Question

The product of two elementary matrices of size 33: (choose ALL correct answers)

A. Is always an elementary matrix
B. Corresponds to a sequence of elementary row operations on a 33 matrix
C. Is never an elementary matrix
D. Is an invertible matrix
E. Is the same in either order of multiplication

Answer 1

Almost by definition, a single elementary matrix corresponds to a single row operation on a 33 matrix. So if you multiply 2 elementary matrices together, you corresponds to a sequence of row operations.

Answer 2

So B is definitely true. However, what about the other ones? You know that every invertible matrix can be broken down into elementary matrices. Is every invertible matrix an elementary matrix?

Answer 3

no every invertible matrix isnt a elementary matrix

Answer 4

Right, so that means that A is not correct. But is it possible to multiply two elementary matrices together to get an elementary matrix?

Answer 5

if you multiply two elementary martrices together dont you get I you dont get another elementary matrix do you

Answer 6

What if you multiply the matrix corresponding to a row operation of 2R(1) with the matrix corresponding to a row operation of 3R(1)?

You'll get an elementary matrix corresponding to 6R(1).

Answer 7

that is true

Answer 8

So you have two elementary matrices multiplied together to get an elementary matrix. (In general, you are correct. Most elementary matrices multiplied together do not give an elementary matrix)

Can you tell me if the product is always an invertible matrix?

Answer 9

no it is only invertible if the determinant is not equal to 1

Answer 10

Almost. It's invertible if the determinant is not equal to 0. So two elementary matrices are always invertible, and if you multiply two invertible matrices together, you also get an invertible matrix. So D is also correct.

Answer 11

oh sorry i ment 0 haha. and then e is not correct because the order of multiplication matters?

Answer 12

Now for the last one. Is \[\left[\begin{matrix}1&0\\2&1\end{matrix}\right]\left[\begin{matrix}1&-1\\0&1\end{matrix}\right]\]\[=\left[\begin{matrix}1&-1\\0&1\end{matrix}\right]\left[\begin{matrix}1&0\\2&1\end{matrix}\right]\]

You're ahead of me with E there :) You're correct in thinking that E is not correct.

Answer 13

So overall, we have that B, C, D are correct, and A, E are incorrect.

Answer 14

Perfect thank you so much!

Answer 15

You're welcome.