh and Aaron Hassan Please be ethical with this resource It is for the use of MathSoc members, so do not repost it on other forums or groups without asking for permission If you appreciate this resource, please consider supporting us by coming to our events and buying our T-shirts Also, happy studying We cannot guarantee that our answers are error-free or would obtain full marks please notify us of any errors or typos at unswmathsoc@gmailcom, or on our Facebook page There are sometimes multiple methods of solving the same question Remember that in the real class test, you will be expected to explain your steps and working out Please note quiz papers that are NOT in your course pack will not necessarily reflect the style or difficulty of questions in your quiz (i) Hints As usual, row-reduce the matrix A, and then the rank will be the number of leading columns in the row-echelon form of A, whilst the nullity is the number of nonleading columns in the row-echelon form You should find that rank (A) and nullity (A) Here is a sample solution Worked Sample Solution We row-reduce A: A 5 5 R R R 6 R 5 R +R 4 9 6

6 R R +4R So there are three leading columns in the row-echelon form, with one non-leading column This means that rank (A) and nullity (A) (ii) Hints Construct the appropriate augmented matrix to represent the corresponding matrix equation Ax Row-reduce this matrix to help find the kernel of A However, instead of performing the calculations over again, notice that having the vector on the right hand side of the matrix doesn t change any calculations/row operations, nor does it change the vector itself So, we can simply use the results from part (i) You should find that a basis for the kernel of A is 8 Worked Sample Solution When we set up the augmented matrix and row-reduced, we achieve the exact same matrix as in part (i) (as the column furthest to the right is all s, which stays unchanged when performing elementary row operations) So when we write Ax where x (x, x, x, x4 )T, as an augmented matrix and row-reduce, we end up with Let x4 α where α R Then from row, we find that x x4 x α From row, we have x x4 x x α 4α x α Now, from row, we have x x + x x4 9α + α α x 8α Hence, our solution is given by 8 x x x x α x4 6

7 Therefore, a basis for the kernel of A is 8 To find the eigenvalues, we know that they are the solutions to the equation det (A λi) So for eigenvalues λ, 7 λ 5 det (A λi) 4 λ (7 + λ) (4 + λ) λ + λ + 8 λ + λ + 8 (λ + 9) (λ + ) Therefore, the eigenvalues are λ 9 and λ For the eigenvectors, they are the basis vectors of the eigenspaces Eλ ker (A λi) For λ 9, Eλ E 9 ker (A + 9I) ker span 5 5 ) ( 5 (by inspection) For λ, Eλ E ker (A + I) ker span 5 5 ( ) (by inspection) 5 Hence the eigenvalues are λ 9 with eigenvectors α for α R, α 6 and λ with eigenvectors β for β R, β 6 (remember, is not an eigenvector, by definition) Hints 7

8 (i) Conditional probability: P (Positive Diseased) P (Positive Diseased) P (Diseased) 54 (ii) To find P (Positive), consider all the different cases (condition it on being diseased and not diseased) P(Positive) (iii) Use Bayes Rule, or note that P(A B)P(B) P(A B) P(B A)P(A) and rearrange with appropriate events A and B P (Diseased Positive) 54 Generally for probability questions, recall the following probability properties/rules P (A B) P (A B) P (B) P (B A) P (A) P (A B) P(A B) P(B) P(B A)P(A) P(B) P (A) P (A B) + P (A B c ) P (A B) P (B) + P (A B c ) P (B c ) Worked Sample Solution (i) We have P (diseased and positive) P (diseased positive) P (positive diseased) P (diseased) Note: a common error in this question is to write P (diseased and positive) P (diseased) P (positive) In general, it is not true that P(A B) P(A) P(B), as this is true if and only if A and B are independent And in this question, the events are not independent, since the probability of testing positive depends on whether the person is diseased (ii) Using the Law of Total Probability, we have P (positive) P (positive diseased) P (diseased) +P (positive not diseased) P (not diseased) {z } 54 from part (i) ( 6) 8

