Step 1: Compute the determinant of matrix \( A \).
\[ |A| = \begin{vmatrix} 1 & \cos \theta & 1 \\ -\cos \theta & 1 & \cos \theta \\ -1 & -\cos \theta & 1 \end{vmatrix}. \]
Using cofactor expansion along the first row:
\[ |A| = 1 \cdot \begin{vmatrix} 1 & \cos \theta \\ -\cos \theta & 1 \end{vmatrix} - \cos \theta \cdot \begin{vmatrix} -\cos \theta & \cos \theta \\ -1 & 1 \end{vmatrix} + 1 \cdot \begin{vmatrix} -\cos \theta & 1 \\ -1 & -\cos \theta \end{vmatrix}. \]
1. Compute the first minor:
\[ \begin{vmatrix} 1 & \cos \theta \\ -\cos \theta & 1 \end{vmatrix} = (1)(1) - (-\cos \theta)(\cos \theta) = 1 + \cos^2 \theta. \]
2. Compute the second minor:
\[ \begin{vmatrix} -\cos \theta & \cos \theta \\ -1 & 1 \end{vmatrix} = (-\cos \theta)(1) - (\cos \theta)(-1) = -\cos \theta + \cos \theta = 0. \]
3. Compute the third minor:
\[ \begin{vmatrix} -\cos \theta & 1 \\ -1 & -\cos \theta \end{vmatrix} = (-\cos \theta)(-\cos \theta) - (1)(-1) = \cos^2 \theta + 1. \]
Substitute back into the determinant:
\[ |A| = 1 \cdot (1 + \cos^2 \theta) - \cos \theta \cdot 0 + 1 \cdot (1 + \cos^2 \theta). \]
Simplify:
\[ |A| = (1 + \cos^2 \theta) + (1 + \cos^2 \theta) = 2 + 2\cos^2 \theta. \]
Step 2: Determine the range of \(|A| \).
Since \(\cos \theta \in [-1, 1]\), we have:
\[ \cos^2 \theta \in [0, 1]. \]
Thus:
\[ |A| = 2 + 2\cos^2 \theta \in [2, 4]. \]
Verification of Assertion (A): The determinant \(|A|\) lies in the interval \([2, 4]\), so the assertion is true.
Verification of Reason (R): The cosine function satisfies \(\cos \theta \in [-1, 1]\) for all \(\theta \in [0, 2\pi]\), so the reason is also true.
Conclusion: Both Assertion (A) and Reason (R) are true, and the Reason (R) correctly explains the Assertion (A).
Determine whether each of the following relations are reflexive, symmetric, and transitive.
Show that the relation R in the set R of real numbers, defined as
R = {(a, b): a ≤ b2 } is neither reflexive nor symmetric nor transitive.
Check whether the relation R defined in the set {1, 2, 3, 4, 5, 6} as
R = {(a, b): b = a + 1} is reflexive, symmetric or transitive.
Evaluate the determinants in Exercises 1 and 2.
\(\begin{vmatrix}2&4\\-5&-1\end{vmatrix}\)
Evaluate the determinants in Exercises 1 and 2.
(i) \(\begin{vmatrix}\cos\theta&-\sin\theta\\\sin\theta&\cos\theta\end{vmatrix}\)
(ii) \(\begin{vmatrix}x^2&-x+1&x-1\\& x+1&x+1\end{vmatrix}\)
Using properties of determinants,prove that:
\(\begin{vmatrix} x & x^2 & 1+px^3\\ y & y^2 & 1+py^3\\z&z^2&1+pz^3 \end{vmatrix}\)\(=(1+pxyz)(x-y)(y-z)(z-x)\)
Using properties of determinants,prove that:
\(\begin{vmatrix} 3a& -a+b & -a+c\\ -b+a & 3b & -b+c \\-c+a&-c+b&3c\end{vmatrix}\)\(=3(a+b+c)(ab+bc+ca)\)