Matrices: Reduced row echelon form 1 | Vectors and spaces | Linear Algebra | Khan Academy

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+ x4 x 2 + x4 x something Well, you see, I have no– –Will– . Surface at r4 How much is x3? How much is this? Axial variables And that’s what I call it Here I have three equations containing four unknowns You can, of course, guess, or you know, of course, if you have Unknowns more than the equations themselves, you probably do not You define it sufficiently In fact you will get An infinite number of solutions And this infinite number of solutions can survive Tied up Let’s assume that we are in four dimensions, in this case Because we have four variables Perhaps we were tied to a four-dimensional surface, or If we were in three dimensions, we might be Bound by a line The streak is an infinite number of solutions, but it is A more restrictive group Let’s solve this set of linear equations We have done this in the past with deletion. What I want to do It is that I want to clarify the idea of \u200b\u200bmatrices Arrays are rows of numbers that are Reduced for this system of equations Let me create an array here I can create a parameter array, where The lab matrix will be, let me write in a more subtle way The parameter matrix is \u200b\u200bparameters that are on The left side of these linear equations The coefficient here is 1 The coefficient here is 2 We have 2, 2, 4 2, 2, 4 1, 2, 0 1, 2, there is no parameter in the expression x3 here, because it is There is no x3 statement here We will say that the coefficient of the expression x3 here is 0 Then we have 1, -1, and 6 And now if you deal with this, then this is considered The coefficient matrix for this system from Equations here What I want to do is I want to increase it, I want to increase it What these equations need to be equal to Let me increase it What I will do is draw Line here, and type 7, 12, and 4 I think you see this Another way to write it And through the site, we know this X1 coefficients, we know these X2 coefficients, and what you’ll do is We have to write x1 and x2 every time We can do the same operations on these Well we will do it on these What we can do is we can replace any equation with that Equation x standard multiplier + Another formula We can divide an equation, or multiply Standard equation We can put them down We can turn them over Let’s do this as an attempt to solve this equation The first thing I want to do, as I did on In the past, I want to get this equation in a picture If I can, I have 1 The main parameter for any one of these rows is 1 And every other entry in that column is 0 In the past, I had verified that every other entry Below is 0 This is what I have been doing in some past shows When we tried to determine if the stuff Linearly independent or not Now I will check if I have 1, if it is 1 major in rows, as everything else is in That column is 0 That model I’m doing is called the low grade class model Let me write that down Low grade class model If we call this the matrix of Z, which is matrix A, then I want Converts it to the low grade grade matrix A And the convention for matrices is, as are vectors Make them good and clear by using big letters, instead Of lowercase letters We’ll talk more about how matrices related to vectors in the future But now let’s solve this system of equations The first thing I want to do is, in this virtual world, I want Get all this here and get 0 Let me replace this with this, i.e. with the first entry – The second entrance Let me do that The first row will not change It will be 1, 2, 1, 1 Then get 7 here This is the first row Now the second row, I’m going to replace it Row – second row What do I get? 1 – 1 = 0 2 – 2 = 0 1 – 2 = -1 Then 1 – -1 = 2 This becomes 1 + 1 Then 7 – 12 = -5 Because I want to get rid of this row here I don’t want to get rid of it I want to get rid of these 2 I see converted to 0 Let’s replace this row with this row – 2 x that row What I will do is, this row – 2 × The first row I will replace this row with that 2 – 2 x 1 = 0 That is all 4 – 2 x 2 = 0 0 – 2 x 1 = -2 6 – 2 x 1 = 6, – 2 = 4 4 – 2 x 7 = 4, – 14 = -10 What can I do now You can see that for a class, we’ll talk more about what This row means When all we have is 0 There is nothing here If I have a nonzero statement here, then I’m going to make this zero Although it is 0 I will move towards this row The first thing I want to do is make this The main parameter becomes 1 What I want to do is, I’ll hit all of this Class B -1 If you multiply all this row by -1 I will not have to rewrite the array This becomes +1, -2, +5 I think you can accept that What can we do now? Well, let’s turn this into 0 Let me rewrite the increase matrix with the new model that I have I have. I will keep the middle class as it is this time The middle row is 0, 0, 1, -2, and then Increase it, and get 5 here What I want to do is get rid of -2 here Why not add this row to 2x that row Then get -2, + 2, and that will work What do I get Well, these are extra suffixes Then I have -2, + 2 x 1 This is equal to 0 4 + 2 x -2 = -4 Whereas 4 + -4 = 0 Then we have -10 + 2 x 5 Well, this is -10 + 10 and equals 0 This 1 has been reset here When I do a regular exclusion, I am Happy to have a condition that contains A group of number 1 Everything below was 0 I was not concerned with what was above 1 What I want to do is, I want to make These also become zeros I want to make this 0 as well What I can do is, I can replace the first row In the first row – the second row What is the result of 1 – 0? Equals 1 2 – 0 = 2 1 – 1 = 0 1 – -2 = 3 7 – 5 = 2 We got it We obtained the matrix with a low grade class model This is the matrix low-grade model – I’ll write it in bold – for matrix A You all know it is