Q.Let $A = \begin{bmatrix} 1 & \sin\theta & 1 \\ -\sin\theta & 1 & \sin\theta \\ -1 & -\sin\theta & 1 \end{bmatrix}$, where $0 \le \theta \le 2\pi$. Then (A) $\det(A) = 0$ (B) $\det(A) \in (2, \infty)$ (C) $\det(A) \in (2, 4)$ (D) $\det(A) \in [2, 4]$

Concept: Determinant Range Analysis — compute the determinant as a function of $\theta$, then find its possible values.

Step 1: Expand the determinant.

Using $A = \begin{bmatrix} 1 & \sin\theta & 1 \\ -\sin\theta & 1 & \sin\theta \\ -1 & -\sin\theta & 1 \end{bmatrix}$, expand along the first row:

Step 2: Compute each minor.

First minor: $1 \cdot 1 - (\sin\theta)(-\sin\theta) = 1 + \sin^2\theta$.

Second minor: $(-\sin\theta)(1) - (\sin\theta)(-1) = -\sin\theta + \sin\theta = 0$, so the term vanishes. …

The key insight here is that the matrix has a clear pattern: it’s skew-symmetric in the off-diagonal entries, with $1$’s on the diagonal. When you compute the determinant of such a matrix, many terms cancel or simplify nicely. Instead of brute-forcing the entire expansion, we can use the structure to reduce work.

Let’s go step by step.

1. Write the determinant explicitly.

2. Expand using the first row (or any row). Expanding along row 1:

3. Compute each 2×2 determinant.

   • First minor: $\begin{vmatrix} 1 & \sin\theta \\ -\sin\theta & 1 \end{vmatrix} = (1)(1) - (\sin\theta)(-\sin\theta) = 1 + \sin^2\theta$
   • Second minor: $\begin{vmatrix} -\sin\theta & \sin\theta \\ -1 & 1 \end{vmatrix} = (-\sin\theta)(1) - (\sin\theta)(-1) = -\sin\theta + \sin\theta = 0$
   • Third minor: $\begin{vmatrix} -\sin\theta & 1 \\ -1 & -\sin\theta \end{vmatrix} = (-\sin\theta)(-\sin\theta) - (1)(-1) = \sin^2\theta + 1$

   Notice the second minor is zero — that’s a nice simplification.

4. Plug back into the expansion.

5. Now find the range of $\det A$ for $\theta \in [0, 2\pi]$. …

Method: Reduce to a Single-Variable Function, Then Find Its Range

This method finds the range of a determinant whose entries depend on one parameter (here $\theta$) by first simplifying the determinant to an explicit function of that parameter, then applying the known range of the trig (or algebraic) expression involved.

Steps

Step 1: Expand the determinant symbolically

Expand along whichever row or column has the simplest entries (often the first row), keeping $\theta$ symbolic throughout. Don't substitute any numeric value yet.

Step 2: Simplify using trig identities

Combine terms using standard identities (e.g. $\sin^2\theta+\cos^2\theta=1$, or noticing a minor vanishes identically). The goal is to collapse the expansion down to a single simple function of $\theta$, ideally something like $k(1+\sin^2\theta)$ or $k+m\cos2\theta$.

Step 3: Identify the range of the underlying trig function

Recall the standard bounds: $-1\le\sin\theta\le1$, so $0\le\sin^2\theta\le1$ (never negative). Apply the given domain restriction on $\theta$ if it doesn't cover a full period.

Step 4: Substitute the bounds into the simplified expression …

Common Mistakes

Mistake 1: Assuming $\sin^2\theta$ can be negative or exceed $1$

Why it's wrong: $\sin\theta$ always lies in $[-1,1]$, so $\sin^2\theta$ is always in $[0,1]$ — never negative, never above $1$, regardless of how large $\theta$ is. Correct approach: once the determinant simplifies to $2(1+\sin^2\theta)$, apply the fixed bounds $0\le\sin^2\theta\le1$ directly to get the range $[2,4]$.

Mistake 2: Missing that the second cofactor term vanishes and re-doing unnecessary algebra …