## Given matrices A, B, find elem. matrices E, F, so EFB = A

Linear spaces and subspaces, linear transformations, bases, etc.
lawrence
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Joined: Wed Apr 01, 2009 10:48 pm

### Given matrices A, B, find elem. matrices E, F, so EFB = A

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.

Martingale
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### Re: Given matrices A, B, find elem. matrices E, F, so EFB = A

lawrence wrote: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.

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?

lawrence
Posts: 15
Joined: Wed Apr 01, 2009 10:48 pm

### Re: Given matrices A, B, find elem. matrices E, F, so EFB = A

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

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
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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.

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 =  32a + 3b = 10

Solve for the values of "a" and "b", and thus for E.

Check my work!

Martingale
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### Re:

stapel_eliz wrote: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.

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 =  32a + 3b = 10

Solve for the values of "a" and "b", and thus for E.

Check my work!

that is not the system you will get...

I think it is quite obvious that we need

$E=\left[\begin{array}{ccc}1&0&0\\0&1&0\\-3&0&1\end{array}\right]$

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
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stapel_eliz wrote:
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 =  32a + 3b = 10
Martingale wrote:that is not the system you will get...

Oops! You're right: I typoed the multiplication. The system should be as follows:

-a + 9b      = 32a +  b + 10 = 4

Where was my head at?

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.

lawrence
Posts: 15
Joined: Wed Apr 01, 2009 10:48 pm
Martingale wrote:I think it is quite obvious that we need

$E=\left[\begin{array}{ccc}1&0&0\\0&1&0\\-3&0&1\end{array}\right]$

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.

It seems to obvious when somebody else does it. thanks!

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