Help with a Linear Algebra problem?

[tfinch2]

Find elementary matrices E1, E2 such that B= E2E1A, where

A:
[1 2 3]
[4 5 6]
[7 8 9]

B:
[1 2 3]
[4 5 6]
[0 6 12]

Solution:

E1:
[1 0 0]
[0 1 0]
[-7 0 1]

E2:
[1 0 0]
[0 1 0]
[0 0 -1]

My book only has the explaination of what an elementary matrix is, but not an example on how to work through a problem like this.

 

[tfinch2]

BUMP

 

[TheoPetro]

makes me wish I didnt sleep though that class. Got an A but honestly couldnt tell ya how to do that. I wanna say something with transpose but not sure. Its been 2 years

 

[exdeath]

I'm assuming you want to learn this and not just get the direct route to the answer =D

Look up "row echelon form" and "row reduction operations" of matrices.

 

[TheoPetro]

try mathworld.com that site got me though all the way to ODE

 

[tfinch2]

Originally posted by: exdeath
I'm assuming you want to learn this and not just get the direct route to the answer =D

Look up "row echelon form" and "row reduction operations" of matrices.

Do you even know what an elementary matrix is? It's a matrix that is one step away from the identity matrix.

If you look at E1 and E2, you see that they are one elementary row operation away from the identity matrix I for a 3x3 matrix:

[1 0 0]
[0 1 0]
[0 0 1]

If you multiply E2 by E1 you get:

[1 0 0]
[0 1 0]
[7 0 -1]

Then multiply that by A to get B like the original question states.

But how did they calculate E1 and E2?

It's not like this is the first week of class and I'm asking what a pivot point is.

 

[UncleWai]

Gaussian-Jordan elimination, then multiply by the identity. I think the solution is just one of the answers since they flipped the sign on the third row.

 

[exdeath]

Originally posted by: tfinch2

Originally posted by: exdeath
I'm assuming you want to learn this and not just get the direct route to the answer =D

Look up "row echelon form" and "row reduction operations" of matrices.

Do you even know what an elementary matrix is? It's a matrix that is one step away from the identity matrix.

If you look at E1 and E2, you see that they are one elementary row operation away from the identity matrix I for a 3x3 matrix:

[1 0 0]
[0 1 0]
[0 0 1]

If you multiply E2 by E1 you get:

[1 0 0]
[0 1 0]
[7 0 -1]

Then multiply that by A to get B like the original question states.

But how did they calculate E1 and E2?

It's a matrix that performs a row operation on another matrix.

What you are doing is coming up with row operations that transform A into B one row at a time in only two operations. Think of it like factoring, but with matrices. You are factoring out two elementary matrices (which represent single row operations) from B until you have B = A*factor1*factor2.

If you look up the things I mentioned, you'll see what row operations are all about.

Then you can see its a simple case of dividing factors out of A[3] until it looks likes B, since row 3 is the only row that is changed.

 

[tfinch2]

Originally posted by: exdeath

Originally posted by: tfinch2

Originally posted by: exdeath
I'm assuming you want to learn this and not just get the direct route to the answer =D

Look up "row echelon form" and "row reduction operations" of matrices.

Do you even know what an elementary matrix is? It's a matrix that is one step away from the identity matrix.

If you look at E1 and E2, you see that they are one elementary row operation away from the identity matrix I for a 3x3 matrix:

[1 0 0]
[0 1 0]
[0 0 1]

If you multiply E2 by E1 you get:

[1 0 0]
[0 1 0]
[7 0 -1]

Then multiply that by A to get B like the original question states.

But how did they calculate E1 and E2?

It's a matrix that performs a row operation on another matrix.

What you are doing is coming up with row operations that transform A into B one row at a time in only two operations. Think of it like factoring, but with matrices. You are factoring out two elementary matrices (which represent single row operations) from B until you have B = A*factor1*factor2.

If you look up the things I mentioned, you'll see what row operations are all about.

Then you can see its a simple case of dividing factors out of A[3] until it looks likes B, since row 3 is the only row that is changed.

Okay I got it now.

So to transform matrix A into B, the first row operation you'd do is R3 <- R3 + -7R1 which gives:

[1 2 3]
[4 5 6]
[0 -6 -12]

Which is like multiplying A by:

[1 0 0]
[0 1 0]
[-7 0 1]

Then you would do R3 <- (-1)R3 to get B which is like multiplying the matrix above by:

[1 0 0]
[0 1 0]
[0 0 -1]

Right?

I have a test on Thursday, and I understand everything but I just couldn't get my head around the process of doing that problem. It's been pissing me off for about an hour now.

 

[exdeath]

Originally posted by: tfinch2

Originally posted by: exdeath

Originally posted by: tfinch2

Originally posted by: exdeath
I'm assuming you want to learn this and not just get the direct route to the answer =D

Look up "row echelon form" and "row reduction operations" of matrices.

Do you even know what an elementary matrix is? It's a matrix that is one step away from the identity matrix.

If you look at E1 and E2, you see that they are one elementary row operation away from the identity matrix I for a 3x3 matrix:

[1 0 0]
[0 1 0]
[0 0 1]

If you multiply E2 by E1 you get:

[1 0 0]
[0 1 0]
[7 0 -1]

Then multiply that by A to get B like the original question states.

But how did they calculate E1 and E2?

It's a matrix that performs a row operation on another matrix.

What you are doing is coming up with row operations that transform A into B one row at a time in only two operations. Think of it like factoring, but with matrices. You are factoring out two elementary matrices (which represent single row operations) from B until you have B = A*factor1*factor2.

If you look up the things I mentioned, you'll see what row operations are all about.

Then you can see its a simple case of dividing factors out of A[3] until it looks likes B, since row 3 is the only row that is changed.

Okay I got it now.

So to transform matrix A into B, the first row operation you'd do is R3 <- R3 + -7R1 which gives:

[1 2 3]
[4 5 6]
[0 -6 -12]

Which is like multiplying A by:

[1 0 0]
[0 1 0]
[-7 0 1]

Then you would do R3 <- (-1)R3 to get B which is like multiplying the matrix above by:

[1 0 0]
[0 1 0]
[0 0 -1]

Right?

I have a test on Thursday, and I understand everything but I just couldn't get my head around the process of doing that problem. It's been pissing me off for about an hour now.

You got it. Just remember matrix multiplication is noncommutative (I actually got the order backwards above with my factoring analogy). The row operations must be executed in order.

 

[FleshLight]

There are 3 different types of elementary matrices:

1. Interchanging 2 rows :
[0 1 0]
[1 0 0]
[0 0 1]

R1 and R2 are switched.

2. Multiplying a row by a nonzero constant:
[1 0 0]
[0 1 0]
[0 0 3]

R3 * 3

3. Adding a multiple of 1 row to another:
[1 0 3]
[0 1 0]
[0 0 1]

R1 + 3R3

So basically you want to find out elementary matrices that give you the R3 in .

 

[exdeath]

Originally posted by: FleshLight
There are 3 different types of elementary matrices:

1. Interchanging 2 rows :
[0 1 0]
[1 0 0]
[0 0 1]

R1 and R2 are switched.

2. Multiplying a row by a nonzero constant:
[1 0 0]
[0 1 0]
[0 0 3]

R3 * 3

3. Adding a multiple of 1 row to another:
[1 0 3]
[0 1 0]
[0 0 1]

R1 + 3R3

So basically you want to find out elementary matrices that give you the R3 in.

Much better... I got this entangled with REF, an application of these row transforms
The two were kind of a packaged deal for me.