Given the L U factorizations you calculated in Exercise 1.3.21, solve the associated linear syst

Given the L U factorizations you calculated in Exercise...

Key Concepts

Backward Substitution

Backward substitution is a method used to solve an upper triangular system of equations U x = y, where U is an upper triangular matrix. The process starts with the last equation and moves upward, solving for each unknown in sequence. This approach is efficient because each equation has fewer unknowns due to the triangular structure of U.

Forward Substitution

Forward substitution is a technique used to solve a system of equations given by L y = b when L is a lower triangular matrix. This method involves solving for the unknowns sequentially from the first equation to the last, taking advantage of the fact that each equation only involves the unknowns that have not yet been computed.

LU Factorization

LU factorization is a method of decomposing a square matrix A into the product of two matrices: a lower triangular matrix L and an upper triangular matrix U. This decomposition simplifies solving linear systems by replacing a single complicated system Ax = b with two simpler systems, namely L y = b and U x = y, which can be solved sequentially using forward and backward substitution.

Solving Linear Systems

The process of solving linear systems involves finding the vector x that satisfies the equation A x = b, where A is a given matrix and b is a given vector. In the context of LU factorization, the solution is obtained by first solving L y = b using forward substitution and then solving U x = y using backward substitution.

Transcript

00:01 So if we want to solve this by an augmented matrix, let's go ahead and get the augmented matrix set up first.

00:09 So remember, we're trying to get this into the form of ax as equal to b.

00:17 And remember, we just kind of pretend like the x isn't there and the equal sign.

00:22 So first we'll rewrite a, so it'll be 1 -2 -1, negative 3, negative 1 -2, 0 -3.

00:28 And then for this equal sign, we just put this bar going straight up and down.

00:32 Then we write b on the other side.

00:34 So it would be 1, or 0 ,1, negative 1.

00:37 Now, if you're allowed to use a calculator to do this row reduction, so normally it's like the rref in a calculator, this should give us the matrix 1 -1 -1 and then a bunch of zeros here.

00:52 And then on the right side, it is 0 .6, negative 0 .8, and 1.

01:00 And remember, the way we read this is x1, is equal to 0 .6, x2 is equal to negative 0 .8, x3 is equal to 1.

01:10 And since we want this as a vector, we would just say x is equal to 0 .6, negative 0 .8, 1.

01:19 So pretty straightforward if you can use a calculator.

01:23 But if for whatever reason you can't, we can still do this by hand pretty easily.

01:29 So let's go ahead and pretend like we didn't do that and just kind of scoot this.

01:35 Down.

01:36 But at least we know, like if for some reason we don't get this, we may have done something weird with one the calculations.

01:41 It'll give us a reason to like go back and check.

01:45 What i would do to start is so this is a non -zero number so we can keep it there.

01:51 And then remember we want to clear all everything below.

01:53 So third row already has zero there, so we just need to clear off the three.

01:58 So if we multiply row one by three and then add that to row two, then that negative three cancels out.

02:06 So first two row.

02:07 Stay the same, so 1, 2, 1, 0, 3, 0, 3, negative 1.

02:16 Now this negative 3 becomes 0, so 2 times 3 plus negative 1 is 5, 1 plus 3 plus 2, or 1 times 3 plus 2 is 5, 0 times 3, so that doesn't change the 1 there.

02:31 Now we could divide all this by 5, but notice that if we actually just did row, 1 or negative row or not negative row 1 negative row 2 plus row 3 then the 5 below it cancels out so let's go ahead and do that so remember first two rows stay the same because we haven't done anything with those so 1 2 1 0 0551 and now this is still going to be 0 that is 0 because it cancels out and then negative 5 plus 2 is negative 2 and then 1 plus negative 1 is negative 2.

03:15 All right.

03:16 And now we can go ahead and divide this first row here by negative 1 1⁄2.

03:26 So we do negative 1 half row 2...

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