Let A be an n n matrix such that |A|=2.

Question

Let $A$ be an $n	imes n$ matrix such that $|A|=2$. If the determinant of the matrix $\operatorname{adj}(2\cdot\operatorname{adj}(2A^{-1}))$ is $2^{84}$, then $n$ is equal to \_\_\_\_\_\_.

Answer: 5

Accepted Answer

Using $|\operatorname{adj}(kM)|=k^{n(n-1)}|\operatorname{adj}M|$ and $|\operatorname{adj}M|=|M|^{n-1}$ repeatedly:
$$\big|\operatorname{adj}(2\operatorname{adj}(2A^{-1}))\big|=2^{n(n-1)}\big|\operatorname{adj}(2A^{-1})\big|^{n-1}=2^{n(n-1)}\big|2A^{-1}\big|^{(n-1)^2}=2^{n(n-1)}\cdot2^{n(n-1)^2}\cdot|A|^{-(n-1)^2}.$$

Simplifying the exponent of 2 and using $|A|=2$: $2^{(n-1)(n^2-n+1)}=2^{84}\Rightarrow(n-1)(n^2-n+1)=84$.

Testing $n=5$: $(4)(21)=84$ .

So $n=5$.

Solutions · 1