Birne Binegar: Well, I guess, so first thing I want to point out is.
Birne Binegar: After today's lecture we're all talk about two applications of the terminal creaminess rule and how to calculate the inverse of the matrix shooting determinants.
Birne Binegar: This material, as well as what we discussed a Spar how to compute determinant that's going to be covered.
Birne Binegar: In some homework problems in homework set eight so i'm going to make that do on Monday, just so that we're all essentially caught up with lecture.
Birne Binegar: Next week, when we start really using determinants to find eigenvectors and eigenvalues so today's plan exactly.
Birne Binegar: Not not not too advanced, what I want to do is give you two formulas that allow you to solve old problems.
Birne Binegar: Using using determinants, the main thing i'm trying to show you in today's lecture is how the algorithm the real reduction algorithm can be avoided, so long as you have a determinant in your hand and.
Birne Binegar: determinants are such an important function yeah any computer software mathematical computer software program is worthy of any respected all will have a function, or you feed in a matrix and it kicks out the determinant so in practice.
Birne Binegar: i've given you a number of formulas, in fact, three formulas for computing the determinant but there, they tend to be hard to apply, simply because a.
Birne Binegar: If you're trying to write out the determinant but an arbitrary matrix it's a polynomial in the entry to the matrix but it's a huge polynomial for any matrix of reaching all sides, I think example I like to give us a simple six by six matrix it's determinant if you.
Birne Binegar: Think of the entries of the matrix of variables is going to be a polynomial up degree six not so bad, but it's going to have 720 distinct terms, so and then, if you use.
Birne Binegar: The permutation formula for computing determinant there again you're you're trying to.
Birne Binegar: express the entire determinant as a function of it, the the entries of a matrix those are end up can be polynomials as well and hard to calculate for matrices of even moderate size.
Birne Binegar: You want to be really efficient way of computing determinism matrices a moderate sized is row reduction, but now so that's kind of contrary to the theme i'm trying to present today and that's.
Birne Binegar: Why we use row reduction sometimes to calculate determined in practice, you don't because you'll any matrix up a moderate to if you would anyway, once we leave this course, be using computers or something else to compute those determinants just like you don't use.
Birne Binegar: Taylor expansions to compute the value of cosine of three rather you plug it into your calculator and get it that's that's essentially.
Birne Binegar: How you'll probably be using determinants in the future.
Birne Binegar: Okay, so the goal for today now is to show you two applications where we don't know how to solve the.
Birne Binegar: problem post except by using row reduction i'm now going to stick give you two rules for calculating, for example, the solution set of a linear system without doing any row reduction or.
Birne Binegar: The process of finding the inverse of the matrix we can carry that out without doing any row reduction at all what's nice about this is.
Birne Binegar: I make a distinction between algorithms and functions algorithms are just a set of rules that you apply, you can reach one stage and then you, you see what you got last stage and you.
Birne Binegar: figure out something new do like the row reduction algorithm you start in the upper corner and you remove to this below, then you.
Birne Binegar: move to the next, lower pivot you move out the entries below etc, but you don't really see the answer until you reach the answer.
Birne Binegar: On the other hand, when things are expressed in terms of formula clothes formulas, you can see some of the structure of the answer before you actually calculated in, for example.
Birne Binegar: they'll be conditions where what we'll see when we actually get to this co-factor matrix is that.
Birne Binegar: The inverse of the matrix it's entries are actually nice rational functions of the of the entries of the original matrix.
Birne Binegar: means just a ratio of polynomials the worst and oftentimes you can actually arrange it so that the.
Birne Binegar: entries of the inverse of the matrix or just polynomials and the entries of the original matrix which makes it even though it's still kind of loose and doesn't specify what the matrix in inverse is exactly at least it shows you the functional form.
Birne Binegar: And, in fact, if you had to integrate the inverse of the matrix it'd be nice to know that the polynomial entry to all polynomials and so are easily integrated.
Birne Binegar: Okay, so let me give you the first rule, it concerns solving linear system.
Birne Binegar: So called kramer's role.
Birne Binegar: and
Birne Binegar: concerns.
Birne Binegar: linear systems so coefficient matrix times the vector variables is equal to another vector but.
