$$ \text { If } y(x)=\left|\begin{array}{ccc} \sin x & \cos x & \sin x+\cos x+1 \\ 27 & 28 & 27 \\ 1 & 1 & 1 \end{array}\right|, x \in \mathbb{R} \text {, then } \frac{d^2 y}{d x^2}+y \text { is equal to } $$

<p>To solve for $ y(x) $, we start with the determinant of the given matrix:</p> <p>$ \begin{vmatrix} \sin x & \cos x & \sin x + \cos x + 1 \\ 27 & 28 & 27 \\ 1 & 1 & 1 \end{vmatrix} $</p> <p>Using the formula for a $3 \times 3$ determinant, we expand along the first row:</p> <p>$ f(x) = \sin x \cdot \begin{vmatrix} 28 & 27 \\ 1 & 1 \end{vmatrix} - \cos x \cdot \begin{vmatrix} 27 & 27 \\ 1 & 1 \end{vmatrix} + (\sin x + \cos x + 1) \cdot \begin{vmatrix} 27 & 28 \\ 1 & 1 \end{vmatrix} $</p> <p>Calculating the smaller $2 \times 2$ determinants:</p> <p>$ \begin{vmatrix} 28 & 27 \\ 1 & 1 \end{vmatrix} = 28 \cdot 1 - 27 \cdot 1 = 1 $</p> <p>$ \begin{vmatrix} 27 & 27 \\ 1 & 1 \end{vmatrix} = 27 \cdot 1 - 27 \cdot 1 = 0 $</p> <p>$ \begin{vmatrix} 27 & 28 \\ 1 & 1 \end{vmatrix} = 27 \cdot 1 - 28 \cdot 1 = -1 $</p> <p>Substituting these into the determinant calculation gives:</p> <p>$ f(x) = \sin x \cdot 1 - \cos x \cdot 0 + (\sin x + \cos x + 1) \cdot (-1) $</p> <p>$ f(x) = \sin x - (\sin x + \cos x + 1) $</p> <p>$ f(x) = \sin x - \sin x - \cos x - 1 $</p> <p>$ f(x) = -\cos x - 1 $</p> <p>Now, calculate the derivatives:</p> <p><p>First derivative $ f'(x) $:</p> <p>$ f'(x) = \frac{d}{dx}(-\cos x - 1) = \sin x $</p></p> <p><p>Second derivative $ \frac{d^2 f}{dx^2} $:</p> <p>$ \frac{d^2 f}{dx^2} = \frac{d}{dx}(\sin x) = \cos x $</p></p> <p>Finally, compute $ \frac{d^2 y}{dx^2} + y $:</p> <p>$ \frac{d^2 f}{dx^2} + f(x) = \cos x - \cos x - 1 = -1 $</p> <p>Thus, $ \frac{d^2 y}{dx^2} + y $ is equal to $-1$.</p>