Let A = [[1, 2], [-2, -5]]. Let α, β ∈ R be such that αA² + βA = 2I. Then α + β is equal to - (A) -1

Q: Why can't I just multiply the matrix $A$ by itself to find $A^2$?

You absolutely could! But in the JEE, time is your most precious resource. Calculating $A^2$ is prone to arithmetic errors, and the Cayley-Hamilton theorem is a much more elegant, robust, and faster path to the solution.

Q: Is the Cayley-Hamilton theorem applicable to all matrices?

It is applicable to all square matrices! Whether it is a $2 \times 2$, $3 \times 3$, or an $n \times n$ matrix, the matrix always satisfies its own characteristic equation. It is a fundamental property of linear algebra.

Q: Why do we replace the constant term with $I$?

In the characteristic equation, the constant term is a scalar. When we transition to the matrix equation, we must maintain dimensional consistency. Since $A^2$ and $A$ are matrices, the constant term must also be a matrix, which is the identity matrix $I$.

Question

Let $A = \begin{bmatrix} 1 & 2 \ -2 & -5 \end{bmatrix}$. Let $\alpha, \beta \in \mathbb{R}$ be such that $\alpha A^2 + \beta A = 2I$. Then $\alpha + \beta$ is equal to -

Options:

(A) -10

(B) -6

(C) 6

(D) 10

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome, future engineers! Today, we are going to dive into the elegant world of matrix algebra. When you are given a matrix $A = \begin{bmatrix} 1 & 2 \ -2 & -5 \end{bmatrix}$ and asked to find scalars $\alpha$ and $\beta$ such that $\alpha A^2 + \beta A = 2I$, your first instinct might be to reach for the brute force method.

While calculating $A^2$ manually is a valid path, it is a dangerous one. In the high-stakes environment of the JEE Advanced, time is your most precious resource, and manual matrix multiplication is a breeding ground for silly arithmetic errors.

Instead, let us embrace a more powerful, elegant, and sophisticated tool: the **Cayley-Hamilton Theorem**.

### The Secret Weapon: Cayley-Hamilton

The Cayley-Hamilton Theorem is one of the most beautiful results in linear algebra. It states that every square matrix satisfies its own characteristic equation. This means we do not need to perform tedious matrix multiplications to find $A^2$.

The characteristic equation is defined by the determinant $|A - \lambda I| = 0$, where $\lambda$ represents the eigenvalues of the matrix. Let us set this up by subtracting $\lambda$ from the diagonal elements of $A$:

$$
\begin{vmatrix} 1 - \lambda & 2 \ -2 & -5 - \lambda \end{vmatrix} = 0
$$

Now, we expand this $2 	imes 2$ determinant using the formula $ad - bc$:

$$
(1 - \lambda)(-5 - \lambda) - (2)(-2) = 0
$$

Expanding this carefully, we get:

$$
-5 - \lambda + 5\lambda + \lambda^2 + 4 = 0
$$

Combining the like terms, we arrive at the quadratic equation:

$$
\lambda^2 + 4\lambda - 1 = 0
$$

### The Transformation

Now, we invoke the magic of the Cayley-Hamilton Theorem. Since every square matrix satisfies its own characteristic equation, we can simply replace the scalar $\lambda$ with the matrix $A$ and the constant term $1$ with the identity matrix $I$.

This gives us the matrix equation:

$$
A^2 + 4A - I = 0
$$

This is a massive shortcut! We have successfully expressed $A^2$ in terms of $A$ and $I$ without ever performing a single matrix multiplication.

Our goal is to match the form $\alpha A^2 + \beta A = 2I$. Let us rearrange our equation:

$$
A^2 + 4A = I
$$

The target equation has $2I$ on the right side, but we currently have $I$. The solution is simple: multiply the entire equation by $2$.

$$
2A^2 + 8A = 2I
$$

### The Final Comparison

Now, we compare our derived equation $2A^2 + 8A = 2I$ with the target equation $\alpha A^2 + \beta A = 2I$. By comparing the coefficients, it is immediately clear that $\alpha = 2$ and $\beta = 8$.

The question asks for the sum $\alpha + \beta$. Therefore:

$$
\alpha + \beta = 2 + 8 = \mathbf{10}
$$

And there you have it! We have solved the problem with precision, elegance, and speed. This is the power of understanding the underlying theory rather than just relying on calculation.

Animated Solution for Mathematics - Matrices and Determinants: Let $A = [1 - 2 2 - 5]$ . Let $α, β \in R$ be such that $α A^{2} + β A = 2 I$ . Then $α + β$ is equal to -

Select Answer:

Visualized Solution

Introduction to Matrix $A$

The Characteristic Equation

Setting up the Determinant

Expanding the Determinant

Simplifying the Polynomial

Applying Cayley-Hamilton Theorem

Rearranging the Equation

Matching the Target Form

Comparing Coefficients

Final Calculation

The Sigma Insight: Adjoint and Inverse of a Matrix

Analyzing the Setup

The Secret Weapon

The Transformation

The Final Comparison

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Animated Solution for Mathematics - Matrices and Determinants: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let A=[1−2​2−5​]. Let α,β∈R be such that αA2+βA=2I. Then α+β is equal to -