Question 1006566: If A is an invertible 2 x 2 matrix,then A^t is invertible and (At)^-1 =
(A^-1)^t.
is this statement true or false and why?
Answer by jim_thompson5910(35256)   (Show Source):
You can put this solution on YOUR website! 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.
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