MHB Inverse of adjoint - where is my mistake ?

[Yankel]

Hello all, I have a matrix A:

\[\begin{pmatrix} 2 &4 &1 \\ -4 &7 &3 \\ 5 &1 &-2 \end{pmatrix}\]

and I need to find the adjoint of the matrix inverse.

I found adj(A) to be:

\[\begin{pmatrix} -17 &9 &5 \\ 7 &-9 &-10 \\ -39 &18 &30 \end{pmatrix}\]

and I found the determinant of A to be -45 and the determinant of adj(A) to be 2025.

Now based on:

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

I tried solving the question, I did:

\[B=adj(A))\]

and looked for:

\[B^{-1}\]

This way:

\[B^{-1}=\frac{1}{\left | B \right |}adj(B)\]

and got:

\[\frac{1}{2025}A\]

which is not the answer. the answer should be:

\[-\frac{1}{45}A\]

And I don't understand what I did wrong here.

Thank you

 

[I like Serena]

Hello all, I have a matrix A:

\[\begin{pmatrix} 2 &4 &1 \\ -4 &7 &3 \\ 5 &1 &-2 \end{pmatrix}\]

and I need to find the adjoint of the matrix inverse.

I found adj(A) to be:

\[\begin{pmatrix} -17 &9 &5 \\ 7 &-9 &-10 \\ -39 &18 &30 \end{pmatrix}\]

and I found the determinant of A to be -45 and the determinant of adj(A) to be 2025.

Now based on:

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

I tried solving the question, I did:

\[B=adj(A))\]

and looked for:

\[B^{-1}\]

This way:

\[B^{-1}=\frac{1}{\left | B \right |}adj(B)\]

and got:

\[\frac{1}{2025}A\]

which is not the answer. the answer should be:

\[-\frac{1}{45}A\]

And I don't understand what I did wrong here.

Thank you

Hi Yankel,

The inverse of $A$ is given by:
$$A^{-1}=\frac{1}{\det A} \text{adj }A$$
See adjugate matrix (as it is called with less ambiguity) on wiki.Speaking about $\det(\text{adj }A)$, it relates to $\det A$ as:
$$\det(\text{adj }A) = (\det A)^{n-1} = (-45)^{3-1} = 2025$$

 

[Yankel]

I like Serena, thank you !

among the formulas out there I mentioned that I did find det(adj(A)) to be 2025, it was easy to miss this line.

This is not what I am asking. Taking the first formula you mentioned, I found the adj(inverse of A), and I was wrong, and can't find my mistake.

 

[I like Serena]

I like Serena, thank you !

among the formulas out there I mentioned that I did find det(adj(A)) to be 2025, it was easy to miss this line.

This is not what I am asking. Taking the first formula you mentioned, I found the adj(inverse of A), and I was wrong, and can't find my mistake.

We have:
$$A^{-1} = \frac 1{\det A} \text{adj }A$$
Therefore we also have:
$$A = \frac 1{\det A^{-1}} \text{adj}(A^{-1}) \quad\Rightarrow\quad \text{adj}(A^{-1}) = \det A^{-1} \cdot A = \frac 1{\det A} \cdot A$$

Now based on:

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

I tried solving the question, I did:

\[B=adj(A))\]

and looked for:

\[B^{-1}\]

This way:

\[B^{-1}=\frac{1}{\left | B \right |}adj(B)\]

Let's substitute $B=\text{adj}(A)$. That gives us:

\[B^{-1}=\frac{1}{\left | \text{adj}(A) \right |}\text{adj}(\text{adj}(A))\]

Did you evaluate $\text{adj}(\text{adj}(A))$?