Let M = [[sin⁴θ, -1-sin²θ], [1+cos²θ, cos⁴θ]] = αI + βM⁻¹, where α = α(θ) and β = β(θ) are real numb

Q: Why can't we just find $M^{-1}$ directly and solve for $\alpha$ and $\beta$?

While you could technically invert the matrix, it would be an algebraic nightmare involving complex trigonometric expressions. The beauty of the Cayley-Hamilton theorem is that it allows us to bypass the inversion entirely, turning a matrix problem into a simple comparison of coefficients. It is the elegant, professional way to handle matrix polynomials.

Q: How do we know $\alpha$ and $\beta$ are just the trace and determinant?

The Cayley-Hamilton theorem tells us that for any $2 \times 2$ matrix $M$, the equation $M^2 - (\text{tr } M)M + |M|I = O$ must hold. When we rearrange the given equation $M = \alpha I + \beta M^{-1}$ into $M^2 - \alpha M - \beta I = O$, we are essentially looking at the same structure. By matching the coefficients of $M$ and $I$, it becomes clear that $\alpha = \text{tr } M$ and $\beta = -|M|$.

Q: Why is $\beta$ negative in the final expression?

This is a common trap! The characteristic equation is $M^2 - (\text{tr } M)M + |M|I = O$. Our rearranged equation is $M^2 - \alpha M - \beta I = O$. When you align these, the coefficient of $I$ in the characteristic equation is $|M|$, while in our rearranged equation, it is $-\beta$. Therefore, $-\beta = |M|$, which means $\beta = -|M|$. Always watch your signs!

Question

Let $M = \begin{bmatrix} \sin^4 	heta & -1-\sin^2 	heta \ 1+\cos^2 	heta & \cos^4 	heta \end{bmatrix} = \alpha I + \beta M^{-1}$, where $\alpha = \alpha(	heta)$ and $\beta = \beta(	heta)$ are real numbers, and $I$ is the $2 	imes 2$ identity matrix. If $\alpha^*$ is the minimum of the set $\{\alpha(	heta) : 	heta \in [0, 2\pi)\}$ and $\beta^*$ is the minimum of the set $\{\beta(	heta) : 	heta \in [0, 2\pi)\}$, then the value of $\alpha^* + \beta^*$ is

Options:

(A) $\frac{37}{16}$

(B) $-\frac{29}{16}$

(C) $\frac{31}{16}$

(D) $-\frac{17}{16}$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing before a matrix equation that looks like a tangled web of trigonometric functions. You are given $M = \alpha I + \beta M^{-1}$.

At first glance, your instinct might be to calculate the inverse of $M$, but stop! In the world of JEE Advanced, brute force is rarely the intended path. Instead, let us look for the hidden structure.

By multiplying the entire equation by $M$, we transform it into $M^2 = \alpha M + \beta I$, or more cleanly:

$$M^2 - \alpha M - \beta I = O$$

This is a quadratic matrix equation, and it is screaming for the Cayley-Hamilton theorem.

### The Cayley-Hamilton Secret

The Cayley-Hamilton theorem is the master key for matrix polynomials. It asserts that every square matrix satisfies its own characteristic equation.

For a $2 	imes 2$ matrix $M$, this equation is $\lambda^2 - (	ext{tr } M)\lambda + |M| = 0$. By replacing $\lambda$ with $M$, we get:

$$M^2 - (	ext{tr } M)M + |M|I = O$$

Now, look at the two equations side-by-side: our derived $M^2 - \alpha M - \beta I = O$ and the characteristic equation $M^2 - (	ext{tr } M)M + |M|I = O$.

The comparison is immediate and powerful: $\alpha = 	ext{tr } M$ and $\beta = -|M|$. We have reduced a complex matrix problem to simple scalar calculations.

### Taming the Trigonometry

Now, we focus on $\alpha(	heta) = 	ext{tr } M$. The trace is simply the sum of the diagonal elements: $\alpha(	heta) = \sin^4 	heta + \cos^4 	heta$.

Using the identity $a^2 + b^2 = (a+b)^2 - 2ab$, we rewrite this as $(\sin^2 	heta + \cos^2 	heta)^2 - 2\sin^2 	heta \cos^2 	heta$. Since $\sin^2 	heta + \cos^2 	heta = 1$, this simplifies to $1 - 2\sin^2 	heta \cos^2 	heta$, which is:

$$1 - \frac{1}{2}\sin^2 2	heta$$

To find the minimum $\alpha^*$, we maximize $\sin^2 2	heta$ to $1$, giving us $\alpha^* = 1 - \frac{1}{2} = \frac{1}{2}$.

### The Final Stretch

Next, we tackle $\beta(	heta) = -|M|$. Calculating the determinant of $M = \begin{bmatrix} \sin^4 	heta & -1-\sin^2 	heta \ 1+\cos^2 	heta & \cos^4 	heta \end{bmatrix}$ requires careful cross-multiplication.

After expanding and simplifying, we find $\beta(	heta) = -(\sin^4 	heta \cos^4 	heta + \sin^2 	heta \cos^2 	heta + 2)$. By substituting $t = \sin^2 	heta \cos^2 	heta = \frac{\sin^2 2	heta}{4}$, where $t \in [0, \frac{1}{4}]$, we get:

$$\beta(t) = -(t^2 + t + 2)$$

To minimize $\beta$, we maximize $t^2 + t + 2$. Since this is an increasing function for $t \ge 0$, the maximum occurs at $t = \frac{1}{4}$.

Thus, $\beta^* = -(\frac{1}{16} + \frac{1}{4} + 2) = -\frac{37}{16}$. Finally, the sum is:

$$\alpha^* + \beta^* = \frac{1}{2} - \frac{37}{16} = \frac{8-37}{16} = -\frac{29}{16}$$

We have navigated the matrix, the theorem, and the trigonometry to arrive at the solution. This is the power of mathematical insight over brute calculation.

Select Answer:

Visualized Solution

Matrix Equation Rearrangement

Cayley-Hamilton Theorem

Comparing Coefficients

Finding $α (θ)$

Simplifying $α (θ)$

Minimum of $α (θ)$

Finding $β (θ)$

Expanding Determinant

Multiplying Terms

Substitution for $β$

Minimum of $β (t)$

Calculating $β^{*}$ and Final Sum

The Sigma Insight: Algebraic Operations on Matrices

Analyzing the Setup

The Cayley-Hamilton Secret

Taming the Trigonometry

The Final Stretch

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Select Answer:

Visualized Solution

Matrix Equation Rearrangement

Cayley-Hamilton Theorem

Comparing Coefficients

Finding .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } α(θ)

Simplifying .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } α(θ)

Minimum of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } α(θ)

Finding .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } β(θ)

Expanding Determinant

Multiplying Terms

Substitution for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } β

Minimum of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } β(t)

Calculating .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } β∗ and Final Sum