Question

Let \(d\in\mathbb{R}\), 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 -

• A. \(-7\)
• B. \(2(\sqrt2+2)\)
• C. \(-5\)
• D. \(2(\sqrt2+1)\)

Hint:

Use \(R_1\to R_1+2R_2\) and \(R_3\to R_3-5R_2\) to reduce the first column to \((0,1,0)\); this gives \(\det A=(d+2)^2-\sin^2\theta\).

Answer 1

The Correct Option is C

Let \(s=\sin\theta\). Apply \(R_1\to R_1+2R_2\) and \(R_3\to R_3-5R_2\) (determinant unchanged):
\(R_1=[0,\ 2s+d+8,\ s+2d-2]\), \(R_2=[1,\ s+2,\ d]\), \(R_3=[0,\ -3s-d-10,\ -s+2-3d]\).
Expanding along column 1 (the entry 1 sits in position (2,1), sign \(-\)): \(\det A=-\left[(2s+d+8)(-s+2-3d)-(s+2d-2)(-3s-d-10)\right]\).
The bracket simplifies to \(s^2-d^2-4d-4\), so \(\det A=(d+2)^2-s^2\).
Minimum over \(s\in[-1,1]\) occurs at \(s^2=1\): \((d+2)^2-1=8\Rightarrow(d+2)^2=9\Rightarrow d=1\text{ or }-5\).
Only \(-5\) is among the options.
Answer: (3) \(-5\)