JEE Challenger
More from Matrices
MathMedAdv 2019 • Paper 1

Minimum Values of Parameters in Matrix Equation with Identity Matrix

Let matrix MM be given by

M=sin⁡4θ−1−sin⁡2θ1+cos⁡2θcos⁡4θ=αI+βM−1,M = \begin{array}{cc} \sin^4 \theta & -1 - \sin^2 \theta \\ 1 + \cos^2 \theta & \cos^4 \theta \end{array} = \alpha I + \beta M^{-1},

where real-valued functions α=α(θ)\alpha = \alpha(\theta) and β=β(θ)\beta = \beta(\theta) are defined, and II denotes the 2×22 \times 2 identity matrix. Let α∗\alpha^* represent the minimum value of {α(θ):θ∈[0,2π]}\{\alpha(\theta) : \theta \in [0, 2\pi]\} and β∗\beta^* represent the minimum value of {β(θ):θ∈[0,2π]}\{\beta(\theta) : \theta \in [0, 2\pi]\}. Determine the value of α∗+β∗\alpha^* + \beta^*.

Options

A

−3716-\frac{37}{16}

B

−3116-\frac{31}{16}

C

−2916-\frac{29}{16}

Correct
D

−1716-\frac{17}{16}

Topics & Concepts

MatricesMatrices

Step-by-Step Solution

To determine the value of α∗+β∗\alpha^* + \beta^*, we begin by analyzing the given matrix equation: M=αI+βM−1M = \alpha I + \beta M^{-1}

Multiplying both sides of the equation by MM on the right gives: M2=αM+βI  ⟹  M2−αM−βI=OM^2 = \alpha M + \beta I \implies M^2 - \alpha M - \beta I = O

By the Cayley-Hamilton theorem, any 2×22 \times 2 matrix MM satisfies its own characteristic equation: M2−tr(M)M+det⁡(M)I=OM^2 - \text{tr}(M) M + \det(M) I = O

Comparing the two matrix equations, we get: α(θ)=tr(M)\alpha(\theta) = \text{tr}(M) β(θ)=−det⁡(M)\beta(\theta) = -\det(M)

1. Finding α(θ)\alpha(\theta) and its minimum value α∗\alpha^*:

The trace of MM is given by the sum of its diagonal entries: α(θ)=tr(M)=sin⁡4θ+cos⁡4θ\alpha(\theta) = \text{tr}(M) = \sin^4 \theta + \cos^4 \theta

Using the identity sin⁡4θ+cos⁡4θ=(sin⁡2θ+cos⁡2θ)2−2sin⁡2θcos⁡2θ\sin^4 \theta + \cos^4 \theta = (\sin^2 \theta + \cos^2 \theta)^2 - 2 \sin^2 \theta \cos^2 \theta: α(θ)=1−2sin⁡2θcos⁡2θ=1−12sin⁡22θ\alpha(\theta) = 1 - 2 \sin^2 \theta \cos^2 \theta = 1 - \frac{1}{2} \sin^2 2\theta

Since 0≤sin⁡22θ≤10 \le \sin^2 2\theta \le 1 for θ∈[0,2π]\theta \in [0, 2\pi], the minimum value α∗\alpha^* occurs when sin⁡22θ=1\sin^2 2\theta = 1: α∗=1−12(1)=12\alpha^* = 1 - \frac{1}{2}(1) = \frac{1}{2}

2. Finding β(θ)\beta(\theta) and its minimum value β∗\beta^*:

The determinant of MM is: det⁡(M)=(sin⁡4θ)(cos⁡4θ)−(−1−sin⁡2θ)(1+cos⁡2θ)\det(M) = (\sin^4 \theta)(\cos^4 \theta) - (-1 - \sin^2 \theta)(1 + \cos^2 \theta) det⁡(M)=sin⁡4θcos⁡4θ+(1+sin⁡2θ)(1+cos⁡2θ)\det(M) = \sin^4 \theta \cos^4 \theta + (1 + \sin^2 \theta)(1 + \cos^2 \theta) det⁡(M)=sin⁡4θcos⁡4θ+1+sin⁡2θ+cos⁡2θ+sin⁡2θcos⁡2θ\det(M) = \sin^4 \theta \cos^4 \theta + 1 + \sin^2 \theta + \cos^2 \theta + \sin^2 \theta \cos^2 \theta det⁡(M)=sin⁡4θcos⁡4θ+sin⁡2θcos⁡2θ+2\det(M) = \sin^4 \theta \cos^4 \theta + \sin^2 \theta \cos^2 \theta + 2

Since β(θ)=−det⁡(M)\beta(\theta) = -\det(M): β(θ)=−(sin⁡4θcos⁡4θ+sin⁡2θcos⁡2θ+2)\beta(\theta) = -\left(\sin^4 \theta \cos^4 \theta + \sin^2 \theta \cos^2 \theta + 2\right)

Let t=sin⁡2θcos⁡2θ=14sin⁡22θt = \sin^2 \theta \cos^2 \theta = \frac{1}{4} \sin^2 2\theta. Since 0≤sin⁡22θ≤10 \le \sin^2 2\theta \le 1, the range of tt is t∈[0,14]t \in \left[0, \frac{1}{4}\right].

In terms of tt, we can write: β(t)=−(t2+t+2)\beta(t) = -(t^2 + t + 2)

The expression g(t)=t2+t+2g(t) = t^2 + t + 2 is strictly increasing for t≥0t \ge 0. Therefore, g(t)g(t) attains its maximum value on the interval [0,14]\left[0, \frac{1}{4}\right] at t=14t = \frac{1}{4}: g(14)=(14)2+14+2=116+416+3216=3716g\left(\frac{1}{4}\right) = \left(\frac{1}{4}\right)^2 + \frac{1}{4} + 2 = \frac{1}{16} + \frac{4}{16} + \frac{32}{16} = \frac{37}{16}

Since β(t)=−g(t)\beta(t) = -g(t), the minimum value β∗\beta^* occurs when g(t)g(t) is maximized: β∗=−3716\beta^* = -\frac{37}{16}

3. Calculating α∗+β∗\alpha^* + \beta^*:

α∗+β∗=12−3716=8−3716=−2916\alpha^* + \beta^* = \frac{1}{2} - \frac{37}{16} = \frac{8 - 37}{16} = -\frac{29}{16}

Thus, the correct option is (C).

PreviousNext