(3) If you replace row i of A by its sum with a multiple of another row, then the determinant is unchanged! Expand across the i th row:

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1 Math Tue Feb 4 Cotiue with sectio 36 Determiats The effective way to compute determiats for larger-sized matrices without lots of zeroes is to ot use the defiitio, but rather to use the followig facts, which track how elemetary row operatios affect determiats: (a) Swappig ay two rows chages the sig of the determiat proof: This is clear for matrices, sice a b c d c d ad bc, a b cb ad For 3 3 determiats, expad across the row ot beig swapped, ad use the swap property to deduce the result Prove the geeral result by iductio: oce it's true for matrices you ca prove it for ay matrix, by expadig across a row that was't swapped, ad applyig the result (b) Thus, if two rows i a matrix are the same, the determiat of the matrix must be zero: o the oe had, swappig those two rows leaves the matrix ad its determiat uchaged; o the other had, by (a) the determiat chages its sig The oly way this is possible is if the determiat is zero (a) If you factor a costat out of a row, the you factor the same costat out of the determiat Precisely, usig i ith row of A, ad writig * i i : i c : i * c : i * proof: expad across the i th row, otig that the correspodig cofactors do't chage, sice they're computed by deletig the i th row to get the correspodig miors: det A j a i j C i j j * c a Ci i j j c j * a Ci i j j c det A * (b) Combiig (a) with (b), we see that if oe row i A is a scalar multiple of aother, the det A 0 (3) If you replace row i of A by its sum with a multiple of aother row, the the determiat is uchaged! Expad across the i th row:

2 i c k k j a c a C a C i j k j i j j i j i j c a C det A j k j i j c k k det A 0 Remark: The aalogous properties hold for correspodig "elemetary colum operatios" I fact, the proofs are almost idetical, except you use colum expasios

3 Exercise ) Recompute 0 3 from yesterday (usig row ad colum expasios we always got a aswer of 5 the) This time use elemetary row operatios (ad/or elemetary colum operatios) Exercise ) Compute Maple check: > with LiearAlgebra : > A Matrix 4, 4,, 0,,,,,, 0,, 0,,,, 0,, ; Determiat A ; 0 A ()

4 Theorem: Let A The A exists if ad oly if det A 0 proof: We already kow that A exists if ad oly if the reduced row echelo form of A is the idetity matrix Now, cosider reducig A to its reduced row echelo form, ad keep track of how the determiats of the correspodig matrices chage: As we do elemetary row operatios, if we swap rows, the sig of the determiat switches if we factor o-zero factors out of rows, we factor the same factors out of the determiats if we replace a row by its sum with a multiple of aother row, the determiat is uchaged Thus, A c A c A c c N rref A where the ozero c k 's arise from the three types of elemetary row operatios If rref A I its determiat is, ad A c c N 0 If rref A I the its bottom row is all zeroes ad its determiat is zero, so A c c N 0 0 Thus A A exists 0 if ad oly if rref A I if ad oly if Remark: Usig the same ideas as above, you ca show that det A B det A det B This is a importat idetity that gets used, for example, i multivariable chage of variables formulas for itegratio, usig the Jacobia matrix (It is ot true that det A B det A det B ) Here's how to show det A B det A det B : The key poit is that if you do a elemetary row operatio to AB, that's the same as doig the elemetary row operatio to A, ad the multiplyig by B With that i mid, if you do exactly the same elemetary row operatios as you did for A i the theorem above, you get A B c A B c A B c c N rref A B If rref A I, the from the theorem above, A c c N, ad we deduce A B A B If rref A I, the its bottom row is zeroes, ad so is the bottom row of rref A B Thus A B 0 ad also A B 0

5 There is a "magic" formula for the iverse of square matrices A (called the "adjoit formula") that uses the determiat of A alog with the cofactor matrix of A I order to uderstad the magic formula for matrix iverses, we first eed to talk about matrix trasposes: Defiitio: Let B m b i j The the traspose of B, deoted by B T is a m matrix defied by etry i j B T etry j i B b j i The effect of this defiitio is to tur the colums of B ito the rows of B T : etry i col j B b i j etry i row j B T etry j i B T b i j Ad to tur the rows of B ito the colums of B T : etry j row i B b i j etry j col i B T etry j i B T b i j Exercise 3) explore these properties with the idetity T

