Let $$d \in R$$, and $$A = \begin{bmatrix} -2 & 4+d & (\sin\theta)-2 \\ 1 & (\sin\theta)+2 & d \\ 5 & (2\sin\theta)-d & (-\sin\theta)+2+2d \end{bmatrix}$$, $$\theta \in [0, 2\pi]$$. If the minimum value of $$\det(A)$$ is 8, then a value of $$d$$ is:

We wish to find that particular real number $$d$$ for which the least possible value of the determinant of

$$A=\begin{bmatrix}-2 & 4+d & \sin\theta-2\\[2pt] 1 & \sin\theta+2 & d\\[2pt] 5 & 2\sin\theta-d & -\sin\theta+2+2d\end{bmatrix},\qquad \theta\in[0,2\pi]$$

is exactly $$8$$.

We start with the standard cofactor expansion of a $$3\times3$$ determinant along the first row:

$$\det(A)=a_{11}M_{11}-a_{12}M_{12}+a_{13}M_{13},$$

where $$M_{1j}$$ is the minor obtained by deleting the first row and the $$j^{\text{th}}$$ column.

Writing the entries explicitly, we have

$$a_{11}=-2,\; a_{12}=4+d,\; a_{13}=\sin\theta-2.$$

First minor

$$M_{11}= \begin{vmatrix}\sin\theta+2 & d\\[2pt] 2\sin\theta-d & -\sin\theta+2+2d\end{vmatrix}$$

Using $$\begin{vmatrix}a & b\\ c & d\end{vmatrix}=ad-bc$$,

$$\begin{aligned} M_{11}&=(\sin\theta+2)(-\sin\theta+2+2d)-d(2\sin\theta-d)\\ &=-\sin^2\theta+4+2d\sin\theta+4d-2d\sin\theta+d^2\\ &=d^2+4d+4-\sin^2\theta\\ &=(d+2)^2-\sin^2\theta. \end{aligned}$$

Second minor

$$M_{12}= \begin{vmatrix}1 & d\\ 5 & -\sin\theta+2+2d\end{vmatrix} =1\bigl(-\sin\theta+2+2d\bigr)-d\cdot5 =-\sin\theta+2-3d.$$

Third minor

$$M_{13}= \begin{vmatrix}1 & \sin\theta+2\\ 5 & 2\sin\theta-d\end{vmatrix} =1\,(2\sin\theta-d)-(\sin\theta+2)\cdot5 =2\sin\theta-d-5\sin\theta-10 =-3\sin\theta-d-10.$$

Substituting these minors into the cofactor expansion, we get

$$\begin{aligned} \det(A)\;&=\;(-2)\bigl((d+2)^2-\sin^2\theta\bigr)\;-\;(4+d)\bigl(-\sin\theta+2-3d\bigr)\;+\;(\sin\theta-2)\bigl(-3\sin\theta-d-10\bigr). \end{aligned}$$

Now we expand each term carefully.

First term

$$-2\bigl((d+2)^2-\sin^2\theta\bigr)=-2(d+2)^2+2\sin^2\theta.$$

Second term

$$\begin{aligned} -(4+d)\bigl(-\sin\theta+2-3d\bigr) &=(4+d)\bigl(\sin\theta-2+3d\bigr) \\ &=(4+d)\sin\theta-2(4+d)+3d(4+d)\\ &=(4+d)\sin\theta-8-2d+12d+3d^2\\ &=(4+d)\sin\theta+10d-8+3d^2. \end{aligned}$$

Third term

$$\begin{aligned} (\sin\theta-2)\bigl(-3\sin\theta-d-10\bigr) &=\sin\theta(-3\sin\theta-d-10)-2(-3\sin\theta-d-10)\\ &=-3\sin^2\theta-d\sin\theta-10\sin\theta+6\sin\theta+2d+20\\ &=-3\sin^2\theta-d\sin\theta-4\sin\theta+2d+20. \end{aligned}$$

Adding all three expanded parts:

$$\begin{aligned} \det(A)=&\;\bigl[-2(d+2)^2+2\sin^2\theta\bigr]\\ &+\bigl[(4+d)\sin\theta+10d-8+3d^2\bigr]\\ &+\bigl[-3\sin^2\theta-d\sin\theta-4\sin\theta+2d+20\bigr]. \end{aligned}$$

Collecting like terms in $$\sin\theta$$, $$\sin^2\theta$$ and $$d$$:

• $$\sin^2\theta$$ terms: $$2\sin^2\theta-3\sin^2\theta=-\sin^2\theta.$$
• $$\sin\theta$$ terms: $$(4+d)\sin\theta-d\sin\theta-4\sin\theta= \bigl[(4+d)-d-4\bigr]\sin\theta=0\cdot\sin\theta=0.$$
• Purely algebraic part in $$d$$ (no $$\theta$$):

$$\begin{aligned} -2(d+2)^2+10d-8+3d^2+2d+20 &=-2(d^2+4d+4)+10d-8+3d^2+2d+20\\ &=-2d^2-8d-8+10d-8+3d^2+2d+20\\ &=( -2d^2+3d^2 ) + ( -8d+10d+2d ) + ( -8-8+20 )\\ &=d^2+4d+4\\ &=(d+2)^2. \end{aligned}$$

Hence the determinant simplifies beautifully to the compact form

$$\det(A)=\boxed{(d+2)^2-\sin^2\theta}.$$

Because $$\sin^2\theta$$ assumes every value in the closed interval $$[0,1]$$ when $$\theta$$ varies through $$[0,2\pi]$$, we have

$$\sin^2\theta_{\text{max}}=1,\qquad \sin^2\theta_{\text{min}}=0.$$

Thus for a fixed $$d$$,

$$\det(A)_{\text{max}}=(d+2)^2-0=(d+2)^2,$$

and

$$\det(A)_{\text{min}}=(d+2)^2-1.$$

The problem states that the minimum value equals $$8$$, so we set

$$ (d+2)^2-1=8 \;\Longrightarrow\; (d+2)^2=9.$$

Taking square roots,

$$d+2=3 \quad\text{or}\quad d+2=-3,$$

which gives

$$d=1 \quad\text{or}\quad d=-5.$$

Among the options provided, the only listed value is $$d=-5$$, which corresponds to Option 3.

Hence, the correct answer is Option 3.