 Hi, I'm Zor. Welcome to a new Zor education Previous lecture was about solution to a system of linear equations two equations with two variables and We came up with the final formula Which looks like this. So if you have a system of Two linear equations We Considered the matrix of coefficients We also considered the matrix which would be The result of replacement of coefficients at x1 with three members of the equations and also Another matrix Which would be the result of a replacement of coefficients at x2 with three members and We came up with a general solution that x1 is equal to determinant of matrix a1 divided by determinant of a and x2 is determinant of a2 Divide this by determinants I Also mentioned that it's very important to look at the problems from above. So this is the view on the problem of solution of this particular equation from the matrix perspective and from the perspective of determinants and these formulas look quite General basically and I also mentioned that the same formulas are true for basically any system of n linear equations with n variables now I'm not going to prove the general case. However, this lecture today's lecture I would like to dedicate to arriving to the same results with the system of three equations with three variables This is a little bit more tedious And tires and process than the previous lecture with two variables. It's just more calculations and my most important goal Is not to make any mistakes on the way. However, I would like actually to illustrate that Solving the general system of three equations with three variables will result in exactly the same type of Solution, I think it's very important So you will feel that the generalization the view from the matrix perspective View from the perspective of determinants of these metrics is very very important because you are completely divorced from any particular Properties of any particular system you are approaching the system From the general standpoint which basically allows you to solve all the systems of linear equations with linear with with corresponding number of variables and By the way, just a side issue obviously the determinant of the a is In the denominator, which just implies which we actually did mention before that solution exists only if this determine Unique solution exists only if the thermo is not equal to zero All right, so this was a preamble and I would like to address now the system of three equations And I would like not to lose my Track and do everything correctly. So the three equations look like this All right, so this is kind of a long-distance Competition with myself because it will be lots of calculations Just bear with me and again consider that this is a good exercise for your mind stamina So you have you know the the muscle stamina and there is a mind stamina. So this is to exercise Your mind stamina. All right, so how can we solve this general? System of three equations with three variables. Well, again, I assume that a one three is not equal to zero because one of these Coefficients at x3 must be non zero otherwise the system would not have a nice and unique solution if all of them are zero so I assume that one three a one three is not equal to zero and My calculations will be based on this I will do exactly the same. I did for Explaining what actually the determinant of the matrix is so I'll try to solve it by basically getting great of x3 and reducing the system to the system of two equations with two variables Now, how can I do this? Well, I will multiply this by a two three multiply this by a one three and subtract and my x3 Would be cancelled out So if I will subtract from this I will subtract this this will be one three two three and this is two three one three It was x3. So it will it will cancel out now. So what I will have my free member will be after the subtraction Be one times a two three minus be two times a One three be one times two three minus be two one three now my coefficient at x1 would be a one one times a two three Minus two one one three That would be x1 and My coefficient at x2 will be one two two three Minus two two one three That would be at x2 Now what I will do I will similarly get rid of x3 in another pair of Equations one and three so one I will multiply by a three three and The third equation I will multiply by a one three and subtract from this I subtract this so what I will have Well, I will have b one a three three Minus b three times a one three equals Now coefficient at x1 would be a one one a three three Minus three one one three That would be at x1 and For x2 will be one two three three Minus three two one three That would be x2 now we have system of two equations with two variables, right now Instead of Solving it again using some kind of a substitution I will use the results of the previous lecture now on the previous lecture. I know how to solve the system of two equations with two Variables using the determinants, right so I can write right now What is x1 and what is x2 and then I will go about one x1 x3 so x1 is equal to Determinant of What matrix in the numerator? I have to substitute the free member instead of coefficients at x1, right? so the matrix would be B one a two three minus b two a One three this is a free member instead of this one But this one in the matrix of coefficients remains the same So this element would be as it is a one two a two three minus a two two a one three and The second row would be substituting this free member instead of this Which means b one a three three minus b three a one three and this remains the same a one two a three three a three two a one three Okay, that would be In the numerator the determinant of this matrix and what is the determinant of this matrix is this multiplied by this minus This multiplied by this main diagonal and this is alternate diagonal with a minus sign now on the bottom in the Denominator, I will have the determinant of the matrix of coefficients, which is a one one a two three minus a two one a one three a one one a three three minus a three one a one three a One two a two three minus a two two a one three and a one two a three three minus a three two a one three That will be on the bottom. So all I have to do right now is do the multiplication and addition and whatever else, right? So let me wipe out this And I will use this space to Basically do the formulas so x1 is equal to It would be a long line I guess So I have four and four I have eight Members So this time this is so let me start with B and then I will use a in the sequence of Increasing the row number and with the same row I will increase the column number so it will be B one Times a one two Times a two three and Times a three three Okay now This and this with a minus sign Minus B one a one three a two three And a three three Now this one times this with a minus sign minus B two A one three Now let me start with a one two because it's smaller a one two first and And then a one three and a three three and Finally this times this with a plus sign One three another one three and Three two okay, unfortunately I have to use the second line All right So we're talking about numerator only now with a minus sign the