9 (iii) We have P (diseased and positive) P (positive) (by definition of conditional probability) P (diseased positive) Alternatively, you can use Bayes Rule to get the same result 9

10 MATH/4 Algebra S Test va Full Solutions August, 7 These solutions were originally written and typed up by Brendan Trinh, and edited and augmented by Aaron Hassan and Henderson Koh Please be ethical with this resource It is for the use of MathSoc members, so do not repost it on other forums or groups without asking for permission If you appreciate this resource, please consider supporting us by liking our Facebook page and coming to our events Also, happy studying We cannot guarantee that our working is error-free, or that it would obtain full marks please notify us of any errors or typos at unswmathsoc@gmailcom, or on our Facebook page There are sometimes multiple methods of solving the same question Remember that in the real class test, you will be expected to explain your steps and working out Please note quiz papers that are NOT in your course pack will not necessarily reflect the style or difficulty of questions in your quiz To prove that T : P R is a linear transformation, we need to show that it satisfies the addition and scalar multiplication conditions Note that from basic calculus, we know that differentiation is a linear operator, ie that (p + q) p + q (in words, the derivative of a sum is the sum of the derivatives) and (λp) λp (in words, constants can come outside the differentiation operator), for any p, q P and scalar λ R We will use these facts in our proof Let p, p P, so note T (p (x)) p () and T (p (x)) p () p () p ()

11 Then, p + p is a polynomial in P and we have, (p + p ) () def T ((p + p ) (x)) (p + p ) () p () + p () (since derivative of a sum is sum of derivatives) p () + p () p () p () (definition of addition of vectors in R ) + p () p () def T (p (x)) + T (p (x)) This implies that T satisfies the addition condition Secondly, let λ R and p P Then λp is a polynomial in P that satisfies def T (λp (x)) λ (λp) () (λp) () λp () λp () λp () (using the facts mentioned at the start about differentiation) λp () p () (definition of scalar multiplication of vectors in R ) p () def λt (p (x)) Thus, T satisfies the scalar multiplication condition Since T satisfies the addition and scalar multiplication conditions, it is a linear transformation (i) To find rank (A), we row-reduce A and find how many leading columns there are in the row-echelon form Recall that A 5 Performing the row operations R A R R, we see R + R and R (Aside: The in this context means is row-equivalent to, which basically just means can be row-

12 reduced to (using elementary row operations) For those of you who have done MATH8 Discrete Mathematics, it can be shown that is an equivalence relation on the set of m n matrices for any given positive integers m, n You can find out more about row-equivalence at its Wikipedia page: Finally, performing R R + R, we have A 4 As there are leading columns in the row-echelon form, there are linearly independent vectors of the 5 column vectors, ie rank (A) By the Rank-Nullity Theorem, we know that rank (A)+nullity (A) # of columns of A Thus, nullity (A) 5 (Alternative (but equivalent) reasoning: the nullity is the number of non-leading columns in the row-echelon form, ie ) (ii) The kernel of A, written as ker (A), is the set of all solutions to Ax, ie x R5 : Ax This means that to find the kernel, we are essentially just solving Ax When we set up the augmented matrix and row-reduced, we achieve the exact same matrix as before (as the column furthest to the right is all s, which stays unchanged when performing elementary row operations) So when we write Ax where x (x, x, x, x4, x5 )T, as an augmented matrix and row-reduce, we end up with 4 As usual, set the non-leading columns variables (in this case x and x5 ) to free parameters So let x β and x5 α, where α, β R are free parameters Then from row, we obtain x4 4x5 x4 4α From row, we have x x + x4 x5 x β + 4α α x β + α Now, from row, we have x x x + x4 x5 (β + α) β + 4α α x β α Hence our solution is given by β α x x β + α x β x 4α x4 α x5

