A = [-1 2 ] B = [ 9 1 ] [ 9 1 ] [-1 2 ] [ 3 4 ] [ 0 10 ]Find elementary matrices E and F such that EFB = A.

I can't think how to get started other than just guessing which probably won't work. A hint would be great. Thank you.

Suppose the matrices A and B are as follows:

I can't think how to get started other than just guessing which probably won't work. A hint would be great. Thank you.

A = [-1 2 ] B = [ 9 1 ] [ 9 1 ] [-1 2 ] [ 3 4 ] [ 0 10 ]Find elementary matrices E and F such that EFB = A.

I can't think how to get started other than just guessing which probably won't work. A hint would be great. Thank you.

- Martingale
**Posts:**333**Joined:**Mon Mar 30, 2009 1:30 pm**Location:**USA-
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to start off with...do you know how to interchange rows using elementary matrices? ie do you know how to switch row 1 and row 2 using a matrix F?Suppose the matrices A and B are as follows:

A = [-1 2 ] B = [ 9 1 ] [ 9 1 ] [-1 2 ] [ 3 4 ] [ 0 10 ]Find elementary matrices E and F such that EFB = A.

I can't think how to get started other than just guessing which probably won't work. A hint would be great. Thank you.

If I were just converting the first two rows, I could do like this:

Thank you for your help.

A = [-1 2 ] B = [ 9 1 ] [ 9 1 ] [-1 2 ] F = [ 0 1 ] [ 1 0 ] FA = [ 0 1 ][-1 2 ] [ 1 0 ][ 9 1 ] = [ 0+9 0+1 ] [ -1+0 2+0 ] = [ 9 1 ] = B [-1 2 ]But there's that third row. What do I do for that?

Thank you for your help.

- stapel_eliz
**Posts:**1628**Joined:**Mon Dec 08, 2008 4:22 pm-
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You'll need to multiply B by a 3-by-3 matrix (being the product of square matrices E and F). I think you can use the 2-by-2 you've provided as part of the 3-by-3 version of F that you need.

Check my work!

F = [ 0 1 0 ] [ 1 0 0 ] [ 0 0 1 ] FB = [-1 2 ] [ 9 1 ] [ 0 10 ]Now you need to find a 3-by-3 matrix E so that EFB equals A. Since the last row of A contains 3 and 4, you need a matrix which is provide the row-operation(s) necessary to create this from what you've already got (if I'm understanding the process correctly).

E(FB) = [ 1 0 0 ][-1 2 ] [ 0 1 0 ][ 9 1 ] [ a b 1 ][ 0 10 ]When you multiply this out, you should get a system of equations:

-a + 9b = 3 2a + 3b = 10Solve for the values of "a" and "b", and thus for E.

Check my work!

- Martingale
**Posts:**333**Joined:**Mon Mar 30, 2009 1:30 pm**Location:**USA-
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that is not the system you will get...

F = [ 0 1 0 ] [ 1 0 0 ] [ 0 0 1 ] FB = [-1 2 ] [ 9 1 ] [ 0 10 ]Now you need to find a 3-by-3 matrix E so that EFB equals A. Since the last row of A contains 3 and 4, you need a matrix which is provide the row-operation(s) necessary to create this from what you've already got (if I'm understanding the process correctly).

E(FB) = [ 1 0 0 ][-1 2 ] [ 0 1 0 ][ 9 1 ] [ a b 1 ][ 0 10 ]When you multiply this out, you should get a system of equations:

-a + 9b = 3 2a + 3b = 10Solve for the values of "a" and "b", and thus for E.

Check my work!

I think it is quite obvious that we need

since if a,b are both non zero we don't have an elementary matrix...at least according to any definition I have ever seen.

- stapel_eliz
**Posts:**1628**Joined:**Mon Dec 08, 2008 4:22 pm-
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E(FB) = [ 1 0 0 ][-1 2 ] [ 0 1 0 ][ 9 1 ] [ a b 1 ][ 0 10 ]When you multiply this out, you should get a system of equations:

-a + 9b = 3 2a + 3b = 10

Oops! You're right: I typoed the multiplication. The system should be as follows:that is not the system you will get...

-a + 9b = 3 2a + b + 10 = 4Where was

Continuing, 9b - 3 = a, so then:

. . . . .2(9b - 3) + b + 10 = 4

. . . . .18b - 6 + b = -6

. . . . .19b = 0

...and so forth, leading to the matrix provided in the previous reply.