Birne Binegar: Unfortunately, because we're going to take some determinants and you can only take determinants of square matrix these.
Birne Binegar: This is this rule is only going to apply to square linear systems where we have as many patients, as we have unknowns, and in addition we're gonna have to require the the inverse of the coefficient matrix exists.
Birne Binegar: alright.
Birne Binegar: So if we have a system of equations and then unknowns and I know that this matrix a has an inverse.
Birne Binegar: Then, in fact, i'm going to need to calculate i'm only going to need to calculate the determinant of a any way to apply this formula.
Birne Binegar: So he doesn't have an inverse what's going to happen is this formula here it's not going to make sense, because you'll be trying to divide by zero remember the determinant.
Birne Binegar: It detects whether or not a matrix has an inverse or not, if it's equal to zero you don't have an inverse if it's not equal to zero, then it will have an inverse so.
Birne Binegar: The fact that this denominator, that the reason why I say in vertical systems is, I want to guarantee this denominator is not equal to zero, so that this formula actually makes sense.
Birne Binegar: But given that.
Birne Binegar: Once I have a linear system, what you can do is you can take the data, the coefficient matrix and the vector on the right hand side, and you can use this to create a bunch of auxiliary matrices so.
Birne Binegar: For example, be one this is going to be the matrix that you formed by taking this vector writing it as a column, and then the remaining columns of this matrix are just going to be the columns a so this second column, of the one is going to be the second column of a.
Birne Binegar: As is and then last call a.
Birne Binegar: Written vertically and these other matrix these I have.
Birne Binegar: As many of these matrix is bi says, I have variables because.
Birne Binegar: Counting cross the top X one X two etc X and the way we usually assign variables two columns we're going to have exactly.
Birne Binegar: we're going to have one variable for each column of this matrix, and that means effectively we're gonna have one component of the solution as well because.
Birne Binegar: We have any of these matrices one for each column of the original matrix and yeah so.
Birne Binegar: That makes it up perfectly good sense and then the statement is, if you want the solution vector.
Birne Binegar: X equals X one X two X and then.
Birne Binegar: The value of the ice component of that is just the determinant.
Birne Binegar: Of the matrix I create by replacing the eighth column of the original matrix by the the vector on the right hand side of the original in your system and I divide that determinant by the determinant a.
Birne Binegar: dope that's the formula it's pretty simple formula right down and it's pretty simple to implement, as well as i'll do with a quick example.
Birne Binegar: Right so discussing here the ladies auxiliary matrices are.
Birne Binegar: written down if you just take the original matrix your race like get a given call I call them and you just replace it with be.
Birne Binegar: So here's here's an example if we have a two by two linear system here's the coefficient matrix to us the coefficient of the first variable in the.
Birne Binegar: First equation, one is the coefficient of the second variable in the SEC, in the first equation, the second row is similarly determine, then you have the column vector on the right hand side that you get simply by.
Birne Binegar: writing the numbers on the right hand side of the original equation writing them as a column.
Birne Binegar: Alright, so let's quickly compute the determinant of the coefficient matrix here it's a two by two matrix so we'll just use the different to the diagonal products two times minus one minus one times two.
Birne Binegar: These two formulas.
Birne Binegar: That are right up on the blackboard here.
Birne Binegar: These you should just know if you have a one by one matrix it's determine is a one determinant of a two by two matrix a B, C D is the difference of the diagonal products.
Birne Binegar: So.
Birne Binegar: always use these two basic formulas throughout the thing, even like computing.
Birne Binegar: determinants of three by three matrix etc what I tend to do is use the Co factor expansion formula to calculate determine so forth, for larger major sees.
Birne Binegar: determine it's a harder to compute but what you use this.
Birne Binegar: generally well there's three different ways to compute them, but let me give you a way that will use today.
Birne Binegar: There are some.
Birne Binegar: co-factor expansion formula.
Birne Binegar: we're in the determinant of the matrix.
Birne Binegar: it's given by some over either roads or columns let's say, well, some of the columns in this first instance.
Birne Binegar: You have the.
Robert Triplett: entry.
Birne Binegar: In the ice row and jake column times assign factor that's the some of the row and column index or a half here, and then you have the determinant.