13 α +β 4, where α, β R Therefore, we see that ker (A) span,, and so a basis for ker (A) is: 4, 4 Hints This whole question revolves around conditional probability, ie P (A B) P (A B) P (B) P (B A) P (A) Also note that by definition P (A B) P(A B) P(B) (if P (B) 6 ) Remember also that (set intersection) generally refers to and for probability (whilst, or set unions, generally refers to or ) So if you see an and being used for probability, it generally refers to the intersection of sets/events We also use the Law of Total Probability Basically, it states that a probability can be broken down into the sum of all its cases Google it and have a read of it as it can be quite tedious to carefully explain mathematically, despite actually being relatively intuitive You can check out an example (and read about the Law) on its Wikipedia page here: Example Sample solutions are below Sample solutions (i) We have P (red die selected and was thrown) P ( was thrown red die selected) P (red die selected)

14 Note: a common error in this question is to write P (red die selected and was thrown) P (red die selected) P ( was thrown) In general, it is not true that P(A B) P(A) P(B), as this is true if and only if A and B are independent events And in this question, the events are not independent, since the probability of a being thrown heavily depends on the colour of the die (ii) Using the Law of Total Probability, we have P ( was thrown) P ( was thrown red die selected) P (red die selected) {z } 8 from part (i) + P ( was thrown blue die selected) P (blue die selected) (iii) We have P (red die was selected was thrown) P (red die selected and was thrown) P ( was thrown) (definition of conditional probability) 8 8/ 4 9/4 4 9 (using our previous results) 6 9 4

15 MATH/4 Algebra S Test vb Full Solutions August, 7 These solutions were originally written and typed up by Brendan Trinh, and edited and augmented with tips by Henderson Koh and Aaron Hassan Please be ethical with this resource It is for the use of MathSoc members, so do not repost it on other forums or groups without asking for permission If you appreciate this resource, please consider supporting us by liking our Facebook page and coming to our events Also, happy studying We cannot guarantee that our working is error-free, or that it would obtain full marks please notify us of any errors or typos at unswmathsoc@gmailcom, or on our Facebook page There are sometimes multiple methods of solving the same question Remember that in the real class test, you will be expected to explain your steps and working out Please note quiz papers that are NOT in your course pack will not necessarily reflect the style or difficulty of questions in your quiz (i) The rank of a matrix is basically the number of vectors required to span the column space of the matrix In order to find this, we need to find how many linearly independent columns there are Equivalently, the rank of A is the number of leading columns in the row-echelon form (in fact, this also always equals the number of leading rows in the row-echelon form) So we row-reduce the matrix A We have A 7 5 5

16 Firstly, we perform R R R and R4 A Next, performing R R4 + R, 5 R R, A 4 (Aside: The in this context means is row-equivalent to, which basically just means can be row-reduced to (using elementary row operations) For those of you who have done MATH8 Discrete Mathematics, it can be shown that is an equivalence relation on the set of m n matrices for given positive integers m, n You can find out more about row-equivalence at its Wikipedia page: Hence, we see that the first, second and fourth columns are linearly independent (as they are the leading columns of this row-reduced matrix) This means that these three columns span the column space of A Thus, rank (A) By the Rank-Nullity Theorem, which states that rank (A) + nullity (A) # of columns of A, we know that nullity (A) 4 Alternative reasoning for rank and nullity part Since A has three leading columns in the row-echelon form, rank (A) Since A has one non-leading column in the row-echelon form, nullity (A) (ii) Note that the image of A refers to the vector space y R4 : Ax y for x R4 So to see if b (,, 6, 7)T is in the image of A, we need to see if there exists a solution for x R4 such that Ax b Constructing an augmented matrix, Recalling from above, we can perform one row operation to make the last row basically all s, we give it a go to see if the last row becomes invalid (in the hope that it does to say that b is not in the image of A) 6

