Question

prove that if matrix A is invertible then A(transpose) is invertible

Answer 1

consider n*n square matrices.
given that, A.B = B.A = I where B ,inverse of A, exists

Now,
\[ A ^{T}.B ^{T}\]
\[=(B .A)^{T}\] by property
\[=(I)^{T}\]
\[=I\]

Answer 2

use the fact that
\[(AB)^t=B^tA^t\] and that you know \[A^t \] exists

Answer 3

then
\[A^t(A^{-1})^t=(A^{-1}A)^t=I^t=I\] and then repeat on there side

Answer 4

thank u sooo much A-clan and satellite 73 both of u

Answer 5

Alternatively, if A is invertible then \( det(A) \neq 0 \) . Now \( det(A^T) = det(A) \) hence \( det(A^T) \neq 0 \) and \(A^T\) is invertible.