Birne Binegar: But not jewelry matrix call the ice gave minor have a what this matrix is it's a with like row.
Birne Binegar: And jade column.
Birne Binegar: deleted.
Birne Binegar: So it's a smaller matrix he has one less row and um one last call, but so in general for a matrix that's larger than a two by two matrix you're going to have to.
Birne Binegar: use a more complicated formula for the determine the one will use mostly today it's co factor expansion formula that.
Birne Binegar: If I had a three by three matrix this formula will allow me to compute the determinant of three by three matrix I summon over a bunch of determines of two by two matrix and then look for these guys i'd be able to use want one of these formulas and, of course.
Birne Binegar: Here i'm keeping the row index sticks and i'm something over the column indices, so this corresponds to moving across the ice grow of a matrix and picking up various terms but there's an alternative formula where instead of something over the column indices you some over the row indices.
Birne Binegar: bye but except for how about some is arrange.
Birne Binegar: The other terms are the thing you some looks has exactly the same form, I play J terminal.
Birne Binegar: Am on Jay.
Birne Binegar: I should remark, also, that when you're competing the determinant of a large matrix.
Birne Binegar: Which which formula which are these formulas you apply and, in particular, which row or which column you keep fix you usually look for rose work lots of entries or zeros because, if I.
Birne Binegar: enter a zero that means this factor is going to be zero that's going to mean it's not going to contribute to the total sum, so you won't have to compute that matrix.
Birne Binegar: Alright, so that's sort of them stuff we've already covered let's.
Birne Binegar: let's now work through this example.
Birne Binegar: System of linear equations coefficient matrix the in homogeneous term on the right hand side expresses a vector.
Birne Binegar: Now we calculated i've calculated determine a and now what i'm going to do is create these auxiliary major cities, so the first one.
Birne Binegar: is where I replaced the first call a with this vector the second column didn't change, but I did play the first one and let's calculate it's determined that.
Birne Binegar: it's two by two so i'll just do the difference of the diagonal product, and then I have a second auxiliary matrix that i'll just replace the second column of a with the specter and i'll compute this determinant and now I have all the information I need to apply kramer's cool.
Birne Binegar: stuff I don't have it written on the board anymore.
Birne Binegar: kramer's world says.
Birne Binegar: Yes, X equals.
Birne Binegar: X one X N, is a solution vector.
Birne Binegar: horizontally.
Birne Binegar: But I could have written it vertically as well if it's a solution vector then each of these components.
Birne Binegar: Is calculable not by doing necessarily doing row reduction but rather by comparing These determinants.
Birne Binegar: The I decided by.
Birne Binegar: The terminal okay.
Birne Binegar: Okay, so let's.
Birne Binegar: we've got all the data computed here we've got.
Birne Binegar: A we've got all these determines computed we've got the terminal a computed will have that denominator is always going to be equal to negative three and then the one is it's determined is negative three be too is it's determinant negative six and so.
Birne Binegar: At first component of the solution vector is one second component of the solution factor is.
Birne Binegar: too.
Birne Binegar: And so that's the solution, the value of X X one X two that will satisfy the original into your system or X equals one X two X equal to two.
Birne Binegar: And indeed here's the original matrix a here's what i'm proposing to be a solution vector vector with component one.
Birne Binegar: and two and I multiply those together, I get a vector for minus one but that's identical to the vector we had originally on the right hand side of the equation, so I verified that this vector athletes satisfies the original equation.
Birne Binegar: Okay, so it's that simple to apply and, in fact.
Birne Binegar: When you're solving linear large linear systems typically what you'll do is once you get past this course is asked computer to solve it for you.
Birne Binegar: This is really fast, you can see, as long as the computer knows how to compute determinants this is really rapid to i'm not for it to carry out so it's typically four square linear systems wait computer algebra system rapidly sold large systems of linear equations.
Birne Binegar: Alright, so.
Birne Binegar: A second thing we use row reduction for.
Birne Binegar: A number prior to this rule you only way we had of solving the system with by writing down an augmented matrix row reducing to row Echelon form.
Birne Binegar: Taking writing down the corresponding equation expanding in terms of the free variables, so that was quite a long process, and in fact you didn't have any idea of what the solution would look like and until you reach the last.