17 So performing the row operations R4 R4 + R, we obtain Now, the last row is a contradictory equation of 6, so b / im (A) as there does not exist x R4 such that this is true Tip Here we really got lucky that we could quickly get to the answer from the original matrix In general though this would not be the case and you may need to end up doing a lot of row-reduction again that you already did in part (i) To overcome this, you should actually augment b when you do the row-reductions initially in question (i), so that you ll have reduced with b present, and will need to do no further row-reductions in part (ii) and can just read off the answer from your hard work row-reducing in part (i) (Eg refer to our Worked Sample Solution to the Question # on page of this document) A lot of students jump into row-reducing A when they look at question (i) and then are forced to do basically the same row-reduction again for part (ii) So make sure you read all parts of the question before starting writing, and if you see a part asking if a vector is in the image of A or something like this, make sure to augment b straightaway before beginning row-reducing (i) The eigenvalues λ are the solutions to the equation det (A λi) So for eigenvalues λ, 7 λ det (A λi) 6 λ (7 λ) ( λ) + 6 λ 9λ λ 9λ + (λ 4) (λ 5) Thus, the eigenvalues are λ 4 and λ 5 For the eigenvectors, they are the basis vectors of the eigenspaces Eλ ker (A λi) For λ 4, Eλ E4 ker (A 4I) ker span 6 ( ) (by inspection, explained below) Tip Explanation of inspection (included for your understanding, you don t need to 7

18 provide this in your working): We can quickly find the eigenvector by thinking about what vector we have to multiply the matrix by to get If you have done things right, we first consider the top row (or whichever row looks easier) Then, we think about what vector it multiplies with to obtain This is then your eigenvector You can double check it by multiplying it with the bottom row too if you wish Since we are dealing with distinct eigenvalues here, the eigenspaces must all be one-dimensional, so we only need to find one eigenvector to get the basis for the eigenspace (for eigenvalues of algebraic multiplicity (multiplicity in the characteristic polynomial) more than, this will not necessarily be the case though) For λ 5, Eλ E5 ker (A 5I) ker span 6 ( ) Hence, the eigenvalues are λ 4 with corresponding eigenvector v λ 5 with corresponding eigenvector v and (ii) Yes, A is diagonalisable as it can be written in the form A P DP where 4 P is invertible and D is diagonal 5 Alternate Solution Yes, A is diagonalisable since its eigenvalues are distinct Further tips When finding the eigenvectors for an eigenvalue λ, when you row-reduce the matrix A λi, this matrix must have at least one zero row in the row echelon form This is because λ being an eigenvalue means that A λi (a square matrix) has determinant, which is equivalent to saying it has a zero row in the row echelon form Therefore, if you end up row-reducing and seeing that there are no zero rows, you must have made a mistake somewhere, and should go back and find it In the matrix case, it is even easier to check your work here In the case, the matrix A λi must be such that one of its rows is a constant multiple of the other (this also applies to the columns), if (and only if) λ is an eigenvalue (This is because a matrix has a zero row in the row echelon form if and only if one of its rows is a multiple of the other) In other words, if when you write the matrix A λi when finding eigenvectors for an eigenvalue λ, and you see that neither rows is a multiple of the other, you must have made a mistake somewhere (because the matrix won t end up with a zero row in the row-echelon form), and should find it Finally, in the Alternate Solution for part (ii) above, it is important to keep in mind 8

19 that while any matrix A with distinct eigenvalues is diagonalisable, the converse is not true In other words, there exist diagonalisable matrices with repeated eigenvalues So if a matrix A has all distinct eigenvalues, then A is diagonalisable, but not necessarily the other way round To show this is a (discrete) probability distribution, we need to show that pk for P pk k,, 4,, and k It is clear that pk for all k,, 4,, because 5k is non-negative for all such k, and P P hence so is 5k Now, consider the sum S pk Note that this is a geometric 5k k series with starting term a k and common ratio r 5 As the geometric series has common ratio smaller than, this series converges and a r 4/5 5 4/5 4/5 S Therefore, the given sequence is a probability distribution 9

