Question:

Step 1: Determine the core condition on matrix A. We are given the matrix equation $(A^2 - 2I)B = O$, where $O$ is the zero matrix and $I$ is the identity matrix. We are also given that $B$ is a non-singular matrix. Since $B$ is non-singular, its inverse $B^{-1}$ exists. We can multiply the equation by $B^{-1}$ from the right: $$ (A^2 - 2I)B B^{-1} = O B^{-1} $$ $$ A^2 - 2I = O $$ This simplifies to the fundamental condition: $$ A^2 = 2I $$ Step 2: Calculate $A^2$ and derive conditions on $a, b, c$. The given matrix $A$ is a symmetric matrix: $$ A = \begin{bmatrix} a & b & c \\ b & c & a \\ c & a & b \end{bmatrix} $$ Now, we compute $A^2 = A \cdot A$: $$ A^2 = \begin{bmatrix} a & b & c \\ b & c & a \\ c & a & b \end{bmatrix} \begin{bmatrix} a & b & c \\ b & c & a \\ c & a & b \end{bmatrix} = \begin{bmatrix} a^2+b^2+c^2 & ab+bc+ca & ac+ba+cb \\ ab+bc+ca & b^2+c^2+a^2 & bc+ca+ab \\ ac+ba+cb & bc+ca+ab & c^2+a^2+b^2 \end{bmatrix} $$ Equating the elements of $A^2$ to the elements of $2I = \begin{bmatrix} 2 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 2 \end{bmatrix}$: From the diagonal elements, we get: $$ a^2+b^2+c^2 = 2 $$ From the off-diagonal elements, we get: $$ ab+bc+ca = 0 $$ Step 3: Analyze the derived conditions and determine $Tr(A)$ and $det(A)$. We have two conditions on $a, b, c$: 1. $a^2+b^2+c^2 = 2$ 2. $ab+bc+ca = 0$ We are also given the constraint $a > b > c > 0$. However, if $a, b, c$ are all strictly positive, then $ab, bc, ca$ must also be strictly positive, which implies $ab+bc+ca > 0$. This directly contradicts the condition $ab+bc+ca = 0$. This suggests an inconsistency in the problem statement's given conditions. In the context of such problems, we generally proceed by prioritizing the algebraic conditions derived from matrix properties. Let's calculate $Tr(A) = a+b+c$. We know that $(a+b+c)^2 = a^2+b^2+c^2 + 2(ab+bc+ca)$. Substituting the derived conditions: $$ (a+b+c)^2 = 2 + 2(0) = 2 $$ So, $a+b+c = \pm\sqrt{2}$. Given $a > b > c > 0$, it must be that $a+b+c > 0$. Therefore, $Tr(A) = a+b+c = \sqrt{2}$. Next, let's find $det(A)$. Since $A^2 = 2I$, we take the determinant of both sides: $$ \det(A^2) = \det(2I) $$ $$ (\det(A))^2 = 2^3 = 8 $$ So, $\det(A) = \pm\sqrt{8} = \pm 2\sqrt{2}$. Since $A$ is a symmetric matrix, its eigenvalues are real. From $A^2=2I$, all eigenvalues $\lambda$ of $A$ must satisfy $\lambda^2=2$, so $\lambda = \pm\sqrt{2}$. The sum of eigenvalues is $Tr(A) = \sqrt{2}$. The product of eigenvalues is $det(A)$. To get a sum of $\sqrt{2}$ from three eigenvalues, each being $\pm\sqrt{2}$, the eigenvalues must be $(\sqrt{2}, \sqrt{2}, -\sqrt{2})$. The determinant is the product of these eigenvalues: $$ \det(A) = (\sqrt{2})(\sqrt{2})(-\sqrt{2}) = -2\sqrt{2} $$ Step 4: Calculate $Tr(AB)$. Matrix $B$ is given as: $$ B = \begin{bmatrix} bc-a^2 & ca-b^2 & ab-c^2 \\ ca-b^2 & ab-c^2 & bc-a^2 \\ ab-c^2 & bc-a^2 & ac-b^2 \end{bmatrix} $$ Let $x = bc-a^2$, $y = ca-b^2$, $z = ab-c^2$. Then $B$ has the same symmetric structure as $A$: $$ B = \begin{bmatrix} x & y & z \\ y & z & x \\ z & x & y \end{bmatrix} $$ Since both $A$ and $B$ have this specific symmetric structure, they commute, i.e., $AB = BA$. The trace of $AB$ is the sum of its diagonal elements. The diagonal elements of $AB$ are: $(AB)_{11} = a(bc-a^2) + b(ca-b^2) + c(ab-c^2)$ $(AB)_{22} = b(ca-b^2) + c(ab-c^2) + a(bc-a^2)$ $(AB)_{33} = c(ab-c^2) + a(bc-a^2) + b(ca-b^2)$ Each diagonal element simplifies to $abc-a^3 + abc-b^3 + abc-c^3 = 3abc-(a^3+b^3+c^3)$. This expression is the formula for $\det(A)$. Thus, $Tr(AB) = 3 \times (3abc-(a^3+b^3+c^3)) = 3 \det(A)$. Using $\det(A) = -2\sqrt{2}$ from Step 3: $$ Tr(AB) = 3(-2\sqrt{2}) = -6\sqrt{2} $$ This matches Option (b). Step 5: Calculate $det(A - \sqrt{2}B)$. Since $A$ and $B$ commute, they are simultaneously diagonalizable. The eigenvalues of $A - \sqrt{2}B$ are $\lambda_k(A) - \sqrt{2}\lambda_k(B)$, where $\lambda_k(A)$ and $\lambda_k(B)$ are the corresponding eigenvalues of $A$ and $B$. The eigenvalues of $A$ are $(\sqrt{2}, \sqrt{2}, -\sqrt{2})$. The trace of $B$ is $Tr(B) = x+y+z = (bc-a^2) + (ca-b^2) + (ab-c^2) = (ab+bc+ca) - (a^2+b^2+c^2)$. Using the conditions from Step 2, $ab+bc+ca = 0$ and $a^2+b^2+c^2 = 2$: $$ Tr(B) = 0 - 2 = -2 $$ For a matrix of this form $M = \begin{bmatrix} x & y & z \\ y & z & x \\ z & x & y \end{bmatrix}$, its eigenvalues are $x+y+z$ and $x - \frac{y+z}{2} \pm \frac{i\sqrt{3}}{2}(y-z)$. Since $B$ is a symmetric matrix, all its eigenvalues must be real. This implies the imaginary part must be zero, so $y-z=0 \implies y=z$. If $y=z$, then $ca-b^2 = ab-c^2 \implies c(a+c) = b(a+b)$. Given $a > b > c > 0$, this equality only holds if $b=c$, which contradicts $b > c$. This confirms another inconsistency in the problem statement, specifically in the constraints `a > b > c > 0` and the symmetric structure of matrix `B`. However, for a JEE problem, there must be an intended calculation. Let's look for a simplification. Consider the properties of the values $x, y, z$ under $ab+bc+ca=0$ and $a^2+b^2+c^2=2$. $x=bc-a^2 = -(ab+ca)-a^2 = -a(b+c)-a^2 = -a(a+b+c)$. Similarly, $y = ca-b^2 = -b(a+b+c)$. And $z = ab-c^2 = -c(a+b+c)$. Since $a+b+c = \sqrt{2}$ (from Step 3), we have: $x = -a\sqrt{2}$ $y = -b\sqrt{2}$ $z = -c\sqrt{2}$ Substitute these into matrix $B$: $$ B = \begin{bmatrix} -a\sqrt{2} & -b\sqrt{2} & -c\sqrt{2} \\ -b\sqrt{2} & -c\sqrt{2} & -a\sqrt{2} \\ -c\sqrt{2} & -a\sqrt{2} & -b\sqrt{2} \end{bmatrix} = -\sqrt{2} \begin{bmatrix} a & b & c \\ b & c & a \\ c & a & b \end{bmatrix} = -\sqrt{2} A $$ So, $B = -\sqrt{2} A$. Now we can calculate $A - \sqrt{2}B$: $$ A - \sqrt{2}B = A - \sqrt{2}(-\sqrt{2}A) = A + 2A = 3A $$ Therefore, $\det(A - \sqrt{2}B) = \det(3A)$. Using the property $\det(kA) = k^n \det(A)$ for an $n \times n$ matrix, and $A$ is $3 \times 3$: $$ \det(3A) = 3^3 \det(A) = 27 \det(A) $$ From Step 3, we found $\det(A) = -2\sqrt{2}$. $$ \det(A - \sqrt{2}B) = 27(-2\sqrt{2}) = -54\sqrt{2} $$ This matches Option (d). Final Answer: The correct statements are (b) and (d). The final answer is $\boxed{\text{BD}}$.