AP EAPCET 15th May 2023 Shift 1 Solved Paper

To solve the problem, we need to find the matrix $S A S^{-1}$ given the matrices $S$ and $A$. Let's proceed step by step.First, let's write down the given matrices:$S = \begin{pmatrix} 0 & 1 & 1 \\ 1 & 0 & 1 \\ 1 & 1 & 0 \end{pmatrix}$$A = \frac{1}{2} \begin{pmatrix} b+c & c-a & b-a \\ c-b & c+a & a-b \\ b-c & a-c & a+b \end{pmatrix}$We need to find $S A S^{-1}$. To do this, we will use the property of similar matrices. Two matrices $A$ and $B$ are similar if there exists an invertible matrix $P$ such that $B = P A P^{-1}$. In our case, $S A S^{-1}$ is similar to $A$ with $P = S$.Let's compute the eigenvalues of $S$. The characteristic polynomial of $S$ is given by:$\det(S - \lambda I) = \begin{vmatrix} -\lambda & 1 & 1 \\ 1 & -\lambda & 1 \\ 1 & 1 & -\lambda \end{vmatrix}$Expanding the determinant, we get:$\det(S - \lambda I) = -\lambda \begin{vmatrix} -\lambda & 1 \\ 1 & -\lambda \end{vmatrix} - 1 \begin{vmatrix} 1 & 1 \\ 1 & -\lambda \end{vmatrix} + 1 \begin{vmatrix} 1 & -\lambda \\ 1 & 1 \end{vmatrix}$$= -\lambda (\lambda^2 - 1) - 1 (-\lambda - 1) + 1 (1 + \lambda)$$= -\lambda^3 + \lambda + \lambda + 1 + 1$$= -\lambda^3 + 2\lambda + 2$The eigenvalues of $S$ are the roots of the characteristic polynomial. By inspection, we can see that $\lambda = 2$ is a root:$-\lambda^3 + 2\lambda + 2 = -(2)^3 + 2(2) + 2 = -8 + 4 + 2 = -2 \neq 0$This is incorrect, so let's recompute the characteristic polynomial correctly:$\det(S - \lambda I) = -\lambda^3 + 3\lambda + 2 = 0$The roots are $\lambda = -1, -1, 3$.Now, let's find the eigenvalues of $A$. Since $S A S^{-1}$ is similar to $A$, the eigenvalues of $S A S^{-1}$ are the same as the eigenvalues of $A$.The matrix $A$ is a symmetric matrix, and its eigenvalues can be found by solving the characteristic equation:$\det(A - \lambda I) = 0$The eigenvalues of $A$ are $a, b, c$.Therefore, the eigenvalues of $S A S^{-1}$ are $a, b, c$. The matrix $S A S^{-1}$ is similar to $A$, so it must be of the form:$S A S^{-1} = \begin{pmatrix} a & 0 & 0 \\ 0 & b & 0 \\ 0 & 0 & c \end{pmatrix}$Thus, the correct answer is:$A$