20 MATH/4 Algebra S Test va Full Solutions August, 7 These solutions were originally written and typed up by Brendan Trinh, and edited and augmented by Aaron Hassan and Henderson Koh Please be ethical with this resource It is for the use of MathSoc members, so do not repost it on other forums or groups without asking for permission If you appreciate this resource, please consider supporting us by liking our Facebook page and coming to our events Also, happy studying We cannot guarantee that our working is error-free, or that it would obtain full marks please notify us of any errors or typos at unswmathsoc@gmailcom, or on our Facebook page There are sometimes multiple methods of solving the same question Remember that in the real class test, you will be expected to explain your steps and working out Please note quiz papers that are NOT in your course pack will not necessarily reflect the style or difficulty of questions in your quiz (i) Anticipating we would need to augment A with b in question (ii), we shall do this now, so that we would not need to do the whole row-reduction again Writing the components of b as separate columns in the right-hand part of the augmented matrix, we row-reduce: R R R 5 R R +R

21 R R 5R 7 5 So the row-echelon form of A (the left-hand matrix above) has three leading columns and two non-leading columns This means that rank (A) and nullity (A) (ii) Since the row-echelon form of the matrix A (found above) has no zero rows, there will b 5 be a solution of x R to the linear system Ax b for any given b b R In other words, im (A) is the whole of R, b so there are no further conditions needed on b, b, b We need to find the eigenvalues and eigenvectors of A The eigenvalues are the solutions λ to the equation det (A λi) But λ 5 det (A λi) 4 λ ( λ) (4 λ) 5 λ 6λ λ 6λ 7 (λ 7) (λ + ) Therefore, the eigenvalues are λ 7 and λ Now, to find the eigenvectors, we compute the eigenspaces Eλ ker (A λi) for each eigenvalue λ Note that since the eigenvalues are distinct (no repeated ones), the eigenspaces are guaranteed to be one-dimensional, so we just need one independent eigenvector for each eigenvalue This makes it relatively easy to find these by inspection from the matrix A λi For λ 7, Eλ E7 ker (A 7I) ker span 5 5 ( ) For λ, Eλ E ker (A + I) ker 5 5 (by inspection)

22 ( span 5 ) (by inspection) That is, the are λ 7 with eigenvector eigenvalues and corresponding eigenvectors 5 and λ with eigenvector v v 5 7, we have A M DM and M Thus, letting D Drawing a tree diagram usually helps organise our thoughts immensely Let H be the event that the customer bought a hamburger and similarly D for the event they bought a drink We then have the following tree diagram based on the given data ham- H 6 Angry P (H D) 6 8 D Bought Dc 7 burger P (H Dc ) Jack s customer Hc 4 P (H c D) 4 8 D Did not buy ham- Dc 8 burger P (H c Dc ) 4 8 (i) Recall that and generally refers to set intersection in probability In terms of our notation, we are asked to find P (H D) We have P (H D) P (D H) P (H) (this corresponds to the top-most node in the tree diagram) 6 8 (ii) We have P (D) P (D H) P (H) + P (D H c ) P (H c ) (Law of Total Probability) (using part (i) s answer for the first summand) 6 (Or using our tree diagram, we can see the answer is 8 + 8, ie the sum of the nodes ending in D, which are the ones where a drink was bought)

23 (iii) We have P (H D) P (D) P (H D) (definition of conditional probability) (from earlier parts) (simplifying) Remark While tree diagrams are very helpful for getting a feel for questions, it may be good, even if you draw a tree diagram, to still try and provide reasoning for your answers in terms of probability rules if you can (like Law of Total Probability etc), in case the marker does not consider just tree diagrams to be rigorous enough working Also, if you can see how to do these questions without drawing a tree diagram (ie just with the reasoning given in our sample answers), by all means do so

24 MATH/4 Algebra S Test va Answers/Hints & Worked Sample Solutions August, 7 These answers/hints were originally written and typed up by Brendan Trinh, with further hints, edits and worked sample solutions provided by Aaron Hassan and Henderson Koh Please be ethical with this resource It is for the use of MathSoc members, so do not repost it on other forums or groups without asking for permission If you appreciate this resource, please consider supporting us by coming to our events and buying our T-shirts Also, happy studying We cannot guarantee that our answers are error-free or would obtain full marks please notify us of any errors or typos at unswmathsoc@gmailcom, or on our Facebook page There are sometimes multiple methods of solving the same question Remember that in the real class test, you will be expected to explain your steps and working out in a clear and logical manner Please note quiz papers that are NOT in your course pack will not necessarily reflect the style or difficulty of questions in your quiz Hints Show that it satisfies the addition and scalar multiplication conditions Worked Sample Solution Let p, q P Then p + q is a polynomial in P and (p + q) () def T ((p + q) (x)) (p + q) () (p + q) () 4

