Matrices & Determinants
Matrices and Determinants
Allen Star Batch
Grade 12
Question:
If maximum and minimum values of the determinant $$\begin{vmatrix} 1+\sin^2 x & \cos^2 x & \sin 2x \\ \sin^2 x & 1+\cos^2 x & \sin 2x \\ \sin^2 x & \cos^2 x & 1+\sin 2x \end{vmatrix}$$ are $\alpha$ and $\beta$, then:
$\alpha + \beta^{90} = 4$
$\alpha^3 - \beta^{17} = 26$
$(\alpha^{2n} - \beta^{2n})$ is always an even integer for $n \in N$
A triangle can be constructed having it sides as $\alpha - \beta$, $\alpha + \beta$ and $\alpha + 3\beta$
Step-by-Step Solution
Key Concept: Apply column/row operations to simplify the determinant containing trigonometric expressions, then use the identity sin²x + cos²x = 1 to reduce it to a function of sin 2x. The determinant evaluates to 1 + 2sin³(2x), whose extrema occur at sin(2x) = ±1, yielding maximum α = 3 and minimum β = -1.
Starting with the determinant $\begin{vmatrix} 1+\sin^2 x & \cos^2 x & \sin 2x \\ \sin^2 x & 1+\cos^2 x & \sin 2x \\ \sin^2 x & \cos^2 x & 1+\sin 2x \end{vmatrix}$, apply column operation $C_1 \to C_1 + C_2$ to get $\begin{vmatrix} 2 & \cos^2 x & \sin 2x \\ 2 & 1+\cos^2 x & \sin 2x \\ 1 & \cos^2 x & 1+\sin 2x \end{vmatrix}$. Factor out and simplify to show that $\max(\sin 2x) = 1$ when $\alpha = 3, \beta = 1, \min(\sin 2x) = -1$. Since $\alpha - \beta = 2$, $\alpha + \beta = 4$, $3\beta = 6$, the values $(\alpha - \beta) + (\alpha + \beta) = \alpha + 3\beta$ cannot simultaneously satisfy triangle inequality conditions.
Correct Answer: 1,2,3