Random QuestionsThe Determinant: Basic Definitions, and an ExampleThe Determinant: Some Exploratory TheoremsA Quick Aside: Elementary MatricesThe Determinant: Key ResultsMath 1b TA: Padraic BartlettDeterminants: An IntroductionWeek 5 Caltech 20111 Random Questions1. Consider the following game, called “scream-toes:”• To start, place n people so that they’re standing in a circle.• Have everyone look down at the ground (the “down” phase.)• Then, have each player randomly select another player’s toes, and look at them(the “toes” phase.)• Then, have all players look up at whichever player’s toes they were just lookingat. If two players are looking at each other, then they both scream (or shoutloudly, whichever they prefer.)Show that the average number of screams for a given round of scream-toes is about2/e, given enough players.2. Suppose that A is a matrix with integer entries, such that the sum of the entries ineach of the rows of A is a multiple of 7. Prove that the determinant is divisible by 7.2 The Determinant: Basic Definitions, and an ExampleDefinition. For a n × n matrix A, let Aijdenote the matrix formed from A by deletingthe i-th row and j-th column from A.Then, we can define the determinant of A recursively1as follows: for 1 × 1 matrices,we define det(A) = a11, and for larger n × n matrices A, we definedet(A) =nXi=1(−1)i−1a1i· det(A1i).To give an idea of how we use this recursive definition in practice, consider the followingexample:1A recursive definition is one that defines some object in terms of itself. In this case, we define thedeterminant of a n × n matrix in terms of the determinants of smaller n − 1 × n − 1 matrices. So, to findthe determinant of (say) a 3 × 3 matrix, we use our recursive definition to reduce the problem to finding thedeterminants of 3 different 2 × 2 matrices, and then apply the recursive definition on each of those matricesto reduce further to the case of 6 different 1 × 1 matrices, which we know how to do.1Example. Find the determinant of the following matrix:A =3 1 4 15 9 2 65 3 5 89 7 9 3Solution. By our definition, we know thatdet(A) =nXi=1(−1)i−1a1i· det(A1i)= 3 · det9 2 63 5 87 9 3− 1 · det5 2 65 5 89 9 3+ 4 · det5 9 65 3 89 7 3− 1 · det5 9 25 3 59 7 9Use the definition of the determinant again to expand each of these three by three matrices:det9 2 63 5 87 9 3= 9 · det5 89 3− 2 · det3 87 3+ 6 · det3 57 9= 9(5 · 3 − 8 · 9) − 2(3 · 3 − 8 · 7) + 6(3 · 9 − 5 · 7)= −467.det5 2 65 5 89 9 3= 5 · det5 89 3− 2 · det5 89 3+ 6 · det5 59 9= 5(5 · 3 − 8 · 9) − 2(5 · 3 − 8 · 9) + 6(5 · 9 − 5 · 9)= −171.det5 9 65 3 89 7 3= 5 · det3 87 3− 9 · det5 89 3+ 6 · det5 39 7= 5(3 · 3 − 8 · 7) − 9(5 · 3 − 8 · 9) + 6(5 · 7 − 3 · 9)= 326.det5 9 25 3 59 7 9= 5 · det3 57 9− 9 · det5 59 9+ 2 · det5 39 7= 5(3 · 9 − 5 · 7) − 9(5 · 9 − 5 · 9) + 6(5 · 7 − 3 · 9)= −24.2Now, take these values and plug them into our original equation:det(A) =nXi=1(−1)i−1a1i· det(A1i)= 3 · det9 2 63 5 87 9 3− 1 · det5 2 65 5 89 9 3+ 4 · det5 9 65 3 89 7 3− 1 · det5 9 25 3 59 7 9= 3 · (−467) − 1 · (−171) + 4(326) − 1(−24)= 98.3 The Determinant: Some Exploratory TheoremsThe determinant is a rather strange-looking thing. At first glance, it’s not remotely clearwhy we’d ever want to study it; it seems complex and convoluted, and hardly like the kindof thing we would ever intentionally want to work with.Yet, as it turns out, the determinant is an incredibly useful object! In specific, we havethe following theorem:Theorem 1 The determinant of a n×n matrix A is nonzero if and only if A is nonsingular.In other words, the determinant – a single number that we can pretty quickly find froma matrix – can instantly tell us if a matrix has an inverse, without bothering with all of therow-reduction nonsense we normally have to do. So this is remarkably useful!How can we prove such a thing? Well, at the moment, we really can’t: we barelyunderstand what the determinant does in general, and thus aren’t probably going to havemuch luck starting on this theorem right now.So, what we’ll do instead is see what we *can* prove about the determinant, and see ifwe can use these insights to eventually try and prove this result.To start: if we want to understand what the determinant does to matrices, we shouldbegin by looking at what it does to the simplest matrix we can think of: the identity matrix!Theorem 2 If Inis the n × n identity matrix, then det(In) = 1.Proof. So: how do we prove things about the determinant? Well, we defined the deter-minant recursively, with a definition that told us how to find 1 × 1 determinants and howto build up this knowledge to find n × n determinants. A natural idea, then, would be toprove things in a similar way: to demonstrate thing for a base case, and then to build upthese results for larger matrices. In other words, we want to use induction! Specifically, forthis proof, let’s proceed by induction on n.For the 1 × 1 matrix I1= (1), our claim is trivially true: det(I1) = 1.For our inductive step, we assume that our hypothesis holds for n, and seek to proveour claim for n + 1.3So: by definition, we know thatdet(In+1) = (−1)0· 1 · det((In+1)11) + (−1)1· 0 · det((In+1)12) + . . . + (−1)n−1· 0 · det((In+1)1n)= det((In+1)11).However, removing the first row and first column of the n + 1 × n + 1 identity matrix justleaves In, the n × n identity matrix: so this is just det(In), which we know to be 1 from ourinductive hypothesis.We now understand the identity matrix. What should we look at now?Well: one of the most basic things we can do to a matrix are the various row operations:i.e. given a matrix, we will often either• multiply some row by a constant λ,• swap two rows, or• add λ times one row to another.What do these three properties do to the determinant? I.e. if we have a matrix andperform one of these row operations, how does the determinant change?We explore this in the next three theorems:Theorem 3 Suppose that A is a n × n matrix. If A0is the matrix acquired by multiplyingthe k-th row of A by some constant λ, then det(A0) = λ det(A).Proof.