Question:
If \(A=\begin{bmatrix}1&\sin\theta&1\\-\sin\theta&1&\sin\theta\\-1&-\sin\theta&1\end{bmatrix}\); then for all \(\theta\in\left(\frac{3\pi}{4},\frac{5\pi}{4}\right)\), \(\det(A)\) lies in the interval :
If \(A=\begin{bmatrix}1&\sin\theta&1\\-\sin\theta&1&\sin\theta\\-1&-\sin\theta&1\end{bmatrix}\); then for all \(\theta\in\left(\frac{3\pi}{4},\frac{5\pi}{4}\right)\), \(\det(A)\) lies in the interval :
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Expand the determinant to get \(2+2\sin^2\theta\), then find the range of \(\sin^2\theta\) on the given interval.
Updated On: Nov 5, 2026
- \(\left(\frac{3}{2},3\right]\)
- \(\left(0,\frac{3}{2}\right]\)
- \(\left[\frac{5}{2},4\right)\)
- \(\left(1,\frac{5}{2}\right]\)
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The Correct Option is A
Solution and Explanation
Let \(s=\sin\theta\). Expanding along the first row:
\(\det A=1(1+s^2)-s(-s+s)+1(s^2+1)=2+2s^2\).
For \(\theta\in\left(\frac{3\pi}{4},\frac{5\pi}{4}\right)\), \(\sin\theta\in\left(-\frac{1}{\sqrt2},\frac{1}{\sqrt2}\right)\) so \(s^2\in\left[0,\frac12\right)\).
Hence \(\det A\in[2,3)\subset\left(\frac32,3\right]\).
Answer: (1) \(\left(\frac{3}{2},3\right]\)
\(\det A=1(1+s^2)-s(-s+s)+1(s^2+1)=2+2s^2\).
For \(\theta\in\left(\frac{3\pi}{4},\frac{5\pi}{4}\right)\), \(\sin\theta\in\left(-\frac{1}{\sqrt2},\frac{1}{\sqrt2}\right)\) so \(s^2\in\left[0,\frac12\right)\).
Hence \(\det A\in[2,3)\subset\left(\frac32,3\right]\).
Answer: (1) \(\left(\frac{3}{2},3\right]\)
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