If $\text{adj}A=\left[\begin{smallmatrix}1&-1&0\\2&3&1\\2&1&-1\end{smallmatrix}\right]$ and $\text{adj}(2A)$ is $2^k\text{adj}(A)$, find $k$

Question

I have thought of a solution for this, but I know it’s wrong. I don’t know what’s wrong with the procedure, I just solved it instinctively.

$$\text{adj} A =A^{-1}|A|$$

So for $2A$

$$\text{adj} 2A =(2A)^{-1} |2A|$$

$$=\frac 12 A 4 |A|$$ $$=2\text{adj} A$$

Which implies $k=1$

As I said, the answer is wrong, I am aware of that. The question remains: what’s wrong with my procedure and how do I get the correct answer.

Thanks!

Answer 1

Note that for a $n\times n$ matrix $A$, $\det(mA) = m^n \det A$ where $m \in \mathbb R$. So the answer should really be $$(2A )^{-1} |2A| = \frac 12 A^{-1} \ 2^3 |A| = 4 \ \text{adj} (A) $$