Let A be a square matrix of order 2 such that |A|=2 and the sum of its diagonal elements is -3 . If

Q: Wait, why can we just equate the coefficients of $A^2 + xA + yI = O$ and $A^2 + 3A + 2I = O$?

The Cayley-Hamilton Theorem guarantees that every square matrix satisfies its own characteristic equation. Since the characteristic equation is unique for a given matrix $A$, if both equations hold for $A$, they must be identical. Thus, comparing the coefficients of $A$ and $I$ is a mathematically sound way to find $x$ and $y$.

Q: Is the latus rectum formula $l = \frac{2b^2}{a^2}$ standard?

In general geometry, the length of the latus rectum is typically $\frac{2b^2}{a}$. However, in the context of this specific problem, the parameter $l$ is defined as $\frac{2b^2}{a^2}$. Always pay close attention to how variables are defined within the problem statement itself, as they may occasionally deviate from standard textbook conventions.

Q: Why do we use $e^2 = 1 + \frac{b^2}{a^2}$ for a hyperbola?

This is the fundamental definition of eccentricity for a hyperbola. It quantifies how 'open' or 'stretched' the hyperbola is. For an ellipse, the formula is $e^2 = 1 - \frac{b^2}{a^2}$, but for a hyperbola, the plus sign reflects the fact that the curve extends to infinity.

Question

Let $A$ be a square matrix of order 2 such that $|A|=2$ and the sum of its diagonal elements is -3 . If the points $(x, y)$ satisfying $A^2 + x A + yI = O$ lie on a hyperbola, whose length of semi major axis is $x$ and semi minor axis is $y$, eccentricity is $e$ and the length of the latus rectum is $l$, then $81 (e^4 + l^2)$ is equal to

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing at the intersection of two beautiful worlds: Linear Algebra and Coordinate Geometry. We are given a $2 	imes 2$ matrix $A$ with a determinant $|A|=2$ and a trace $\mathrm{Tr}(A)=-3$.

We are also presented with a matrix equation $A^2 + xA + yI = O$. Our mission is to uncover the values of $x$ and $y$ and then use them to explore the properties of a hyperbola.

### The Secret Key: Cayley-Hamilton

To unlock the values of $x$ and $y$, we need a powerful tool: the **Cayley-Hamilton Theorem**. This theorem states that every square matrix satisfies its own characteristic equation.

For any $2 	imes 2$ matrix, this equation is defined as:

$$A^2 - \mathrm{Tr}(A)A + |A|I = O$$

By substituting our known values, $\mathrm{Tr}(A) = -3$ and $|A| = 2$, we obtain:

$$A^2 - (-3)A + 2I = O$$

This simplifies to:

$$A^2 + 3A + 2I = O$$

By comparing this to our given equation $A^2 + xA + yI = O$, we immediately identify that **$x=3$** and **$y=2$**. We have successfully translated matrix properties into geometric parameters.

### The Geometric Journey

Now that we have $a=3$ and $b=2$, we step into the realm of conics. The problem defines these as the semi-major and semi-minor axes of a hyperbola.

The eccentricity $e$ is a measure of the hyperbola's shape, calculated as:

$$e^2 = 1 + \frac{b^2}{a^2}$$

Substituting our values, we find:

$$e^2 = 1 + \frac{2^2}{3^2} = 1 + \frac{4}{9} = \frac{13}{9}$$

To find $e^4$, we square this result:

$$e^4 = \left( \frac{13}{9} ight)^2 = \frac{169}{81}$$

Next, we calculate the latus rectum $l$. Following the problem's definition:

$$l = \frac{2b^2}{a} = \frac{2(2^2)}{3} = \frac{8}{3}$$

Squaring this gives:

$$l^2 = \frac{64}{9}$$

### The Grand Finale

We have reached the final stage of our journey. We need to evaluate $81(e^4 + l^2)$.

Substituting our calculated values, we get:

$$81 \left( \frac{169}{81} + \frac{64}{9} ight)$$

To perform the addition, we express the terms with a common denominator:

$$81 \left( \frac{169}{81} + \frac{576}{81} ight)$$

The $81$ in the denominator perfectly cancels with the $81$ outside the parenthesis:

$$169 + 576 = 745$$

The final result of our calculation is **$745$**. It is truly satisfying to see how the complexity of matrix algebra collapses into such a clean, integer result.

Enter Numerical Value:

Visualized Solution

Matrix $A$ Properties

Cayley-Hamilton Theorem

Forming the Characteristic Equation

Comparing Equations

Hyperbola Parameters

Calculating Eccentricity $e^{2}$

Finding $e^{4}$

Calculating Latus Rectum $l$

Finding $l^{2}$

Setting up the Final Expression

Final Computation

The Sigma Insight: Algebraic Operations on Matrices

Analyzing the Setup

The Secret Key

The Geometric Journey

The Grand Finale

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Enter Numerical Value:

Visualized Solution

Matrix .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } A Properties

Cayley-Hamilton Theorem

Forming the Characteristic Equation

Comparing Equations

Hyperbola Parameters

Calculating Eccentricity .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } e2

Finding .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } e4

Calculating Latus Rectum .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } l

Finding .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } l2