We are given with two invertible matrices A and B, how to prove that $${(AB)^{ - 1}} = {B^{ - 1}}{A^{ - 1}}$$ ? We know that if A.B $$ = I$$ then it means B is inverse of matrix A where $$I$$ is an identity matrix. If, we can prove that … [Continue reading]
We are given an invertible matrix A then how to prove that (A^T)^ – 1 = (A^ – 1)^T?
How to prove that $${({A^T})^{ - 1}} = {({A^{ - 1}})^T}$$ where A is an invertible square matrix, T represents transpose and $${A^{ - 1}}$$ is inverse of matrix A. In other words we want to prove that inverse of $${A^T}$$ is equal to $${({A^{ - … [Continue reading]
How to prove that transpose of adj(A) is equal to adj(A transpose)?
If, we have invertible square matrix A, then how to prove that $${(adj(A))^T} = adj({A^T})$$ ? adj(A) is adjoint of A and T represents transpose of matrix. … [Continue reading]
How to prove that adj(AB)=adj(B).adj(A)?
How to prove that adjoint(AB)= adjoint(B).adjoint(A) if its given that A and B are two square and invertible matrices. Using formula to calculate inverse of matrix, we can say that $${(AB)^{ - 1}} = adj(AB)/\det (AB)$$ … [Continue reading]
How to prove that A.adj(A)= adj(A).A=det(A).I ?
Note: This property holds for square matrices. If, we are given matrix A then How to prove that $$A.adj(A) = adj(A).A = \det (A)I$$? where adj(A) is adjoint of A and det(A) is determinant of A. … [Continue reading]
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