1. Write the determinant

Given

$$y(x)=\begin{vmatrix} \sin x & \cos x & \sin x+\cos x+1\\ 27 & 28 & 27\\ 1 & 1 & 1 \end{vmatrix}$$

We need to find:

$$\frac{d^2y}{dx^2}+y$$

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2. Use column operation to simplify the determinant

Notice that the third column is:

$$C_3 = C_1 + C_2 + \begin{bmatrix}1\\27\\1\end{bmatrix}$$

But a more direct simplification is to do:

$$C_3 \to C_3 - C_1 - C_2$$

Then the determinant becomes:

$$y(x)=\begin{vmatrix} \sin x & \cos x & 1\\ 27 & 28 & -28\\ 1 & 1 & -1 \end{vmatrix}$$

since

$$(\sin x+\cos x+1)-\sin x-\cos x=1,$$ $$27-27-28=-28,$$ $$1-1-1=-1.$$

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3. Expand along the third column

Now compute

$$y=1\cdot \begin{vmatrix}27&28\\1&1\end{vmatrix} -(-28)\cdot \begin{vmatrix}\sin x&\cos x\\1&1\end{vmatrix} +(-1)\cdot \begin{vmatrix}\sin x&\cos x\\27&28\end{vmatrix}$$

Let us evaluate each minor carefully with signs.

Using cofactor expansion along column 3:

$$y = a_{13}C_{13}+a_{23}C_{23}+a_{33}C_{33}$$

• For $a_{13}=1$:

$$C_{13}=(-1)^{1+3}\begin{vmatrix}27&28\\1&1\end{vmatrix}=\begin{vmatrix}27&28\\1&1\end{vmatrix}=27-28=-1$$

• For $a_{23}=-28$:

$$C_{23}=(-1)^{2+3}\begin{vmatrix}\sin x&\cos x\\1&1\end{vmatrix} =-\big(\sin x-\cos x\big)$$

So contribution is

$$(-28)\cdot \big[-(\sin x-\cos x)\big]=28(\sin x-\cos x)$$

• For $a_{33}=-1$:

$$C_{33}=(-1)^{3+3}\begin{vmatrix}\sin x&\cos x\\27&28\end{vmatrix} =28\sin x-27\cos x$$

So contribution is

$$(-1)(28\sin x-27\cos x)=-28\sin x+27\cos x$$

Therefore,

$$y=-1+28(\sin x-\cos x)-28\sin x+27\cos x$$

Simplify:

$$y=-1+28\sin x-28\cos x-28\sin x+27\cos x$$ $$y=-1-\cos x$$

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4. Differentiate twice

Since

$$y=-1-\cos x$$

we get

$$y' = \sin x$$ $$y'' = \cos x$$

Thus,

$$y''+y=\cos x+(-1-\cos x)=-1$$

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5. Match with the options

$$\frac{d^2y}{dx^2}+y=-1$$

So the correct option is:

C: $-1$

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6. Comparison with stored answer

Stored correct answer: C

Our derived answer: C

They agree.