Q.Assertion (A): For the matrix $A = \begin{pmatrix} 1 & \cos\theta & 1 \\ -\cos\theta & 1 & \cos\theta \\ -1 & -\cos\theta & 1 \end{pmatrix}$, where $\theta \in [0, 2\pi]$, $|A| \in [2, 4]$. Reason (R): $\cos\theta \in [-1, 1],\ \forall\, \theta \in [0, 2\pi]$. (A) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A). (B) Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A). (C) Assertion (A) is true, but Reason (R) is false. (D) Assertion (A) is false, but Reason (R) is true.

The problem asks us to evaluate an Assertion-Reason pair. This requires us to perform three distinct checks:

1. Determine if Assertion (A) is true.
2. Determine if Reason (R) is true.
3. If both are true, determine if Reason (R) provides a correct explanation for Assertion (A).

The core concept here involves calculating the determinant of a $3 \times 3$ matrix and then finding the range of a trigonometric expression. The intuition is that once we express the determinant in terms of $\cos\theta$, the known bounds of $\cos\theta$ (which is what Reason (R) states) will directly help us find the bounds of the determinant.

1. Evaluate Reason (R):

   Reason (R) states: $\cos\theta \in [-1, 1],\ \forall\, \theta \in [0, 2\pi]$.

   This is a fundamental property of the cosine function. For any real value of $\theta$, the cosine function's output is always between $-1$ and $1$, inclusive. The interval $\theta \in [0, 2\pi]$ covers all possible values of $\cos\theta$.

   Therefore, Reason (R) is true.

2. Calculate the Determinant of Matrix A:

   The given matrix is $A = \begin{pmatrix} 1 & \cos\theta & 1 \\ -\cos\theta & 1 & \cos\theta \\ -1 & -\cos\theta & 1 \end{pmatrix}$.

   We calculate the determinant $|A|$ using cofactor expansion along the first row:

Recall that for a $2 \times 2$ matrix $\begin{pmatrix} a & b \\ c & d \end{pmatrix}$, its determinant is $ad - bc$.
Applying this:

3. Determine the Range of $|A|$ using Reason (R): We have found that $|A| = 2 + 2\cos^2\theta$. From Reason (R), we know that $\cos\theta \in [-1, 1]$. To find the range of $\cos^2\theta$: Since $\cos\theta$ can be any value between $-1$ and $1$, its square, $\cos^2\theta$, will be non-negative. The minimum value of $\cos^2\theta$ occurs when $\cos\theta = 0$, giving $\cos^2\theta = 0$. The maximum value of $\cos^2\theta$ occurs when $\cos\theta = 1$ or $\cos\theta = -1$, giving $\cos^2\theta = 1$. …

The key idea here is simple: a square matrix is called non-singular (or invertible) exactly when its determinant is not zero. If the determinant is zero, the matrix is singular — it has no inverse, and its rows (or columns) are linearly dependent.

So the problem reduces to: find all real numbers $a$ for which the determinant of the given matrix is non-zero. Then the set $A$ is precisely that collection of $a$ values.

Let’s compute the determinant.

1. Write the matrix:

2. Compute $\det(M)$ using expansion along the first row (or any row/column). I’ll expand along the first row:

3. Evaluate each $2 \times 2$ determinant:

   • $\begin{vmatrix} 3 & 1 \\ a & 1 \end{vmatrix} = (3)(1) - (1)(a) = 3 - a$
   • $\begin{vmatrix} 2 & 1 \\ 3 & 1 \end{vmatrix} = (2)(1) - (1)(3) = 2 - 3 = -1$
   • $\begin{vmatrix} 2 & 3 \\ 3 & a \end{vmatrix} = (2)(a) - (3)(3) = 2a - 9$

4. Substitute back:

Simplify:

Combine like terms:

$\det(M) = (3 + 2 - 9) + (-a + 2a) = (-4) + (a) = a - 4$ …