Example solve the equations ax b where a beginbmatrix 2 1 1...

Example: Solve the equations $Ax = b$, where $A = \begin{bmatrix} 2 & -1 & & \\ -1 & 2 & -1 & \\ & -1 & 2 & -1 \\ & & -1 & 2 \end{bmatrix}$, $b = \begin{bmatrix} 0 \\ 0 \\ 0 \\ 0 \end{bmatrix}$

Transcript

00:01 Hello, today we're going to find the solution set for ax equals b when we already know a, americanob, b.

00:09 So first we need to augment this matrix, as you see i've already done to the right, so that we can apply row operations simultaneously to our matrix.

00:20 So looking at this original matrix, we see that we have a leading one in the first row first column.

00:26 So that's good.

00:27 So we're going to preserve that in our first operations.

00:38 And then we see that we have a zero right below it.

00:41 So we're going to look at that third row and see that we have a four in the first column.

00:44 So we're going to want to get rid of that.

00:46 So we're going to do row three minus four times row one.

00:53 And we already said we're preserving the second column or the second row because of that first column right there with the zero.

01:03 And then four minus four, two minus four is negative two.

01:10 Negative three plus four.

01:13 1, 13 minus 5 times 4 is 20.

01:19 So, sorry, 13 minus 20 is negative 7.

01:27 And then we have our 0, because anything times 0 is 0.

01:31 All right, so now we look at it and we see that the third row is basically the same as the second row.

01:39 So we can instantly go and negate that by doing.

01:46 Row 3, row 2.

01:52 And preserving the first row again, we now have 0, 0, 0, 0, 0, 0, 0, 0, 0.

02:03 That's our 0, 2, negative 1, 7, and 0.

02:08 So now we look at the second row, and we see that we have a 2 there.

02:12 So we're going to want to get that to a 1.

02:16 So that we have the leading 1 in the next column to the right of the.

02:21 First leading one.

02:23 So to do that, all we have to do is one -half times row two.

02:30 That gives us 1 -1 -negative 1 -5 because we're preserving the first row.

02:37 And then 0, 2 divided by 2 is 1.

02:40 Negative 1 divided by 2 is negative 1 1⁄2 and then we have our 7 divided by 2 and then our zeroed out third row.

02:50 So now we're getting pretty close to reduced row, etchalant, form.

02:55 We have zeros all under that first one.

02:59 We have zero under that second pivot point, that second leading one, but we have a one above that second leading one.

03:07 So we want to clear that.

03:09 So what we're going to do is do row one minus row two.

03:20 So we have one...

Question

Example: Solve the equations $Ax = b$, where $A = \begin{bmatrix} 2 & -1 & & \\ -1 & 2 & -1 & \\ & -1 & 2 & -1 \\ & & -1 & 2 \end{bmatrix}$, $b = \begin{bmatrix} 0 \\ 0 \\ 0 \\ 0 \end{bmatrix}$

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