25 p () + q () p () + q () p () + q () q () p () p () + q () p () (definition of p + q) (definition of vector addition in R ) q () def T (p (x)) + T (q (x)) Thus T satisfies the addition condition Now, let p P and α R Then αp is a polynomial in P with αp () def T (αp (x)) αp () αp () p () α p () (definition of scalar multiplication in R ) p () def αt (p (x)) Thus T also satisfies the scalar multiplication condition Therefore, T is linear QED (i) Hints As usual, row-reduce the matrix A, and then the rank will be the number of leading columns in the row-echelon form of A, whilst the nullity is the number of nonleading columns in the row-echelon form You should find that rank (A) and nullity (A) Here is a sample solution Try to do both parts (i) and (ii) at the same time for efficiency 5

26 Worked Sample Solution We row-reduce A: A R R R 5 4 R R R 6 8 R R +R 6 So there are three leading columns in the row-echelon form, with one non-leading column This means that rank (A) and nullity (A) (ii) Hints Recall that the kernel of a matrix is simply the set of vectors that multiply the matrix to yield the zero vector Mathematically, ker (A) x R4 : Ax, where is the zero vector of R In other words, we are solving for x in the equation Ax So as usual, to find the basis for the kernel, we just row-reduce A (conveniently done in part (i)) and find the solutions from that (using back-substitution etc) You may think that normally to solve Ax b, we would augment A with the vector b too The reason we don t need to do this here is that when finding the kernel, the vector b is just, and no matter what elementary row operations we perform, this vector would stay as, so there is no need to actually augment it You can just imagine it is there in your mind It can be found that a basis is: 5 Here is a worked sample solution Worked Sample Solution x x Let x x be a vector in ker (A), ie let Ax Then from our row-echelon x4 form of A in part (i), we see column three is non-leading, so we set x α, where α R is a free parameter From row 4, we see that x4 Now from row, we have x x + x4 6

27 x α +, so x α Now, from row, we have x x x x4 ( α) α 6α α 5α, so x 5α Thus x is in ker (A) if and only if x 5α x α x x α x4 5 α, for some α R Thus a basis for the kernel of A is 5 General Hints These questions test your understanding of the rules of basic probability So make sure to keep in mind these rules, such as the definition of conditional probability, the multiplication rule P (A B) P (A) P (B A), the rule P (A) P (A B) + P (A B c ), etc The aforementioned ideas are the main ones needed to do this particular problem, as the worked sample solutions will show In general though for probability questions, other important rules you may find useful to be familiar with include Bayes Rule, the Law of Total Probability, and for harder questions, the Inclusion/Exclusion Principle You can read up on these various rules online or from your course pack if you are unsure of them 7

28 before tackling this question Numerical Answers (i) P (car was red and caught speeding) (ii) P (car caught speeding) 5 (iii) P (car was red caught speeding) 8 Worked Sample Solutions We will use the following notation Let R be the event that the car was red and S the event that the car was caught speeding Then we are given the following data: P (R) 4, P (S R), P (S Rc ) 5 Note that P (Rc ) P(R) 4 6 (i) We are asked to find P (R S) We have P (R S) P (R) P (S R) 4 (given data) (ii) We are asked to find P (S) Using the probability rule P (A) P (A B)+P (A B c ) (which follows from the facts that A (A B) (A B c ), A B and A B c are disjoint and the axiom of probability that states that the probability of a union is the sum of probability for pairwise disjoint events; it is easily seen from a Venn diagram too), we have P (S) P (S R) + P (S Rc ) ( ) Now, S R and R S are the same event, so we know from part (i) that P (S R) To find P (S Rc ), we have P (S Rc ) P (Rc ) P (S Rc ) 6 5 (using given data) Putting these findings into Equation ( ), we have P (S) + 5 (iii) We are asked to find P (R S) We have P (R S) P (R S) P (S) (definition of conditional probability) 8