Birne Binegar: The last step when you expand in terms of the free variables, but this calculation.
Birne Binegar: Effectively, gives you an explicit formula, this is going to be a polynomial in the entries of a but it's going to be linear in be because each column it's been only going to have one entry from be.
Birne Binegar: So you can actually extract certain formal information about the solutions of this from this formula of these are just polynomials in the entries of a oh this guy yeah it's going to actually be linear in this variable there.
Birne Binegar: So what I want to do next, then, is i'll give you another formula that allows us to avoid row reduction and.
Birne Binegar: A calculation previously, we can only carry out using row reduction.
Birne Binegar: let's see that up here.
Birne Binegar: Let me remind you of the co factor expansion.
Birne Binegar: determinant of the matrix that determinant of a matrix a is equal to a some override equals one and let's do columns G equals one the end a iga minus one the I plus J times the determinant of the IGF minor.
Birne Binegar: or summation.
Birne Binegar: equals one to n.
Birne Binegar: H I J minus one day I played a determinant Am I J don't hear the two formulas, in the first version, what we do is we keep a row index sticks and then we move across a row grabbing the entry.
Birne Binegar: assign factor that really only depends on the position of this entry is like the matrix and then the determinant of the matrix smaller than the original matrix that's obtained by removing the ICO and jake calm.
Birne Binegar: But there's an alternative formula, instead of moving across the ice row you could move down the jade column, as you move down to a column you change the row index.
Birne Binegar: But you have a similar formula, and if you just ignore the some these two factors are exactly the same in both formulas exactly but i'm just how you carry out the sun, the changes and so these guys all right see my J defined as.
Birne Binegar: minus one to the I J.
Birne Binegar: determinant Am I J.
Birne Binegar: This common factors in both formulas bigger called co factors.
Birne Binegar: And it's not there for each row index in each column index you're going to have a distinct minor, to which you can compute it to determine and so you'll have.
Birne Binegar: n squared at these numbers to compute if you want to compute all the Co factors is going to range between one and and and the.
Birne Binegar: column index J is going to range between one and all, right now, these guys, which are computed you might take them determine to minors we're going to use those to calculate the inverse of the matrix.
Birne Binegar: So the first thing we do before I even give you the formula for the inverse is we take these numbers.
Birne Binegar: let's as soon as I give you a matrix a.
Birne Binegar: You can compute all these numbers explicitly so let's assume we've computed all these numbers, I have n squared numbers.
Birne Binegar: labeled by two indices a row index in a column index I can use those numbers and I can just arrange them in the matrix in a natural way here through index and here's the column index by simply taking this set of numbers and arranging them according to the labeling up there.
Birne Binegar: What well the way they're labeled we can arrange them in a n by n matrix that matrix is called the co-factor matrix of a.
Birne Binegar: But what's going to happen is you're going to form a matrix see where you have seen one one in the first column C one, two, in the second column C one and last column of the first row and then you'll have several entry, see no one to see and.
Birne Binegar: Generally speaking, there's a lot of these numbers to compute you have a three by three matrix you have nine numbers to compute.
Birne Binegar: But let's assume that we have a good computer software program and for which, once we feed in a these determines are all readily evaluated and we just use them, the plugin to create a new matrix see.
Birne Binegar: Alright, so here's the formula.
Birne Binegar: If a isn't in verbal matrix than the entry in the I throw and jade column of it inverse is well there's one determinant one over the determinant of a and then we're going to use these numbers.
Birne Binegar: That we created here, however.
Birne Binegar: important thing to note about this formula, and this is not a typo the row indexing column index have been reversed that's essential to get this formula work, but with the small caveat here that you have to reverse the rules of the.
Birne Binegar: Road and column index there, this gives you a way of calculating the inverse of a matrix using only determinants.
Birne Binegar: This is clearly a determinant and those numbers The co factors well they're defined as determinants of miners have the original matrix.
Birne Binegar: And so, this formula then allows me to calculate all the entries of the inverse of a so, in fact, a minus first itself like simply using determinants.
Birne Binegar: All right now.
Birne Binegar: that's the way I like to explain it, but the textbook energy does it instead of one one fell swoop is I did it they introduce an auxiliary matrix called the yet another matrix called the matrix as you get this to a what that is it's the.
