Consider the following matrix A =

Consider the following matrix \begin{equation*} A = \begin{pmatrix} 2 & 1 & 3 \\ 6 & 7 & 10 \\ 10 & 41 & 26 \end{pmatrix} \end{equation*} Decompose the matrix A into LU where L has the form \begin{equation*} L = \begin{pmatrix} 1 & 0 & 0 \\ l_{21} & 1 & 0 \\ l_{31} & l_{32} & 1 \end{pmatrix} \end{equation*} The coefficient $l_{21}$ is equal to Answer:

Transcript

00:01 For this problem, we're given a linear system of equations, and we're asked to find an lu decomposition for the coefficient matrix, this 3 by 3 matrix right there.

00:10 And then we'll use that result to solve the system.

00:14 So let a be this 3 by 3 matrix, and then we're going to perform elementary row operations on a to make it an upper triangular matrix.

00:23 So we need this block to be all zeros, and then we'll be able to find that lu decomposition.

00:30 All right, so first of all, i want a 0 in this entry.

00:36 So i can get a 0 by taking 1 3rd times row 1, adding that to row 2, and then making that the new row 2.

00:50 So here's the elementary row operation, and then the elementary matrix that represents that row operation is this matrix right here.

01:01 All right, then we consider the product of e1 and a.

01:15 We get the following matrix.

01:17 So the first row stays the same.

01:19 This is negative 3, 12, negative 6.

01:23 The last row stays the same as well, 0, 1, 1.

01:28 And this entry is 0 by construction, and this entry is 2, and this one is 0.

01:36 So we have zeros there.

01:39 And lastly, we just need this entry to be a 0.

01:42 So we can do that by taking minus 1 half row 2, add that to row 3, make that the new row 3.

01:57 And then the elementary matrix that does that is this one.

02:04 Let's call it e2.

02:07 This is 1 0 0 0 1, and then minus 1 0 0 1.

02:17 And the inverse of e2 exists, and it's this matrix right here.

02:28 And then the product of e2, e1, and then a is this 3 by 3 matrix.

02:39 So the first two rows, they stay the same.

02:45 And then the last row, this is 0, 0, and then 1.

02:51 So this is what we wanted.

02:52 We have an upper triangular matrix right here, the matrix that's circled.

02:58 And then what we can do is multiply this whole equation by e2 inverse, and then multiply by e1 inverse.

03:11 And then eventually, we'll get that a.

03:17 So this is e1 inverse times e2 inverse, and then times, call this matrix u for upper triangular.

03:30 And this matrix right here, this is l, which is e1 inverse times e2 inverse.

03:41 And this is equal to the following matrix...

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