Question 1006566
Note: the notation "[a,b;c,d]" means "the matrix with the first row being a,b and the second row being c,d"


Matrix A is some 2 x 2 matrix, so let's set it up as
A = [a,b;c,d]
where a,b,c,d are constants


A is invertible ---> determinant D = ad-bc is nonzero


A^T = [a,c;b,d]
determinant of A^T = ad-bc = determinant of matrix A


Note: only b and c swap places when we go from A to A^T. So both A and A^T have the same determinant


Since A has a nonzero determinant, so does A^T.


If A is invertible, then A^T is definitely invertible.


So the first part of the claim has been proven true.


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Let's compute the inverse of matrix A


A = [a,b;c,d] 
A^(-1) = 1/D * [d,-b;-c,a] 


Now let's transpose A^(-1) to get
(A^(-1))^T = 1/D * [d,-c;-b,a] 


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Now compute the inverse of matrix A^T


A^T = [a,c;b,d]
(A^T)^(-1) = 1/D * [d,-c;-b,a]


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So,


(A^T)^(-1) = 1/D * [d,-c;-b,a]
and
(A^(-1))^T = 1/D * [d,-c;-b,a] 


we can see that (A^T)^(-1) = (A^(-1))^T is true.


The second part of the statement has been proven true.


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Both parts of the statement are true. A^T is invertible and the equation given holds true.