Question:

Step 1: Given $(A^2 - 2I)B = O$. Since $B$ is a non-singular matrix, $B^{-1}$ exists. Multiplying both sides by $B^{-1}$ from the right, we obtain $A^2 - 2I = O$, which implies $A^2 = 2I$. Step 2: The matrix $A$ is given by $A = \begin{bmatrix} a & b & c \\ b & c & a \\ c & a & b \end{bmatrix}$. The cofactor matrix $C$ of $A$ has entries $C_{11} = bc-a^2$, $C_{12} = ac-b^2$, $C_{13} = ab-c^2$, and so on. Since $A$ is a symmetric matrix, its cofactor matrix $C$ is also symmetric, so $\text{adj}(A) = C^T = C$. The matrix $B$ is given by $B = \begin{bmatrix} bc-a^2 & ca-b^2 & ab-c^2 \\ ca-b^2 & ab-c^2 & bc-a^2 \\ ab-c^2 & bc-a^2 & ac-b^2 \end{bmatrix}$. By direct comparison of entries, it is observed that $B = -\text{adj}(A)$. Step 3: From $A^2 = 2I$, taking the determinant yields $\det(A^2) = \det(2I)$. This implies $(\det(A))^2 = 2^3 \det(I) = 8$, so $\det(A) = \pm 2\sqrt{2}$. For the given matrix $A$ with $a > b > c > 0$, the determinant is positive. Thus, $\det(A) = 2\sqrt{2}$. Using the property $A \cdot \text{adj}(A) = \det(A) \cdot I$ and the relation $B = -\text{adj}(A)$ from Step 2, we have $AB = A(-\text{adj}(A)) = -A \cdot \text{adj}(A) = -\det(A) \cdot I$. Step 4: The trace of $AB$ is $Tr.(AB) = Tr.(-\det(A) \cdot I)$. Since $I$ is a $3 \times 3$ identity matrix, $Tr.(-\det(A) \cdot I) = -3\det(A)$. Substituting $\det(A) = 2\sqrt{2}$ from Step 3, we get $Tr.(AB) = -3(2\sqrt{2}) = -6\sqrt{2}$. Step 5: We need to compute $\det(A - \sqrt{2}B)$. From $A^2 = 2I$, we can multiply by $A^{-1}$ (since $A$ is non-singular as $\det(A) \neq 0$) to get $A = 2A^{-1}$, or $A^{-1} = \frac{1}{2}A$. Using the property $\text{adj}(A) = \det(A) \cdot A^{-1}$, and substituting $\det(A) = 2\sqrt{2}$ from Step 3 and $A^{-1} = \frac{1}{2}A$, we get: $$ \text{adj}(A) = (2\sqrt{2}) \cdot \frac{1}{2}A = \sqrt{2}A $$ Step 6: Substitute $B = -\text{adj}(A)$ from Step 2 and $\text{adj}(A) = \sqrt{2}A$ from Step 5 into the expression $A - \sqrt{2}B$: $$ A - \sqrt{2}B = A - \sqrt{2}(-\text{adj}(A)) $$ $$ A - \sqrt{2}B = A + \sqrt{2}(\text{adj}(A)) $$ $$ A - \sqrt{2}B = A + \sqrt{2}(\sqrt{2}A) $$ $$ A - \sqrt{2}B = A + 2A $$ $$ A - \sqrt{2}B = 3A $$ Step 7: Therefore, $\det(A - \sqrt{2}B) = \det(3A)$. For a $3 \times 3$ matrix, $\det(3A) = 3^3 \det(A) = 27\det(A)$. Substituting $\det(A) = 2\sqrt{2}$ from Step 3, we get $\det(A - \sqrt{2}B) = 27(2\sqrt{2}) = 54\sqrt{2}$.