Matrix proof...

[Orsorum]

Original question:
a11x1 + a12x2 = b1
a21x1 + a22x2 = b2

If a11a22 - a12a21 != 0, then prove that this system is equivalent to:

c11x1 + c12x2 = d1
c22x2 = d2

where c11 and c22 != 0.

I solved it by setting the cross product equal to a constant(cA) that's not equal to 0, then doing a row reduction where a21 is set equal to a constant (cB) times a11; substitute that into the cross product, you get c22 = a22 - cAa12; then c11 and c12 are just a11 and a12, respectively.

Dunno if it's the actual solution. I'm kind of lost. I think I proved it, but I'm missing something...

 

[gopunk]

what are x1 and x2?

actually.... is this a book problem? what number....

 

[rival]

yea, i thought it was a good movie, makes you think

 

[Orsorum]

Originally posted by: gopunk
what are x1 and x2?

actually.... is this a book problem? what number....

Haha... section 1.1, question #38.

 

[GoodToGo]

Umm I dont even know whether I am on the right track or not but if a11a22-a12a21 =0, then the system cannot be solved. In linear algebra, Ax=b where A is the coeffcient, x are the variables and b is the constant. To solve for x,

x = (A^-1)b (DO NOT write x = b/A)

If A has to be inversed, the determinant must not be zero. But in this case, the determinant is zero. So the system has no solution. Kinda similar to the other system also which has no solution(x2 = divide by zero and x1 cannot be solved). Hope this makes some sense

 

[gopunk]

Originally posted by: Zakath15

Originally posted by: gopunk
what are x1 and x2?

actually.... is this a book problem? what number....

Haha... section 1.1, question #38.

doh different book! mine only goes up to 33 or something.

 

[Orsorum]

Originally posted by: gopunk

Originally posted by: Zakath15

Originally posted by: gopunk
what are x1 and x2?

actually.... is this a book problem? what number....

Haha... section 1.1, question #38.

doh different book! mine only goes up to 33 or something.

Bah!

 

[Orsorum]

Originally posted by: GoodToGo
Umm I dont even know whether I am on the right track or not but if a11a22-a12a21 =0, then the system cannot be solved. In linear algebra, Ax=b where A is the coeffcient, x are the variables and b is the constant. To solve for x,

x = (A^-1)b (DO NOT write x = b/A)

If A has to be inversed, the determinant must not be zero. But in this case, the determinant is zero. So the system has no solution. Kinda similar to the other system also which has no solution(x2 = divide by zero and x1 cannot be solved). Hope this makes some sense

In this case, the determinant cannot equal zero... != == not equal to.

 

[GoodToGo]

Originally posted by: Zakath15

Originally posted by: GoodToGo
Umm I dont even know whether I am on the right track or not but if a11a22-a12a21 =0, then the system cannot be solved. In linear algebra, Ax=b where A is the coeffcient, x are the variables and b is the constant. To solve for x,

x = (A^-1)b (DO NOT write x = b/A)

If A has to be inversed, the determinant must not be zero. But in this case, the determinant is zero. So the system has no solution. Kinda similar to the other system also which has no solution(x2 = divide by zero and x1 cannot be solved). Hope this makes some sense

In this case, the determinant cannot equal zero... != == not equal to.

Oops sorry, I am not a comp major. In this case, just take opposite of what I said, the two systems have unique solutions for x1 and x2.

 

[Orsorum]

I'm thinking it has something to do with what can and cannot equal zero, and how it ties in with the original equation... hmm.