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# Գծային համակարգերի լուծում մատրիցային հավասարումների օգնությամբ

## Տեսանյութի սղագրությունը

in the last video we saw that we could take a system of two equations with two unknowns and represent it as a matrix equation where the matrix a is are the coefficients here on the left hand side the column vector X has our two unknown variables SNT and then the column vector B is essentially representing the right hand side over here and what was interesting about it well then that would be the equation a the matrix a time's the column vector X being equal to the column vector B and what was interesting about that as we saw well look if a is invertible we can multiply both the both the left and the right hand sides of the equation and we have to multiply them on the left hand sides of their respective sides by a inverse because remember matrix multiplication order matters so we're multiplying the left hand side of both sides of the equation if we do that then we can get to essentially solving for the unknown column vector if we know what column vector X is then we know what s and T are and we've essentially solved the system of equations so now let's actually do that let's actually figure out what a inverse is and multiply that times the column vector B to figure out what the column vector X is and what s and T are so a inverse a inverse is equal to 1 over the determinant of a the determinant of a it for a for a 2x2 here is going to be 2 times 4 minus negative 2 times negative 5 so it's going to be 8 minus positive 10 8 minus positive 10 which would be negative 2 so this would become negative 2 right over here once again 2 times 4 is 8 minus negative 2 times negative 5 so minus positive 10 which gets us negative 2 and you multiply 1 over the determinant times what is sometimes called the adjoint of a which is essentially swapping the top left and bottom right at least for a 2x2 matrix so this would be a 4 this would be a 2 notice I just swapped these and making these 2 negative the negative of what they already are so this is from a negative 2 that's going to become a positive 2 and this right over here is going to become a positive five if all of this looks completely unfamiliar to you you might want to review the tutorial on inverting matrices because that's all I'm doing here and so a inverse is going to be equal to a inverse is going to be equal to let's see this is negative 1/2 times 4 is negative 2 negative 1/2 negative 1/2 times 5 is negative 2 point 5 negative 2 point 5 and negative 1/2 times 2 is negative 1 negative 1/2 times 2 is negative 1 so that's a inverse right over here so now let's multiply a inverse times our column vector 7 negative 6 so let's do that so this is a inverse I'll rewrite it negative 2 negative 2 point 5 negative 1 negative 1 times 7 and negative 6 times I'll just write them all in white here now 7 negative 6 we've had a lot of practice multiplying multiplying matrices so what is this going to be equal to so the first entry is going to be negative 2 times 7 which is negative 14 plus negative 2 point 5 times negative 6 so let's see that's going to be positive that's going to be 12 plus another 3 so it's going to be plus 15 plus 15 negative 2 point 5 times negative 6 is positive 15 and then we're going to have negative 1 times 7 which is negative 7 plus negative 1 times negative 6 well that is positive 6 and so our the product a inverse B which is the same thing as a column vector X is equal to we deserve a little bit of a drumroll now the column vector 1 negative 1 so we have just shown that this is equal to 1 negative 1 or that X is equal to 1 negative 1 or we could even say that the column vector a column vector st s column vector with the entries s and T is equal to is equal to 1 negative 1 is equal to 1 negative 1 which is another way of saying that s is equal to 1 and T is equal to negative 1 and I know what you're saying I said this in the last video and I'll say it again in this video you're like well you know it was so much easier to just solve the system directly just with using elimination or using substitution and I agree with you but it is this is a useful technique because when you are doing problems in computation there may be situations where you have the left hand side of the system is stays the same but there are many many many different values for the right hand side of the system and so it might be easier to just compute the inverse once and just keep multiplying keep multiplying this inverse times the different the what we have on the right hand side and you know you probably are familiar with you know sometimes you know you have graphics processors and graphics cards on computers and they talk about special graphics processors but these are really all about are the hardware that is special-purpose for really fast matrix multiplication because when you're doing graphics processing when you're thinking about modeling things in three dimensions and you're doing all these transformations you're really just doing a lot of matrix multiplications really really really fast in real time so that to the user playing the game or whatever they're doing it feels like they're in some type of a you know 3d real-time reality so anyway I just want to point that out this wouldn't be you know if I if the if I saw this just randomly and my instincts would do be to solve this with elimination but this this ability to think of this as a as a matrix equation is is a very very useful concept one actually not just in computation but also as you go into higher level Sciences especially physics you will see a lot of matrix vector equations like this that kind of speak in generalities and it's really important to think about what these actually represent and how they can actually be solved