2 10 marks consider the matrices a beginbmatrix 2 1 1 1 1 2...

(2) (10 marks) Consider the matrices $$A = \begin{bmatrix} 2 & 1 & 1 \\ -1 & 1 & 2 \\ 2 & -1 & 1 \end{bmatrix}, B = \begin{bmatrix} 2 & 3 & 1 \\ -1 & 1 & 2 \\ 3 & 2 & 1 \end{bmatrix}$$ (a) Find a series of elementary row operations through which one can turn A to B (list the ERO in the correct order). (b) Show that any matrix that is row equivalent to an invertible matrix is invertible. (Hint: Use the Fundamental Theorem Invertible Matrices). (c) Prove that any two invertible matrices are row equivalent. Make sure your proof explains each step properly and clearly. (d) Find two $3 \times 3$ matrices that are not row equivalent and provide all the details why they are not row equivalent.

Transcript

00:01 In this problem, our job is to answer each of the given questions concerning matrices.

00:08 In the first question, we are asked to find a 4x4 elementary matrix corresponding to taking row 2 in a matrix and replacing it by 4 times row 3 plus row 2.

00:23 So we can think of that operation as 4 times row 3 plus row 2.

00:29 We're going to replace the old row 2 with this, using this as the new.

00:34 Second row.

00:38 So the key, since we want an elementary matrix, is to start with the 4x4 identity.

00:44 That has ones on the main diagonal, and 0s everywhere else.

00:51 So here is the identity matrix in the 4x4 case.

00:57 We're starting with that, and we want to replace the old row 2 by 4 times row 3 plus row 2.

01:13 So we're going to use row 3, but we will keep that in its place.

01:18 So row 1 is not affected, row 4 is not affected, and row 3 will remain the same.

01:28 We're simply going to use it to get a new row 2.

01:33 So let's do that.

01:35 When we multiply row 3, here's our row 3 in the original identity matrix, we multiply through by 4 and then add the result to row 2.

01:44 We get 0 plus 0 or 0 in the first column.

01:48 We get 0 plus 1 or 1 in the second column.

01:53 In the third column, we get 4 plus 0 or 4, and in the last column we get 0 plus 0 or 0.

02:01 So we notice that the only change here is we now have a 4 in the second row and third column.

02:09 So here's our elementary matrix.

02:11 We'll call it e.

02:13 Then we are asked to find the inverse of that matrix.

02:18 And so it seems reasonable to think that if we multiply row 4 by, excuse me, row 3 by 4, multiply row 3 by 4 and add row 2 to get a new row 2.

02:28 Then to undo this, we would want to multiply by negative 4.

02:32 In fact, that turns out to be the case.

02:34 Let's make a little bit more room here.

02:37 In fact, it does turn out that e inverse is the same as e, except we're going to have a negative 4 instead of a positive 4 in the second row, third column.

02:52 So the first and second columns are the same.

02:55 Third column has 0 negative 4, 1 -0, and then the last column is the same.

03:05 So here are the elementary matrix e, and here's its corresponding inverse.

03:16 So next, let's move on to part b.

03:20 In part b, we are dealing with 2 n -by -in matrices, a and b.

03:26 We are told that they are both invertible.

03:30 So both a inverse and b inverse exist, and by definition, that means that a times a inverse is the identity matrix, and b times b inverse is the identity matrix.

03:53 In fact, in the special case of inverses, we have that the matrices commute.

03:58 A inverse times a is the identity, and b inverse times b is the identity.

04:02 So the product of a with its inverse is the identity, and the product of b with its inverse is the identity.

04:09 That's the identity matrix i, the n by n identity matrix.

04:14 So we claim that a, b, is also invertible...

Question

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