a low grade class model because all The basic 1 in each class – what it is Next 1 in each row? I have this 1 and I have that 1 They are non-zero pickles in their columns This is called the axial input Let me name this Let’s call it a pivot entrance Axial entrance It is the only non-zero input found in Its columns If I have any zeros outside the rows, already I have 0 outside The class, he’s here This zero outside According to the agreement, meaning For a low grade class model, it must be the last row Our main inputs are – all 1 This is a case We shouldn’t have these 5 We’ll divide that equation by 5 if it’s 5 The basic entries in each row are 1 As the main entries in each successive row are to the right The main entrance to the class that comes before it This is to the right of this This is the agreement For the low grade class model If I had any 0 outside of the class, it would be in the last row Finally, of course, and I think I’ve said this for several Of times, this is the only non-zero entry in the class What did this do to me? Now I can go back from here, come back To linear equations We remember that these were the coefficients x1, and these Coefficients were x2 These were the coefficients x3, and x4, then those were Constants I can rewrite the system of equations using The low grade score is as follows: x1 + 2×2 There is no x3 here 3×4 = 2 + This equation, there is no x1, there is no x2, I have x3 I have x3 – 2×4 = 5 I have no other equation here These have been reset I could reduce this system of equations to This system of equations Variables that you share with Pivot inputs, we will call those pivot variables x1 and x3 are pivot variables Variables that do not share with Pivotal entry, we can call them free variables So x2 and x4 are free variables Now let’s find, you can just find it Axial variables Free variables can be set to any variable I said it at the beginning of this equation We have equations less than variables This would not be a good restricted solution You won’t get just one point in R4 to solve This equation We will get several points Let’s find pivot variables, because that’s all What we can find This equation shows, here, it turns out that x3 – let me Use a good color – x3 = 5 + 2×4 Then we get x1 = 2 – x2, 2 – 2×2 2 – 2×2, sorry, – 3×4 I have proposed this from both sides of the equation This is as far as we can move to Solve this system of equations I can choose any values \u200b\u200bfor the free variables I can choose any values \u200b\u200bfor x2 and x4 and I can I find x3 And what I want to do now is to write this In a different way, so that we can visualize it a little better Of course, it is always difficult to imagine things Four dimensions We can visualize things a little better, as well This solution set Let me write this way If you want to write it vector, the solution is Vector x1, x3, x3, x4 How much is this? Well, it’s equal – let me write it like this – Equal – I’m rewriting, I’m Rewrite this set of solution vectorly So x1 = 2 – let me write a column Here – + x2 Let me write this way + x2 x something + x4 x something x1 = 2 – 2 x x2, or + x2 – 2 -2 was placed there I can say x4 x -3 + I can put -3 in there This here, the first inputs of these vectors This equation represents here. x1 Equals 2 + x2 x -2 + x4 x -3 How much is x3? x3 = 5 We put these 5 here + x4 x 2 x2 does not apply to it We can just put 0 0 x x2 + 2 x x4 Now how much is x2? You could say, x2 = 0 + 1 x x2 + 0 x x4 x2 is x2 It’s a free variable Similarly, how much is x4? x4 = + 0 x x2 + 1 x x4 What did this do to us? Well, we’ve clarified the set of solution As a linear component Of the linear component of the three vectors It’s vector You can consider it as a coordinate Vector in any direction It’s vector in R4 You can consider it a vector or coordinate position in R4 You can say, see, that the solution set – This is in R4– Each of these contains four ingredients, but you can Imagine it was found in r3 That is the set of solution and equal A vector, a vector here That is the vector Think of it as a vector site The coordinate will be 2, 0, 5, 0 And it’s clearly – these are four dimensions here – Equal to the multiples of these two vectors Let’s call this vector here, let’s We call this vector a Let’s call this vector here the vector b The solution group is all of this point, Almog Here, or I think we can call the vector location The vector location that will look like this From where we started from the original point here, + Multiples of these two If this vector is a– let me put the vector a in a different color– The vector a looks like this Suppose that the vector a looks like this, then that vector is b looks like this This is vector b, and this vector a I don’t know if this will be easier or more difficult To visualize for you, that we are clearly dealing in Four dimensions here, and I draw on Two dimensions are dimensional What you can click is, this set of solution is equal That fixed point, i.e. location vector, + Linear components of a and b We are of course in R4 now Let me write that down We are in R4 now But the linear components of a and b Will be a surface You can multiply a x 2, b x 3, or a x -1, and b x-10 You can continue to add and subtract these Linear components of a and b It will create a surface that contains The location vector, or will contain point 2, 0, 5, 0 Solve these three equations containing four unknowns The surface is at R4 I know this is hard to imagine, and maybe I will Another one with three dimensions I hope this at least gave you an understanding Why is the increase matrix, what is the low grade score model? What are the valid operations to apply to Matrix without creating clutter in the system

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