Q.Let $A = \begin{bmatrix} 1 & -\cos\theta & -1 \\ \cos\theta & 1 & -\cos\theta \\ 1 & \cos\theta & 1 \end{bmatrix}$ where $0 < \theta < \frac{\pi}{2}$. Then ___.

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:

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