TT's Maths Thread

[TrueTears]

thought so but here you're assuming doesn't that mean you also have to prove uniqueness too? Or are you assuming uniqueness too

[m@tty]

lol I don't know. This is only what I remember from the start of the year in uni maths.

[/0]

Proof by example should work just fine. Alternatively,



Right multiplying by :



Right multiplying by



(Repeat for right inverse)

By construction, is invertible. Alternatively, it is invertible because

Uniqueness can be proven by assuming there are two inverses: ,

If and then and so .

(Repeat for right inverse)

[kamil9876]

Well your guess is good. Remember the defn of an inverse of is a matrix such that . Using the associativity of matrix multiplication, you can show that the you guessed works and hence is an inverse => the inverse by uniquness. ie it isn't good to write if you don't know if it exists, maybe that's what TT was worried about.

[TrueTears]

Ahh yup cool thanks for filling in the last loophole, thanks guys

[itolduso]

Show that if A and B are invertible matrices of the same size, then AB is invertible and

How do I prove this?

I was thinking that if we can show then we are done... but how?

ABX=Y, BX=A^-1(Y), X=B^-1(A^-1(Y))=(B^-1A^-1)Y
.: (AB)^-1 exists and (AB)^-1=B^-1A^-1

[TrueTears]

I don't get this example, why does ? If it's reduced to row-echelon form, can't also satisfy the condition of the augmented matrix being in row echelon form?

[/0]

But if won't the last row be 0+0+0=1?

[TrueTears]

o i see haha, true that

[TrueTears]

Prove that for the 3 x 3 matrix, , where denotes the cofactor entry of

Any ideas?

I've think I've heard of a more general result, where if one multiplies the entries in any row (or column) of any matrix by the corresponding cofactors from a different row (or column), the sum of these products is always zero. I don't know the proof of this result and hopefully will know in the near future, but I suspect this question is a more specific case?

[kamil9876]

Looks like just an algebraic manipulation exercise. If the generalisation is indeed true, it looks like something that can be proven by induction (as is the case with a lot of things when it comes to determinants).

[TrueTears]

mind showing me ?

[kamil9876]

actually, I just thought of a really elegant solution! but i will post it 2moro as i am going to sleep now. The idea is that that sum calculates the determinant of a matrix with two equal rows (and we know such matrices have 0 determinants). In our case the matrix is that where you replace the last row with the first row. (though I think there may be some sign change or something going on, need to sort out those details.

[Ahmad]

Kamil's right if you copy and paste the first row of the matrix to replace the third row then it's just computing the determinant using cofactor expansion on the 3rd row.

[TrueTears]

actually, I just thought of a really elegant solution! but i will post it 2moro as i am going to sleep now. The idea is that that sum calculates the determinant of a matrix with two equal rows (and we know such matrices have 0 determinants). In our case the matrix is that where you replace the last row with the first row. (though I think there may be some sign change or something going on, need to sort out those details.

Kamil's right if you copy and paste the first row of the matrix to replace the third row then it's just computing the determinant using cofactor expansion on the 3rd row.

Oh yes that's right, very smart, thanks guys!