Question

If A is invertible, then Adj(A) is invertible and
[Adj(A)]^-1 = (1/det(A)) A = adj(A^-1)

Answer 1

@Lyrae help me

Answer 2

If A is invertible, then Adj(A) is invertible and

\[[Adj(A)]^{-1} = \frac{ 1 }{ \det(A) } A = adj (A^{-1})\]

Answer 3

prove it

Answer 4

take small 2*2 matrix for ex this one

\(\large\tt \color{black}{\left[\begin{matrix} 2 & 2 \\ 3 & 2 \end{matrix}\right]}\)

Answer 5

See ... The Laplace expansion

Answer 6

i'm confused :(

Answer 7

\[A^{-1} = \frac{ 1 }{ \det(A) }A \times adj (A)\]
\[A A^{-1} = \frac{ 1 }{ \det(A) } A \times adj(A)\]
\[I = \frac{ 1 }{ \det(A) } A \times A adj(A)\]
\[I [adj(A)]^{-1} = \frac{ 1 }{ \det(A) } A\]
So
\[[adj(A)]^{-1} = \frac{ 1 }{ \det(A) } A\]

Answer 8

What you have typed is not correct (small errors)

Answer 9

look at the first line again

Answer 10

ahhh the first line would be

\[A^{-1} = \frac{ 1 }{ \det(A) } adj(A)\]

Answer 11

just typo :)

Answer 12

\[A \space adj(A) = \det(A)\space I\]Proofs for this property can be fount on the internet http://thejuniverse.org/PUBLIC/LinearAlgebra/LOLA/laplace/adjoint.html for example.

From this we can derive

If and only if A is invertible
\[A^{-1}A \space adj(A) = adj(A) = \det(A) \space A^{-1}\]\[\frac{ 1 }{ adj(A) } = adj(A)^{-1} = \frac{ 1 }{\det(A) A^{-1}} = \frac{ 1 }{ \det(A) } A\]
Similarly (from the same property) \[A^{-1}adj(A^{-1}) =\det(A^{-1}) \space I = \frac{ 1 }{ det(A) } I\]\[AA^{-1}adj(A^{-1}) = adj(A^{-1}) = \frac{ 1 }{ \det(A) } \space A\]
thus \[adj(A)^{-1} =\frac{ 1 }{ \det(A) } = adj(A^{-1})\]

Answer 13

yes

Answer 14

which is given by the The Laplace expansion

Answer 15

now replace \(A^{-1}\) by \(A\)

then

\[A^{-1} = \frac{ 1 }{ \det(A) } adj(A)\]

becomes

\[A = \frac{ 1 }{ \det(A^{-1}) } adj(A^{-1})\]

and then

\[adj(A^{-1})=det(A^{-1})A\]

\[adj(A^{-1})=\frac{1}{det(A)}A\]