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

Linear spaces and subspaces, linear transformations, bases, etc.
lawrence
Posts: 15
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
Posts: 333
Joined: Mon Mar 30, 2009 1:30 pm
Location: USA
Contact:

### Re: 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.
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
Posts: 1628
Joined: Mon Dec 08, 2008 4:22 pm
Contact:
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 =  3
2a + 3b = 10
Solve 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
Contact:

### Re:

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 =  3
2a + 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
Posts: 1628
Joined: Mon Dec 08, 2008 4:22 pm
Contact:
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
that is not the system you will get...
Oops! You're right: I typoed the multiplication. The system should be as follows:
-a + 9b      = 3
2a +  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
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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