Birne Binegar: You calculate that matrix see and then you take it transpose so you convert rose in the columns what that will do.
Birne Binegar: Is.
Birne Binegar: Here the roles of rows and columns is a row here, but I, as a column index for see jays are a column index for a but it's going to be.
Birne Binegar: A row index for this matrix see so there's a discrepancy between what we interpret as a row index and the column index depending on which side of the equation, we look at.
Birne Binegar: Well, we can get rid of that discrepancy by simply taking the matrix see interchanging it's rows and columns and now everything will match up.
Birne Binegar: The textbook in web assign they refer to the transpose of this matrix see that we're creating this way they're calling that the matrix educate the matrix aggregate today.
Birne Binegar: But in terms of that matrix, then I can express a formula directly in terms of the matrix these rather than in terms of the entries of the matrix, and so the.
Birne Binegar: inverse of the matrix a is one over the determinant of eight times.
Birne Binegar: The transpose the aggregate metric or the transpose of this matrix.
Birne Binegar: again.
Birne Binegar: i'm only using determinants this ad junk matrix it just the transpose of C and C itself is computed by taking a bunch of determinants of miners of the original matrix so.
Birne Binegar: You calculate in applying this formula, all I have to do is compute some determinants a lot of them, but it's only involving computation to that determines and, in fact, let me see if I can point on ella feature of this.
Birne Binegar: This um.
Birne Binegar: These are co factors there determinants of.
Birne Binegar: miners have a each entry of this matrix is an entry of the original matrix a just that i've deleted a certain role in a certain call, so these matrices their entries are all entries in a they're determined then it's just going to be a polynomial in the entries of a so, in fact.
Birne Binegar: These entries see one one see want to see one and in the end, these are each polynomials in the entry of age, so we see that now taking at least computing this co-factor matrix nothing more complicated from going from the entries of a two it's The co factor matrix it's just.
Birne Binegar: The individual entries of this matrix or just polynomials in the entries of the original matrix that's not quite true of this inverse matrix we didn't have this factor here.
Birne Binegar: Taking the transfer.
Birne Binegar: Taking the transpose of a matrix is just putting them in different places it's not changing how they depend on the entries of a you'll still be polynomials where the.
Birne Binegar: singularities where the bloody non singularities in this inverse occurred a occur because of this denominator determine if a so.
Birne Binegar: This is going to be a polynomial entries of a the entries of this matrix are all going to be polynomials in the entries of a, but when I combine them in this fashion, what I get is a rational function right because I have a bunch.
Birne Binegar: a bunch of a matrix of bunch of polynomial entries divided by a polynomial polynomial divided by a polynomial, generally speaking, a rational function but still.
Birne Binegar: Not too bad pretty pretty fairly simple function to to deal with not like we have a complicated transcend transcendental functions relating non entries of a matrix to its inverse.
Birne Binegar: Okay, so let me apply this formula with an example.
Birne Binegar: here's a two by two matrix I want to show you the the polynomial structure explicitly here the rational function explicitly here so let's start with a two by two matrix a B, C D.
Birne Binegar: For this we can do everything symbolically, we can these first co-factor it's going to be the one one co factor, so we take.
Birne Binegar: i'm just applying this formula I look at where the entry occurs and i'm grabbing a column index a row index and a column index and then i'm computing flying that's going to.
Birne Binegar: be the.
Birne Binegar: evening or on depending on how the row index in column index and up and then i'll have a determinant of a minor I get by removing the I throw in jake all from the original matrix.
Birne Binegar: So here we are applied here zero indexing column index or both one there's the sign factor.
Birne Binegar: And then, when I remove the first row and I removed the first column unless before one by one matrix but the single entry D.
Birne Binegar: So this is just going to work in the determinant of a one by one matrix is just the single entry, so I have a bathroom one because it's minus one to an even power and determinant of a one by one matrix.
Birne Binegar: We can do the same for each of the other co factors for this entry here the entry is be.
Birne Binegar: be assigned if this isn't the one to position first row second column, so the sign is going to be minus one to an odd number, and then the.