6 Theorem: Let A, ad deote its cofactor matrix by cof A C i j, with C i j i j M i j, ad M i j the determiat of the matrix obtaied by deletig row i ad colum j from A Defie the adjoit matrix to be the traspose of the cofactor matrix: Adj A cof A T The, whe A exists it is give by the formula A det A Adj A Exercise 4) Show that i the case this reproduces the formula a b d b c d ad bc c a Exercise 5) For our fried A 0 3 we worked out cof A ad det A 5 Use the Theorem to fid A ad check your work Does the matrix multiplicatio relate to the dot products we computed betwee various rows of A ad rows of cof A?

7 Exercise 6) Cotiuig with our example, 5 6 A 0 3 cof A Adj A a) The, etry of A Adj A is Explai why this is det A, expaded across the first row 6b) The, etry of A Adj A is Notice that you're usig the same cofactors as i (4a) What matrix, which is obtaied from A by keepig two of the rows, but replacig a third oe with oe of those two, is this the determiat of? 6c) The 3, etry of A Adj A is What matrix (which uses two rows of A) is this the determiat of? If you completely uderstad 6abc, the you have realized why A Adj A det A I for every square matrix, ad so also why A det A Adj A Precisely, etry i i A Adj A row i A col i Adj A row i A row i cof A det A, expaded across the i th row O the other had, for i k, etry k i A Adj A row k A col i Adj A row k A row i cof A This last dot produce is zero because it is the determiat of a matrix made from A by replacig the i th row with the k th row, expadig across the i th row, ad wheever two rows are equal, the determiat of a matrix is zero

8 There's a related formula for solvig for idividual compoets of x whe A x b has a uique solutio ( x A b ) This ca be useful if you oly eed oe or two compoets of the solutio vector, rather tha all of it: Cramer's Rule: Let x solve A x b, for ivertible A The x k det A k det A where A k is the matrix obtaied from A by replacig the k th colum with b proof: Sice x A b the k th compoet is give by x k etry k A b etry k A Adj A b Notice that col k cof A A row k Adj A b A col k cof A b b is the determiat of the matrix obtaied from A by replacig the k th colum by b, where we've computed that determiat by expadig dow the k th colum! This proves the result 5 x Exercise 7) Solve 7 4 y 7a) With Cramer's rule 7b) With A, usig the adjoit formula

9 Math Week 7 otes: Sectios 4-43 vector space cocepts Tues Feb Fiish sectio 36 o Determiats ad coectios to matrix iverses Use last week's otes The if we have time o Tuesday, begi: 4-43 The vector space m ad its subspaces; cocepts related to liear combiatios of vectors We ever wrote it dow carefully i Chapter 3, but for ay atural umber m,, 3 the space m may be thought of i two equivalet ways I both cases, m cosists of all possible m tuples of umbers: (i) We ca thik of those m case we ca write (ii) We ca thik of those m case we ca write tuples as represetig poits, as we're used to doig for m,, 3 I this m x,, x m, st x,, x m tuples as represetig vectors that we ca add ad scalar multiply I this x m x : x m, st x,, x m Sice algebraic vectors (as above) ca be used to measure geometric displacemet, oe ca idetify the two models of m as sets by idetifyig each poit x,x m i the first model with the displacemet vector T x x,x m from the origi to that poit, i the secod model, ie the positio vector (Notice we just used a traspose, writig a colum vector as a traspose of a row vector) Oe of the key themes of Chapter 4 is the idea of liear combiatios These have a algebraic defiitio (that we've see before i Chapter 3 ad repeat here), as well as a geometric iterpretatio as combiatios of displacemets, as we will review i our first few exercises Defiitio: If we have a collectio of vectors v, v, v i m, the ay vector v m that ca be expressed as a sum of scalar multiples of these vectors is called a liear combiatio of them I other words, if we ca write v c v v c v, the v is a liear combiatio of v, v, v The scalars c,,, c are called the liear combiatio coefficiets