Matrix, Prove of matrix theorem

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The discussion focuses on proving two matrix theorems involving nonsingular matrices: (AB)^{T} = B^{T}A^{T} and (ABC)^{-1} = C^{-1}B^{-1}A^{-1}. One participant initially attempted to solve these by creating random matrices, but was advised to prove the statements for any matrices without specific examples. For the first theorem, it was suggested to consider the ijth element of the matrix and utilize properties of matrix dimensions. The second theorem can be approached by letting D = (ABC)^{-1} and manipulating the equation through multiplication. The conversation emphasizes the importance of general proofs in linear algebra.
nekteo
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prove that if ABC are nonsingular matrices,
A) (AB)^{T} = B^{T}A^{T}
B) (ABC)^{-1} = C^{-1}B^{-1}A^{-1}

I attempted to solve it by creating a random matrices by my self and solved it, however, my teacher demand an answer without "creating" a new matrices by our self...
 
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Try considering the ijth element of the matrix. That is, for the first one, consider (AB)^T_{ij} and expand.
 
o! i got the 1st question... Thx!
is the 2nd ques also using the same method?

Cheers!
Kenneth
 
nekteo said:
prove that if ABC are nonsingular matrices,
A) (AB)^{T} = B^{T}A^{T}
B) (ABC)^{-1} = C^{-1}B^{-1}A^{-1}

I attempted to solve it by creating a random matrices by my self and solved it, however, my teacher demand an answer without "creating" a new matrices by our self...
Do you understand what your teacher was saying? If you "create a random matrix" and do the calculations for that matrix, then you have proved the statement is true for that matrix. You are asked to prove it is true for any matrix.
 
For the 2nd one let D=(ABC)^{-1} and then go from there
mutilyply by ABC

ABCD=(ABC)^{-1}(ABC)
ABCD=I
then proceed to multiply by A^{-1} and so forth

for the first one, I think you need to use the property that if A is a mxn matrix and B is a nxs matrix then AB is mxs matrix..then you need to say what kind of matrix would A^{T} would be.(nxm)
 
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Since ##px^9+q## is the factor, then ##x^9=\frac{-q}{p}## will be one of the roots. Let ##f(x)=27x^{18}+bx^9+70##, then: $$27\left(\frac{-q}{p}\right)^2+b\left(\frac{-q}{p}\right)+70=0$$ $$b=27 \frac{q}{p}+70 \frac{p}{q}$$ $$b=\frac{27q^2+70p^2}{pq}$$ From this expression, it looks like there is no greatest value of ##b## because increasing the value of ##p## and ##q## will also increase the value of ##b##. How to find the greatest value of ##b##? Thanks
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