29 (from parts (i) and (ii)) 9

30 MATH/4 Algebra S Test vb Hints & Worked Sample Solutions August, 7 These hints were originally written and typed up by Brendan Trinh and edited by Henderson Koh, with worked sample solutions provided by Aaron Hassan Please be ethical with this resource It is for the use of MathSoc members, so do not repost it on other forums or groups without asking for permission If you appreciate this resource, please consider supporting us by coming to our events and buying our T-shirts Also, happy studying We cannot guarantee that our answers are error-free or would receive full marks please notify us of any errors or typos at unswmathsoc@gmailcom, or on our Facebook page There are sometimes multiple methods of solving the same question Remember that in the real class test, you will be expected to explain your steps and working out Please note quiz papers that are NOT in your course pack will not necessarily reflect the style or difficulty of questions in your quiz Hints (i) It can be found that rank(a) and nullity(a), by row-reducing A and looking at the number of leading and non-leading columns in the row-echelon form (ii) It can also be found that b is not in the image of A Worked Sample Solutions (i) Anticipating we would need to augment A with b in question (ii), we shall do this now, so that we would not need to do the whole row-reduction again Writing the Refer to Question in Test S 9 vb

31 components of b as separate columns in the right-hand part of the augmented matrix, we row-reduce: 8 +R R R 4 R R R,R4 R4 R 7 R R R So the row-echelon form of A (the left-hand matrix above) has three leading columns and no non-leading columns This means that rank (A) and nullity (A) (ii) From the row-echelon form above, we see that the right-hand column is leading This means that b is not in the image of A Hints (i) It can be found that the eigenvalues are λ and 5 withcorresponding eigenvectors and respectively being (up to constant non-zero multiples) 4 that the matrix A is diagonalisable, with A P DP, where (ii) Also, this implies D and P 5 4 Worked Sample Solutions (i) We have A λi 7 λ 8 λ For the eigenvalues, we set the determinant of this equal to and solve for λ We have det (A λi) (7 λ)( λ) + 8 λ 8λ + 5 (λ ) (λ 5) Hence the eigenvalues are λ and λ 5 To find the corresponding eigenvectors, we find bases for the eigenspaces Eλ ker (A λi) for each eigenvalue λ We have Eλ E ker (A I) Refer to Question in Test S vb

32 7 ker ker span ( ) 4, (by inspection) and so a corresponding eigenvector for eigenvalue λ is v Also, 4 Eλ E5 ker (A 5I) ker ker span ( ), (by inspection) In summary, the eigenvalues of A are λ with corresponding eigenvectors being (non-zero multiples of) v and λ 5 with corresponding eigenvectors being (non-zero 4 multiples of) v and so a corresponding eigenvector for eigenvalue λ 5 is v (ii) Yes, A is diagonalisable, because it is and has two linearly independent eigenvectors v, v Alternative answer Yes, A is diagonalisable This is implied by the fact that its eigenvalues are distinct Hints This is a probability distribution if pk for k,, and if P pk It can be k shown that these results indeed hold And so, the sequence is a probability distribution Worked Sample Solution Clearly, pk for every k,, 4,, because for all such k, k > and also 6 > Furthermore, we have X k X 6 pk k k Refer to Question in Test S vb

33 This is an infinite geometric series with common ratio r and starting term a 6 Since the common ratio satisfies r <, the series converges and is equal to a r Hence the sum of all the pk s is, and the pk s are all non-negative Therefore, the given sequence defines a probability distribution

