Linear Algebra Problem... Need some help quick please!

[krwlz]

Ok, I just dont know how to solve this...

We want to solve for X, when AX=B
A= (-1 2)
(2 -4)
(A is 2x2 matrix)

B=(1 2)
(3 5)
(B is 2x2 matrix)

So I need to find X, which as far as I can tell is a 2x2 matrix. Every problem we ever did in class, had X and B as a single column matirx, so I'm stumped as hell.

I've tried using x=(x1 x2)
(x3 x4)

But when you get the equations, and sub through them, you get 0=0. If you set up the equations in two seperate augmented matrices (Using the coefficients on the X's), then reduce (In an attempt to get to reduced row echolon form) it ends up being a matrix of 0's with real numbers in the augmented spot. Im so helplessly, and entirly confused...

[a-j]

It is obviously clear that matrix A is singular (row 2 = -2*row 1), that is its columns are not linearly independent. This means that there is no inverse to A and hence trying X=A^(-1)*B won't work, which is probably the "normal" way of doing things. This either means that there is no solution to the system, or infinitely many solutions. I tried myself to solve it and ended up getting -2=3, which means there are no solutions to that equation.

[Thermopyle]

hmmm, a long time ago i studied lin alg but a (2x2)- times a (2x2)-matrix shouldn't produce another (2x2)-matrix.

[GeePeeS'r]

Originally Posted by Thermopyle

hmmm, a long time ago i studied lin alg but a (2x2)- times a (2x2)-matrix shouldn't produce another (2x2)-matrix.

Actually, thats not correct. a 2x2 matrix times a 2x2 matrix will equal a 2x2.

The sizes of the final matrix is easy - Drop the 2 inside numbers of the matrices and only use the outer two numbers (rows from the first matrix and columns from the second). Example: a 1x4 matrix times a 4x6 matrix will result in a 1x6 matrix. (keep in mind that the 2 inside numbers must be the same).

Other than that, i think a-j is right. The determinate of A is 0 therefore it does not have an inverse - which is what you would need to properly solve it. I believe there is no solution.

[Thermopyle]

eeerrr.....




my mistake, as I said, it was 6 years ago.
*crying and sobbing in the shower for hours*

[GeePeeS'r]

Originally Posted by Thermopyle

eeerrr.....




my mistake, as I said, it was 6 years ago.

Yeah, Its only been about 1 year for me - so its still peudo-fresh in my mind - not as if that is stuff I will ever use again.

[Thermopyle]

Originally Posted by GeePeeS'r

Yeah, Its only been about 1 year for me - so its still peudo-fresh in my mind - not as if that is stuff I will ever use again.

I've actually used it times to times, unlike other mathematical knowledge I 've acquired. Mostely in computer related stuff. It's nice to see that 6 years of university wasn't all in vain...

[Corneo]

Well you can get an approximated answer using the least squares solutions. If your interested in that, so solve (A^T)(A)x=A^T(b)

A^T = transpose(A)

[jaded]

krwlz, this question is relatively easy to solve. since you probably wanna try by yourself, here\'s some hint:

you know x is a 2x2 matrix, because A and B both are: [2x2] [2x2] = [2x2].

Now let the elements of X be:
(a b)
(c d)

now we have:

(-1 2) (a b) = (1 2)
(2 -4) (c d) (3 5)

use the multiplication rules for matrix multiplication and you will have four equations with four unknowns to solve (a, b, c, d), which is very easy to do. Try it, and let me know if you still have trouble.

[krwlz]

THats what I thought. I used the 5 part theorem to prove it does not have a solution (kinda...)

Hopefully it was right.

[phukraut]

What is the 5 part theorem?

[krwlz]

First, the rules for use
A is a matrix, and it is square (n x n)
1. A ^ (-1) exists
2. AX=O has one solution
3. AX=B has one solution per choice of B
4. A~I
5. det(A) does not equal zero

Either they are all true, or they are all false, for any n x n matrix

[phukraut]

Ah I see now. Actually 2 is that AX=O only has the <em>trivial</em> solution. An old textbook I have also lists two more after 4 and before 5:

A is expressible as a product of elementary matrices, and
AX=B is consistent for every n by 1 matrix B.

In fact, that theorem can also be extended to at least 17 parts after you include things about linear transformations, spanning, basis, and rank/nullity. Pretty cool.

[pan6467]

We want to solve for X, when AX=B
A=
(-1 2)
(2 -4)
(A is 2x2 matrix)

B=
(1 2)
(3 5)
(B is 2x2 matrix)


AX=B
X= (AI)*B

GAUSS-JORDAN

(-1 2|1 0)
(2 -4|0 1)

R2+R1
(1 -2|1 1)
(2 -4|0 1)

-2R1+R2
(1 -2|1 1)
(0 -8|-2 -1)

-1/4R2+R1
(1 0| 1.5 .75)
(0 -8|-2 -1)

1/8R2
(1 0|1.5 .75)
(0 1| -.25 -.125)

X=
(1.5 | .75)*B
(-.25 | -.125)*B

(1.5 |.75)*(1 2)
(-.25|-.125)*(3 5)

1.5+2.25 | 3+ 3.75 = X
-.25+-.375|-.5+-.625 = X

(3.75 |6.75) or (3 3/4| 6 3/4)=X
(-.625|-1.125) or (-5/8 |-1 1/8)=X

This is how I would solve.

*Edited because I had forgotten to multiply identity A with B.

[jaded]

good krwlz, i think you got the right answer!

Originally Posted by pan6467

We want to solve for X, when AX=B
A

GAUSS-JORDAN

(-1 2|1 0)
(2 -4|0 1)

R2+R1
(1 -2|1 1)
(2 -4|0 1)

-2R1+R2
(1 -2|1 1)
(0 -8|-2 -1)

it\'s ok to use GJ, but you made a mistake in the last step above:

-2R1+R2 should give you a zero row, therefore suggesting no solution.

-2R1+R2
(1 -2|1 1)
(0 0 |x x)

[krwlz]

Originally Posted by jaded

good krwlz, i think you got the right answer!



it\'s ok to use GJ, but you made a mistake in the last step above:

-2R1+R2 should give you a zero row, therefore suggesting no solution.

-2R1+R2
(1 -2|1 1)
(0 0 |x x)

My problem exactly.
And that 5 part theorem is only the things that are completley logically equivalent... Anything you can derive from there needs not be listed, because it is implied. Such as spanning is essentially, #3, and liniear indepenence is #2.