Let $A$ be a $3\times3$ matrix. Given $\mathrm{adj}(A)$, find $\det(A)$.

Question

Let $A$ be a $3\times3$ matrix such that $$\mathrm{adj}(A) = \begin{pmatrix}3 & -12 & -1 \\ 0 & 3 & 0 \\ -3 & -12 & 2\end{pmatrix}.$$Find the value of $\det(A)$.

I know that the adjoint of A is the transpose of the cofactors of $A$, but how do I find A or $\det(A)$ in reverse for a particular matrix? all I can think of is using the $A^{-1} = \det(A)/\mathrm{adj}(A)$ but I have no idea how to apply this for a direct value question.

Answer 1

$$A^{-1} = \dfrac{\text{adj} A}{\det A} \Rightarrow\\ \det(A^{-1}) = \dfrac{\det(\text{adj} A)}{(\det A)^n} \Rightarrow\\ \dfrac{1}{\det A } = \dfrac{\det(\text{adj} A)}{(\det A)^n} \Rightarrow\\ (\det A)^{n-1} = \det(\text{adj} A)\Rightarrow\\ \large\boxed{\det A = \sqrt[\leftroot{-3}\uproot{1}n-1]{\det (\text{adj}A)}}$$

Answer 2

We know that if $A \in \mathbb{R}^{n \times n}$ is an invertible matrix, then

$$A^{-1}=\frac{1}{\det(A)}\cdot \operatorname{adj}(A).$$

Multiplying from left with $A$ and cross-multiply with scalars we get

$$A \cdot \operatorname{adj}(A)=\det(A) \cdot I_n$$

By using property $(4)$ and $(5)$ from here and taking determinants for both side we get that

$$\det(A) \cdot \det(\operatorname{adj}(A))=(\det(A))^n \cdot \det(I_n)$$

So

$$\det(\operatorname{adj}(A))=(\det(A))^{n-1}.$$

Answer 3

Let's denote the adjoint of an $n\times n$ matrix by $\ ^\tilde\ $. We have the equalities:

\begin{eqnarray} A \cdot \tilde A = \det A \cdot I \end{eqnarray}

From here taking $\det$'s we get

$$\det A \cdot \det \tilde A = \det A^{n}$$

and so

$$\det \tilde A = \det A^{n-1}$$

Writing the equality

$$ \tilde A \cdot \tilde{ \tilde A} = \det \tilde A \cdot I$$ we get

$$\tilde {\tilde A}= \det A^{n-2} \cdot A$$

Now to our problem:

You can find $\det A^2$ and even $\det A \cdot A$ ( $= \text{adj}(\text{adj}(A))$ but $\det A$ or $A$ only up to a sign.

Answer 4

We have following identities involving determinants and adjugates.

• $\det(AB)=\det(A)\det(B)$
• $\det(I)=1$
• $A \operatorname{Adj}(A)=\det(A)I$

Finally, because of multilinearity of the determinant, if we multiply one column by a constant, the determinant scales by that same constnat, and so if we multiply an $n \times n$ matrix by a constant, we are multiplying $n$ columns by constnats, generating a factor of $c^n$.

Putting everything together, if we take the determinant of our third identity, we get

$$\det(A) \det(\operatorname{Adj}(A))=\det(A)^n\det(I)=\det(A)^n$$

Therefore, $\det(\operatorname{Adj}(A))=\det(A)^{n-1}$. For a $3\times 3$ matrix, $n=2$.