Find the value of $k$ if $\det(A^{–1}) = (\det A)^k$ where $A$ is a $3 \times 3$ matrix

Question

Now I'm getting two answers to this question i.e. $k = -1$ and $k = 1$

Reason i'm getting $k = 1$ :

$$A(A^{-1}) = I$$

Taking determinant on both sides we get

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

$$\det(A)(\det(A^{-1})) = 1$$

Dividing both sides with $\det(A)$ we get

$$\det(A^{-1}) = \det(A)^{-1}$$

Hence, $k = -1$

Reason I'm getting $k = 1$ :

$$A^{-1} = \frac1{\det(A)}Adj(A)$$

Applying determinant on both sides we get

$$\det(A^{-1}) = \det{[\frac1{\det(A)}]Adj(A)}$$

$$\det(A^{-1}) = \frac1{\det(A)}\det[Adj(A)]$$

because $\det[Adj(A)] = \det(A)^{n-1}$

i.e. $\det[Adj(A)] = \det(A)^{3-1}$

Or, $$\det[Adj(A)] = \det(A)^2$$

Thus, $$\det(A^{-1}) = \frac1{\det(A)}[\det(A)^2]$$

Or, $$\det(A^{-1}) = \det(A)^1$$

Hence $k = 1$

Now my Question is which one is the right answer And why am I getting two answers, have i neglected a condition or what Any kind of help will be much appreciated.

P.S. $k = -1$ is the answer given in all books I've come across.

Answer 1

First the mistake:

$$\det(A^{-1}) = \frac{1}{\det(A)^\color{red}{3}}{\det[Adj(A)]}$$

because $\det[Adj(A)] = \det(A)^{n-1}$

i.e. $$\det[Adj(A)] = \det(A)^{3-1}$$

Or, $$\det[Adj(A)] = \det(A)^2$$

Thus, $$\det(A^{-1}) = \frac1{\det(A)^3}[\det(A)^2]=\frac{1}{\det(A)}$$

which is a known result.

Next, this is how I would approach the problem:

$$\det(A^{-1})= \det(A)^k$$

Hence we have $$1=\det(A)^{k+1}$$

If we want the equality to hold as long as $\det(A) \neq 0$, we require $k+1=0$.

Notice that I did not use the propery of the size of $A$.

Answer 2

Well. Observe that if $A$ is $3\times 3$ matrix then \begin{align} \det(c A)= c^3 \det(A). \end{align}

Example: Consider \begin{align} \det \begin{pmatrix} c & 0 & 0\\ 0 & c & 0\\ 0 & 0 & c \end{pmatrix} = c\det \begin{pmatrix} 1 & 0 & 0\\ 0 & c & 0\\ 0 & 0 & c \end{pmatrix} = c^2\det \begin{pmatrix} 1 & 0 & 0\\ 0 & 1 & 0\\ 0 & 0 & c \end{pmatrix} = c^3\det \begin{pmatrix} 1 & 0 & 0\\ 0 & 1 & 0\\ 0 & 0 & 1 \end{pmatrix} \end{align}

Answer 3

Suppose that $A$ has the property $det(A)=1$, then $det(A^{-1})=1$, hence

$det(A^{-1})= ( \det(A))^k$ for all(!) $k \in \mathbb Z$.