Find the eigenvalue of the matrix. (Hint: one of the eigenvalues is 4. What are the eigenvalues of ?)

Find the corresponding eigenvectors. Fortunately, for this problem there are three independent eigenvectors so the matrix is "diagonalizable" (every symmetric matrix is diagonalizable). That is, if P is the matrix having the eigenvectors as columns, where A is the given matrix and P is the diagonal matrix having the eigenvalues on the main diagonal.

Multiply the entire equation by P to get Py'= (Py)'= PAy. Let so that . Then the equation becomes . X'= DX is easy to solve and then y= P^{-1}X.