Birne Binegar: Minor associated with this entry is the one by one by one matrix with salt entry see until it works out that and, similarly, you get us to one entry, the two to entry corresponds to this font.
Birne Binegar: there's the minor and the sign is because this is in the two to spot is minus one to the two plus two, and now we have all the ingredients that we computed all the entries of this matrix.
Birne Binegar: There it is I just take.
Birne Binegar: This natural ordering here and i've insert.
Birne Binegar: The results and the corresponding slot the matrix see.
Birne Binegar: Now the adjunct matrix it's going to be the transpose of this matrix so to write it down, I take the first column of see.
Birne Binegar: To get his transpose I write that first column is the first row you get the second row I take something home and write it in the same the same numbers in the same order but I write them horizontally alright, so now i've got.
Birne Binegar: i've got everything except the formula that we need.
Birne Binegar: A formula is the inverse of a.
Birne Binegar: is equal to one over the determinant of a times this adjunct matrix which all right, a C transpose.
Birne Binegar: and
Birne Binegar: Here it is and so.
Birne Binegar: there's the formula that will apply.
Birne Binegar: For a two by two matrix the determinant it's going to be a times D minus B times see explains that factor and then there's the aggregate matrix and we're done and what you, what do you see about this inverse it is.
Birne Binegar: This matrix see as advertised it's a nice haul normally on jackets linear and the entries of the original matrix and.
Birne Binegar: The inverse though has this additional factor coming from the determinant a that appears in denominator, so if you have problems with the individual entries, they have a inverse of the matrix it's they're probably coming from this denominator.
Birne Binegar: that's where you can have singularities but everywhere else it's a perfectly good function.
Birne Binegar: All right, and then i'll prove to you that, in fact.
Birne Binegar: The matrix given by the formula is indeed the inverse of a by simply multiplying the things together.
Birne Binegar: When you multiply these two matrices together, this is what result notice that the off diagonal entries cancel minus C plus ch going to equal zero this.
Birne Binegar: db minus B D is going to equal zero, and then the things that along the main diagonal they're actually going to be canceled by this factor, a D minus B, C is exactly that factor, as is minus C B plus ad until we get exactly the inverse of the matrix that way.
Birne Binegar: All right now what's nice about.
Birne Binegar: about this formula, if you can it's not so hard to memorize and it's really useful even outside linear algebra.
Birne Binegar: Two by two linear systems.
Birne Binegar: As mx plus B y is equal to some number D C D y is equal to see those are fairly common when you have systems have two equations and to unknown.
Birne Binegar: Here you get the inverse matrix of the coefficient matrix from which you can use to solve the linear system and it's not that hard to even memorize.
Birne Binegar: You need to get a need to use the determinant of a well that's not hard to write down for two by two matrix a D minus BC and then what happens to a itself is two things the on this main diagonal they switched positions and along the opposite diagonal they just switch songs but.
Birne Binegar: This formula is really not that hard to memorize and it gives you if you've long as you got it memorized you can almost instantly calculate the determinant of any to calculate the inverse of any two by two matrix.
Birne Binegar: So, for example, let me see if I can work on that.
Birne Binegar: i'll do it right here, because it's so simple to pose I want to know, I have a is equal to one two.
Birne Binegar: To one what is.
Birne Binegar: A minor spurs well.
Birne Binegar: I can yeah i'm just i'm this formula is given in terms of arbitrary injuries, all I have to do is plug into this formula so it's going to equal a the minus first it's going to be one over.
Birne Binegar: eight times D, which is one minus two times two, which is four.
Birne Binegar: And then.
Birne Binegar: Then, what I do is I switch the positions of the diagonal elements well they're the same solo, even though I switched positions they look the same and then what you do to the off diagonal elements you switch their song, so this matrix I could write down really quickly.
Birne Binegar: Using this formula is going to be the inverse of a a the minus first that can be looks like.
Birne Binegar: This, it can be a factor of negative one third so it's going to be minus one third.
Birne Binegar: plus two thirds.
Birne Binegar: plus two third minus one third.
Birne Binegar: main main point of this is, I calculated this inverse without doing any row reduction.
Birne Binegar: Now.
Birne Binegar: These formulas that like thus far, only applied for two by two matrices can.