34 MATH/4 Algebra S Test va Hints/Answers & Worked Sample Solutions August, 7 These hints/answers were originally written and typed up by Brendan Trinh and edited by Henderson Koh, with worked sample solutions provided by Aaron Hassan Please be ethical with this resource It is for the use of MathSoc members, so do not repost it on other forums or groups without asking for permission If you appreciate this resource, please consider supporting us by coming to our events and buying our T-shirts Also, happy studying We cannot guarantee that our answers are error-free or would receive full marks please notify us of any errors or typos at unswmathsoc@gmailcom, or on our Facebook page There are sometimes multiple methods of solving the same question Remember that in the real class test, you will be expected to explain your steps and working out Please note quiz papers that are NOT in your course pack will not necessarily reflect the style or difficulty of questions in your quiz Hints/Answers (i) It can be found that rank(a) and nullity(a), by row-reducing A looking at the number of leading and non-leading columns (ii) b always belongs to im(a), since rank(a) and dim R Refer to Question in Test S 9 vb 4

35 Worked Sample Solutions (i) As usual, we row-reduce A We have A R R +R 5 R R +R R R R 5 As we can see, there are three leading columns in the row-echelon form, so rank(a) Also, there are two non-leading columns in the row-echelon form, so nullity(a) (ii) No extra conditions are required on the components of the vector b R This is because A has rank (and its image is a subspace of R as it has three rows), so its columns span the entire R, ie every vector in R belongs to the image of A Hints/Answers It can be found thatthe eigenvalues are λ 5 and λ 4, with corresponding 5 4 and respectively And so, D and v eigenvectors v 4 4 M Note that your columns of matrix M may differ by non-zero multiples For example, M 4 could also be M where the first column is multiplied by, and the second 6 column is multiplied by But either answer would be correct You could also have written your diagonal matrix in the opposite order This is fine too, but you would also need to write the M matrix s columns in the opposite order to what we have done then (Remember, the order of the columns of M going from left to right must match the order of the eigenvalues placed into the diagonal entries in D) Worked Sample Solution We are being asked to diagonalise A We have A λi λ 4 4 λ Hence det (A λi) ( λ)(4 λ) 8 λ λ Refer to Question in Test S 9 vb 5

36 (λ 5) (λ + 4) So the eigenvalues of A are λ 5, λ 4 Now we find corresponding eigenvectors We have ker (A λ I) ker (A 5I) ker ker So we may take v (by inspection) as the corresponding eigenvector for eigenvalue λ 5 Also, ker (A λ I) ker (A + 4I) ker ker span 4 4 ) ( span ( ) 4 (by inspection) as the corresponding eigenvector for eigenvalue λ 4 So putting the eigenvalues into a diagonal matrix D and the corresponding eigenvectors as So we may take v columns of the matrix M (in the order they appear from top to bottom in the diagonal of D), we can write A M DM, where M 4 4, D 5 4, your M would need to have its columns in the opposite 5 order to what we wrote, ie it would be (up to constant non-zero multiples for the columns) 4 M If instead you used D Hints/Answers Using probability theory concepts like the definition of conditional probability, law of total probability and Bayes Rule, it can be found that Refer to Question in Test S 9 v7a 6

37 (i) P (hamburger and drink) 8 (ii) P (drink) 6 (iii) P (hamburger drink) 9 Worked Sample Solutions We introduce some notation Let H be the even that the customer bought a hamburger and D the event they bought a drink Translating the given data into our notation, we are thus given that P (H) 6, P (D H), PD H c (i) We are asked to find P (H D) (remember, in probability essentially corresponds to and, and formally refers to the intersection of sets or events) We have P (H D) PHPD H (multiplication rule) 6 (given data) 8 (ii) We are asked to find P (D) We know from basic probability theory that P (D) P (D H) + PD H c (?) We found in (i) already that P (D H) 8 (remember, D H and H D are the same event) Also, we have P (D H c ) P (H c ) P (D H c ) (multiplication rule) (given data, noting P (H c ) P (H)) ( 6) 8 So from (?), we have P (D) (iii) We are asked to find P (H D) We have P (H D) P (D) P (H D) (definition of conditional probability) (using answers to (i) and (ii)) 7