Let $$A$$ be a $$3 \times 3$$ real matrix. If $$\det(2\text{Adj}(2\text{Adj}(\text{Adj}(2A)))) = 2^{41}$$, then the value of $$\det(A^2)$$ equals _________

We work with a $$3\times 3$$ real matrix $$A$$. Throughout, we shall recall three standard facts about determinants and adjugates:

1. For any scalar $$k$$ and matrix $$M$$ of order $$3$$, $$\operatorname{Adj}(kM)=k^{3-1}\operatorname{Adj}(M)=k^{2}\operatorname{Adj}(M).$$

2. For any invertible matrix $$M$$ of order $$3$$, $$\det(\operatorname{Adj}(M))=\det(M)^{3-1}=\det(M)^{2}.$$ (This remains true even if $$M$$ is singular, by continuity.)

3. For order $$3$$, $$\operatorname{Adj}(\operatorname{Adj}(M))=\det(M)^{3-2}\,M=\det(M)\,M.$$

We now unravel the given expression step by step. Put $$B=2A$$ so that $$\det(B)=\det(2A)=2^{3}\det(A)=8\det(A).$$

First inner adjugate:
$$\operatorname{Adj}(B)=\operatorname{Adj}(2A)=2^{2}\operatorname{Adj}(A)=4\,\operatorname{Adj}(A).$$

Next, applying fact 3 to $$B$$:

$$\operatorname{Adj}(\operatorname{Adj}(B))=\det(B)\,B=\det(2A)\,(2A)=8\det(A)\,(2A)=16\det(A)\,A.$$

Multiply this by the scalar $$2$$ that sits immediately outside it:

$$D=2\,\operatorname{Adj}(\operatorname{Adj}(B))=2\,(16\det(A)\,A)=32\det(A)\,A.$$

We must now take the adjugate of $$D$$ and afterwards multiply once more by the leading scalar $$2$$ present in the original problem. Using fact 1 on $$D=kA$$ with $$k=32\det(A)\,,$$ we have

$$\operatorname{Adj}(D)=\operatorname{Adj}(32\det(A)\,A)=(32\det(A))^{2}\operatorname{Adj}(A)=1024\,\det(A)^{2}\operatorname{Adj}(A).$$

Placing the outermost factor $$2$$ finally gives the matrix whose determinant is prescribed:

$$M=2\,\operatorname{Adj}(D)=2\left(1024\,\det(A)^{2}\operatorname{Adj}(A)\right)=2048\,\det(A)^{2}\operatorname{Adj}(A)=32\,\det(2A)^{2}\operatorname{Adj}(A).$$

Observe that $$2048=2^{11}$$ and, crucially, $$\det(2A)=8\det(A)$$; so in determinant form we write succinctly

$$M=32\,\bigl(\det(2A)\bigr)^{2}\operatorname{Adj}(A).$$

We now evaluate $$\det(M)$$. Because $$M$$ is a product of a scalar matrix factor and $$\operatorname{Adj}(A)$$, we use $$\det(kM)=k^{3}\det(M)$$:

$$ \det(M) =\bigl(32\,\det(2A)^{2}\bigr)^{3}\,\det\!\bigl(\operatorname{Adj}(A)\bigr). $$

Insert the separate determinants one by one:

• $$32=2^{5} \Longrightarrow 32^{3}=2^{15}.$$br> • $$\det(2A)^{2}=(8\det(A))^{2}=64\det(A)^{2} \Longrightarrow (\,\det(2A)^{2}\,)^{3}=64^{3}\det(A)^{6}=2^{18}\det(A)^{6}.$$br> • By fact 2, $$\det\!\bigl(\operatorname{Adj}(A)\bigr)=\det(A)^{2}.$$

Putting all these together,

$$ \det(M)=2^{15}\;\times\;2^{18}\,\det(A)^{6}\;\times\;\det(A)^{2} =2^{33}\,\det(A)^{8}. $$

The problem states $$\det(M)=2^{41}$$, so we equate:

$$2^{33}\,\det(A)^{8}=2^{41}\quad\Longrightarrow\quad \det(A)^{8}=2^{41-33}=2^{8}.$$

Taking the real eighth root gives $$\det(A)=2$$(the sign is immaterial here, because we soon square the determinant).

Finally, to reach the required quantity, recall $$\det(A^{2})=\bigl(\det(A)\bigr)^{2}$$ for any square matrix. Hence

$$\det(A^{2})=\Bigl(2\Bigr)^{2}=4.$$

So, the answer is $$4$$.