Birne Binegar: Also work equally well for larger matrices.
Birne Binegar: tried to keep the formulas up.
Birne Binegar: And just give you some more examples so here's a.
Birne Binegar: Three by three systems.
Birne Binegar: To will use kramer's rules to solve it so we're going to use this point.
Birne Binegar: Well, what we need to apply that formula we need to construct these auxiliary matrices and take their determinant.
Birne Binegar: we're also going to need to determine a.
Birne Binegar: So we started off writing down converting the equations into the matrix seats that encapsulate the data that goes into specifying those equations and now let's see i'm going to compute the determinant of a eventually.
Birne Binegar: Ah yeah so the determinant a pretty easy way of calculating it.
Birne Binegar: looks like what I chose to do here is an expansion go factor expansion along the bottom row, so I have a zero entry, then I have a one.
Birne Binegar: And I have a sign factor associated with this position as well it's going to be minus one.
Birne Binegar: race, the power three plus two because this entry is in the third row and second call well minus one to an odd power it's going to give you a minus sign.
Birne Binegar: So I have the entry assign and then I have the determinant of the corresponding minor when I remove the second column in the second row and left with a matrix 1001 that's this matrix and similarly.
Birne Binegar: i'll have the contribution from the three entry of this matrix there's the entry itself the sign wealth minus, this is the three three entry, so the sign associated with this position.
Birne Binegar: is going to be minus one to the three plus three minus one to the six is one so there's no sign change here, and then the corresponding minor is.
Birne Binegar: One minus 112.
Birne Binegar: All right, until we quickly calculate determine of A to B for.
Birne Binegar: Then we have these are jewelry matrices so in each case i'm replacing one of the columns of a with the.
Birne Binegar: column vector 602 that's coming from the original system.
Birne Binegar: So there's the first one first column has been replaced second one, and the third one.
Birne Binegar: And now, like compute some determinants.
Birne Binegar: compute the one i'll do a Co factor expansion along the second row.
Birne Binegar: there's the.
Birne Binegar: first term in the last term i'm not going to contribute, because the entry 00 and I only have to compute one determinant to get the determinant of this three by three matrix.
Birne Binegar: The second one here's another nice convenient row I can do it to co factor expansion, on the second row so only have to compute one determine it works out like that.
Birne Binegar: And then, for this third matrix well there's no row or column that only has a single non zero entry well I have maybe a couple choices for.
Birne Binegar: co-factor expansions that will require me to calculate to determinant I think what i've done here is i've done a Co factor expansion along.
Birne Binegar: The bottom row there's 01 with the right sign and then the two with the right sign as well, so.
Birne Binegar: Anyway, i've reduced computation of the determinants of V1 V2 v3 to the complication of a bunch of two by two matrix ease and I carry that out, and then I can write down the solution now here my labeling of variables as a bit different.
Birne Binegar: And he originally in your system, I called my variable X, Y and Z instead of X one X two X three so when I write down the solution i'm going to use the original sort of nomenclature.
Birne Binegar: So the first X one it's going to be that first dog determinant that first auxiliary matrix divided by determining a.
Birne Binegar: And that's going to work out to be according to our competitions at by two by four or two want the second variable it's going to be minus 44.
Birne Binegar: And the value of the third variable in the solution is going to be 12 divided by four so very rapidly and using only determinants you can solve square linear systems.
Birne Binegar: Alright let's do an example of a inverse calculation for a three by three matrix as well.
Birne Binegar: This is going to be a little bit more Labor intensive because we're going to have to compute all the Co factors that matrix and there's going to be nine of them, one for each slot of the original matrix.
Birne Binegar: So let's start computing then.
Birne Binegar: The first one going to correspond to the Co factor associated with this position, we have.
Birne Binegar: not compete these co factors i'm ignoring the actual entry here i'm really just interested in.
Birne Binegar: The position here and the associated minor so here I get a sign factor from the position and also a minor I get like the leaving the row and entry, corresponding to the one one entry of the original matrix.
Birne Binegar: don't get that calculate like calculate all three of the co factors associated with the entries in the first round i'm going to pause here because.
Birne Binegar: let's see because.
Birne Binegar: In my here's my formula i'm going to do apply this formula to get the inverse i'm going to anyone also need to have the determinant of a.
