Let I be the identity matrix of order $$3 \times 3$$ and for the matrix $$A = \begin{bmatrix} \lambda & 2 & 3 \\ 4 & 5 & 6 \\ 7 & -1 & 2 \end{bmatrix}$$, $$|A| = -1$$. Let B be the inverse of the matrix $$\text{adj}(A \cdot \text{adj}(A^2))$$. Then $$|(\lambda B + I)|$$ is equal to ________.

The determinant of the given matrix $$A=\begin{bmatrix}\lambda&2&3\\4&5&6\\7&-1&2\end{bmatrix}$$ is

$$|A|=\lambda\bigl(5\cdot2-6\cdot(-1)\bigr)-2\bigl(4\cdot2-6\cdot7\bigr)+3\bigl(4\cdot(-1)-5\cdot7\bigr)$$
$$\;\;=\lambda(10+6)-2(8-42)+3(-4-35)=16\lambda-49.$$

Given $$|A|=-1,$$ we get $$16\lambda-49=-1\; \Longrightarrow\; \lambda=3.$$

Case 1: Simplifying $$A\cdot\text{adj}(A^{2})$$

For any non-singular matrix, $$\text{adj}(A)=|A|\,A^{-1}.$$ Hence

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

Therefore

$$A\cdot\text{adj}(A^{2})=A\,\bigl(|A|^{2}A^{-2}\bigr)=|A|^{2}A^{-1}.$$

With $$|A|=-1,\quad |A|^{2}=1,$$ so

$$A\cdot\text{adj}(A^{2})=A^{-1}.$$ Denote $$M=A^{-1}.$$

Case 2: Finding the matrix $$B$$

First observe

$$\text{adj}(M)=\text{adj}(A^{-1})=|A^{-1}|\,A=(|A|)^{-1}A=(-1)^{-1}A=-A.$$ (The determinant $$|A^{-1}|=(|A|)^{-1}=-1.$)

By definition, $$B=\bigl($$\text{adj}$$(M)\bigr)^{-1}=(-A)^{-1}=-A^{-1}.$$

Case 3: Evaluating $$|($$\lambda$$ B+I)|$$

Put $$$$\lambda$$=3$$ and $$B=-A^{-1}$$:

$$$$\lambda$$ B+I=3(-A^{-1})+I=I-3A^{-1}.$$

Factor out $$A^{-1}$$: $$I-3A^{-1}=A^{-1}\,(A-3I).$$

Hence, for a $$3$$\times$$3$$ matrix,

$$|\,I-3A^{-1}\,|=|A^{-1}|\,|A-3I|.$$

We already know $$|A^{-1}|=(|A|)^{-1}=-1.$$ So we only need $$|A-3I|$$.

Compute $$A-3I$$ when $$$$\lambda$$=3$$:
$$A-3I=$$\begin{bmatrix}$$0&2&3\\4&2&6\\7&-1&-1\end{bmatrix}.$$ Its determinant is

$$0\bigl(2$$\cdot$$(-1)-6$$\cdot$$(-1)\bigr)-2\bigl(4$$\cdot$$(-1)-6$$\cdot$$7\bigr)+3\bigl(4$$\cdot$$(-1)-2$$\cdot$$7\bigr)=38.$$

Therefore

$$|\,$$\lambda$$ B+I\,|=|\,I-3A^{-1}\,|=(-1)$$\times$$38=-38.$$

The numerical value (modulus) of the determinant is $$38.$$

Final answer = 38