Let A be a 2 × 2 real matrix and I be the identity matrix of order 2. If the roots of the equation |

Q: Wait, why can't we just find matrix $A$ first? Can't we assume $A = \begin{pmatrix} a & b \\ c & d \end{pmatrix}$ and solve for the variables?

You absolutely could, but that is the 'brute force' trap! In JEE Advanced, time is your most precious resource. Solving for $a, b, c, d$ involves solving a system of equations that is often underdetermined. Always look for the 'eigen-shortcut' first.

Q: Is the trace always the sum of eigenvalues? Does this hold for any matrix?

Yes! This is a fundamental property of the trace. For any $n \times n$ matrix, the trace is the sum of its eigenvalues. It is one of the most elegant invariants in linear algebra.

Q: Does this property hold for any power of $A$? What if the question asked for $A^3$ or $A^{-1}$?

Excellent question! Yes, if $\lambda$ is an eigenvalue of $A$, then $\lambda^k$ is an eigenvalue of $A^k$ for any integer $k$ (provided $A$ is invertible for negative $k$). This property is incredibly powerful for high-degree polynomial matrices.

Question

Let $A$ be a $2 	imes 2$ real matrix and $I$ be the identity matrix of order 2. If the roots of the equation $|A - xI| = 0$ be -1 and 3, then the sum of the diagonal elements of the matrix $A^2$ is.

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing before a $2 	imes 2$ matrix $A$. To most, it is just a grid of four numbers, but to a mathematician, it is a transformation—a way to stretch, rotate, or shear space.

The problem provides the characteristic equation: $|A - xI| = 0$. This equation is the DNA of the matrix, and its roots, denoted as $\lambda_1$ and $\lambda_2$, are the **eigenvalues**.

These scalars dictate how the matrix behaves and remain invariant under the transformation. In this problem, the roots are given as $\lambda_1 = -1$ and $\lambda_2 = 3$.

### The Power of Eigenvalues: Bypassing Brute Force

The question asks for the sum of the diagonal elements of $A^2$. In linear algebra, the sum of the diagonal elements is known as the **Trace**, denoted as $	ext{Trace}(A)$.

A student might be tempted to assume $A = \begin{pmatrix} a & b \ c & d \end{pmatrix}$, square it, and then find the trace. However, this is the "brute force" trap.

Instead, we use the fundamental theorem: If $\lambda$ is an eigenvalue of $A$, then $\lambda^k$ is an eigenvalue of $A^k$. We can find the eigenvalues of $A^2$ without knowing the individual elements of $A$.

Given the eigenvalues of $A$ are $\lambda_1 = -1$ and $\lambda_2 = 3$, the eigenvalues of $A^2$ are:

$$\lambda_1^2 = (-1)^2 = 1$$

$$\lambda_2^2 = (3)^2 = 9$$

### The Trace Connection: The Final Elegance

We have determined the eigenvalues of $A^2$ to be $1$ and $9$. The final step relies on the property that the trace of any matrix is equal to the sum of its eigenvalues.

This is one of the most elegant invariants in linear algebra. Therefore, the trace is calculated as follows:

$$	ext{Trace}(A^2) = \lambda_1^2 + \lambda_2^2$$

$$	ext{Trace}(A^2) = 1 + 9 = 10$$

**The final answer is 10.** By utilizing the "eigen-shortcut," we have solved the problem by focusing on the structure of the transformation rather than getting lost in tedious arithmetic.

Animated Solution for Mathematics - Matrices and Determinants: Let $A$ be a $2 \times 2$ real matrix and $I$ be the identity matrix of order 2. If the roots of the equation $∣ A - x I ∣ = 0$ be -1 and 3, then the sum of the diagonal elements of the matrix $A^{2}$ is.

Enter Numerical Value:

Visualized Solution

The Characteristic Equation

Defining Eigenvalues

Given Roots of the Equation

The Goal: Matrix $A^{2}$

Property of Matrix Powers

Eigenvalues of $A^{2}$

Computing the New Eigenvalues

Trace as Sum of Eigenvalues

Final Calculation

The Sigma Insight: Algebraic Operations on Matrices

Analyzing the Setup

The Power of Eigenvalues

The Trace Connection

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