Birne Binegar: We can compute determine choosing co factors, and so I can actually use this calculation, not only to partially create this matrix see, but I can also use these three co factors to compute the determinant a.
Birne Binegar: Because I can do a Co factor expansion of this matrix along that first row, what do I get.
Birne Binegar: On that first row, we would fix the row index to be one, and then we we compute The co factors associated on that first row well it's exactly these numbers.
Birne Binegar: This this is this going to be the Co factor see on one J.
Birne Binegar: we've computed all these numbers, now we just put them together and we see that the determinant a is minus two.
Birne Binegar: Alright, so what i've got so far as i've got.
Birne Binegar: Determined determine up a that's been computed.
Birne Binegar: And i've got.
Birne Binegar: The first row see one one if you want to see one three at the Co factor matrix computed, but I still have.
Birne Binegar: More entries of that co-factor matrix to compute before I can apply this formula.
Birne Binegar: let's go ahead and get those your second row cool factors.
Birne Binegar: Well, look at this guy.
Birne Binegar: yeah I guess I did out so in this entry here i'm going to need a sign that sign is going to be.
Birne Binegar: This is in the two one positions, so the sign it's going to be minus one to the two plus one or minus one to an odd number or simply negative one.
Birne Binegar: And then the corresponding minor delete the column and row corresponding that entry and you're left with a matrix to 111 there were there, it is, and we can do it for that entry, you can get a similar expression for the two to entry and entry in the second row and third call multi.
Birne Binegar: Factor matrix is going to end up being what minus one now I got the second row computed.
Birne Binegar: See one to see to to see two, three and what renamed.
Birne Binegar: bars computations go as I need to get the Co factors and last row now the row index is three and I compute.
Birne Binegar: Alright, so what i've done.
Birne Binegar: Is I computed nine numbers and now I placed them in a matrix in their natural slot in the first row first column i'm going to use the number see one etc.
Birne Binegar: All right, but this is not quite the matrix we use in this formula we use its transpose rather than see itself, so what I have to do next is take the transpose of this matrix which means.
Birne Binegar: Taking the first column in writing it as the first row of the transpose so minus one minus one minus one minus one minus 111 minus 111 i'm just converting columns in droves, to get a new matrix.
Birne Binegar: And now I can write down.
Birne Binegar: Have all the data computed i've got this matrix I computed the determinant of a I guess it worked was out worked out to be too, and so I simply take this matrix and it worked out to be minus two, I think.
Birne Binegar: What was it.
Birne Binegar: back.
Birne Binegar: yeah the terminator a itself with negative two.
Birne Binegar: And so, all I have to do is take this transpose of the co factor matrix.
Birne Binegar: divided by negative two and then i've got the inverse no real reduction at all.
Birne Binegar: Okay, so those are the.
Birne Binegar: two.
Birne Binegar: Nice.
Birne Binegar: algorithm for solving linear systems for computing inverse of the matrix each without doing any row reduction and it just involves this calculation of the determinant which.
Birne Binegar: You should think of their complexity, the fact that you need 720 separate terms to write down that determine have a six by six matrix you can think of that complexity that being related to the the the difficulty of the problem they solve.
Birne Binegar: Solving a system of 60 equations six unknowns is generally going to involve a lot of reduction.
Birne Binegar: But if you simply have formulas for determinants such that, like a computer software program might already know the Nice these complications are really rapid.
Birne Binegar: And in fact it's typically the way maple or mathematics or drive or any of the other common mathematical software pro which is typically the way they invert matrices and sold linear systems.
Birne Binegar: Okay, so i'll stop here um one thing I want to.
Birne Binegar: Ask if there's anyone who has not.
Birne Binegar: taken back there either first exam or second exam i've got them here.
Birne Binegar: So you can come pick it up.
Birne Binegar: I do want to in sort of insist that you.
Birne Binegar: At some point, pick up your exam before the final because what I typically do on the final exam is I look for the problem that caused the most difficulty on the first two exam and I give you a second chance on the final.
Birne Binegar: But if you don't know what you did wrong then you're sort of in a bad position so do pick up your your exam and.
Birne Binegar: With that i'll just let you go for today.