product of these two So minus sign, okay this times this B one a one two a two three And a three three Now this times this now this is a minus sign, but there is a minus in front of this product, so it's plus B three a one two a One three and a two three Now this times this it's a minus, but there is a minus in front of this product, so it's plus B one a one three a Two two and a three three And finally this times this it's minus and minus it's plus, but there is a minus in front of it so it's minus B three a One three another a one three and a two two So this is my denominator By the way, I Wanted to cancel out something here, right? This cancels out that's interesting great The rest stays Now let's talk about denominator now denominator, so let me put the Line and this is my x one is equal now denominator Is A determinant of the matrix of the coefficients, right? So it's this times this minus this times this All right, let's do the same thing So it's a one one a one two a Two three and a three three Okay This is this times this a One one one two two three three three now times this One one one three and this is the minus sign one one one three two three and Three two one one one three two three three two, okay Now this times this minus sign Start with a one two a one three a Two one and a three three and final this one and this with a plus A one three another a one three To one and three two minus minus Product of this times this so it's a one one two a two three and A three three That's four minus now this time this Now it's negative, but there is a minus in front so it will be plus This times this A one two A one three a two three and a three one Okay, now this times this minus sign but minus in front so it would be plus a one one a One three a two two and a three three And the last one would be this time this it's minus because it's minus minus and minus in front so it's One three would be twice Than two two and three one and Incidentally the first two also are cancelled out Now what else is interesting is that all members contain A one three So I assume in the very beginning if you remember that a one three is not equal to zero because some coefficient at At the x3 should not be equal to zero so I can just reduce it So a one three Here goes out Here goes out Here goes out out out out That's my numerator now the nominator Cancel cancel cancel Cancel cancel and cancel So what's the result so I take the liberty to wipe out this because if I made a mistake well I made a mistake So be it So let's try to Rerite this thing X1 is equal to Okay, let me first collect all the pluses. So the pluses are this this and this and I will put it in a Proper order B1 a 2 2 a 3 3 plus B2 I will order by B A 1 3 a 3 2 and the third one B3 a 1 2 a 2 3 Now with a minus sign B1 with a minus sign B1 2 3 3 3 Now B2 with a minus sign is this one a 1 2 a 3 3 and B3 with a minus sign 1 3 2 Okay, that's my numerator my denominator Okay, again pluses This plus 1 1 let's arrange it by the first one a 2 2 a 3 3 Plus 1 3 and plus 1 2 so let's do it plus 1 2 first a 2 3 a 3 1 and 1 3 Now with a minus sign Minus a 1 1 2 3 3 2 minus a 1 2 a 2 1 a 3 3 and minus a 1 2 Okay, so this is the answer for x 1 now I would like to check if my initial statement that all the Linear systems basically can be solved using the determinant Let's just check my my rule. My rule was that if I have a matrix of Coordinates sorry coefficients for this particular equation and Matrix looks like a 1 1 a 1 2 a 1 3 a A 2 1 a 2 2 a 2 3 a 3 1 a 3 2 a 3 3 Then in in the denominator, I will have the determinant of this matrix Well, let's just see the terminate of the matrix 3 by 3 Is very easily Can be constructed by multiplying first the main diagonal Which is this one and then two triangles with a short side parallel to it this one a 1 2 3 3 1 and This one also side parallel to the main diagonal, which is a 3 2 1 and 3 2 now the negative members are correspondingly with the alternate diagonal, so this is my alternate diagonal, which is One three two two three one one three two two three one and two triangles here here With a short side parallel to alternate diagonal, which is one one two three three two one one two three three two and this one Which is one two two one? and three three so as you see at the bottom we do have the terminate of the a matrix Of coefficients Now how about the top? Well, you remember that again the rule was take the three members Which are b1 b2 b3 and substitute instead of the coefficients at x1 So let's substitute So that would be b1 b2 and b3 and this is called matrix a1 not a11 a1 Well, let's see. What's the? Determinant of this thing main diagonal b1 a to 2 a 3 3 Then the triangle was a parallel Now the triangle was Okay, this triangle right with this side, so it's b2 a1 3 and a 3 2 b 2 a 1 3 2 and another triangle this one Which is b3 a 2 a 1 2 and a and a 2 3 b3 a 1 2 and a 2 3 so all the positive members negative members minus So the main diagonal, which is b3 a 2 2 and a 1 3 b3 a 2 2 and a 1 3 and With a minus sign should be this b 2 a 1 2 and a 3 2 a 3 3 b 2 and a 1 2 a 3 3 correct And this triangle b1 a 2 3 a 3 2 b 1 a 2 3 a 3 2 sir B1 Okay, this is my mistake. Not yet. 3 3 is here. So this will be this is supposed to be 3 2. I'm sorry alright, so Basically as I was saying the formula is very Much similar to the one which we obtained for the two by two matrix and two equations with two variables so my purpose was to Basically go through the same calculations to come up with the same result. What was out a small mistake? I'm sorry about that But but any case this is just an illustrative Lecture that the rules the laws of these matrix manipulations are the same Regardless of whether it's two-dimensional three-dimensional, etc This is again the illustration of how the generalized view on the problem helps you to solve Not just one problem which you have but many other problems So not only the system of three equations with three two variables You can solve but you can solve the system of ten equations with with ten variables The problem is it's kind of a tedious and obviously these systems with ten whatever number of equations are solved by computers because computers are well That's what they're good for basically stupidly make calculations As long as they as long as they know how as long as we program them correspondingly, but that's easy All right, so thanks very much for listening to me and bearing through all these tedious calculations Again, let me remind you that this is the development of your brain stamina your mind stamina stamina You really have to be able to do something which is tedious because you know the real life unfortunately requires this every once in a while I Do suggest you to go to unizor.com to the lecture about matrices and the solution for trip trip a tree I'm pretty sure that I did not make that mistake similar to this one on the website So among notes everything seems seems to be fine. Just follow the calculations and again try to to develop your Patience if you wish Patience and precision to do these type of things. That's a very good qualities for for the real life All right. Thanks very much and good luck