Let A = [[2, a, 0], [1, 3, 1], [0, 5, b]]. If A³ = 4A² - A - 21I, where I is the identity matrix of

Q: Why can't we just compute $A^2$ and $A^3$ directly? It seems like brute force would work.

While brute force is a valid path, it is a dangerous one in a JEE Advanced exam. Calculating $A^2$ and $A^3$ involves massive algebraic expansion where a single sign error will ruin your result. The Cayley-Hamilton Theorem is the 'elegant' path—it is designed to save you time and minimize error. Always look for the structural shortcut first!

Q: How do I know the given equation is the characteristic equation?

Great question! The Cayley-Hamilton theorem states that $A$ satisfies its characteristic equation. If the problem gives you a polynomial equation that $A$ satisfies, and that polynomial is of degree 3 (matching the dimension of the matrix), it is almost certainly the characteristic equation. We are essentially 'reverse-engineering' the matrix's properties from its behavior.

Q: Wait, why is the constant term in the characteristic equation $-\det(A)$?

That is a standard property of the characteristic polynomial for a $3 \times 3$ matrix. The characteristic polynomial is defined as $\det(A - \lambda I) = 0$. If you expand this determinant, the constant term (where $\lambda = 0$) is exactly $\det(A)$. However, the standard form is written as $\lambda^3 - \text{tr}(A)\lambda^2 + \dots - \det(A) = 0$. Always watch your signs carefully!

Question

Let $A = \begin{bmatrix} 2 & a & 0 \ 1 & 3 & 1 \ 0 & 5 & b \end{bmatrix}$. If $A^3 = 4A^2 - A - 21I$, where $I$ is the identity matrix of order $3 	imes 3$, then $2a + 3b$ is equal to:

Options:

(A) -9

(B) -13

(C) -10

(D) -12

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome, future engineers. Today, we are not just solving a matrix problem; we are embarking on a journey to understand the hidden language of linear algebra. When you look at a matrix like 
$$A = \begin{bmatrix} 2 & a & 0 \ 1 & 3 & 1 \ 0 & 5 & b \end{bmatrix}$$
do you see just a grid of numbers? Or do you see a transformation waiting to be understood?

The problem presents us with a cubic equation: $A^3 = 4A^2 - A - 21I$. At first glance, this looks like a nightmare of matrix multiplication. But in the world of JEE Advanced, whenever you see a matrix satisfying a polynomial equation, your intuition should immediately scream: **"Cayley-Hamilton!"**

### The Bridge Between Worlds

The Cayley-Hamilton Theorem is one of the most beautiful bridges in mathematics. It tells us that every square matrix satisfies its own characteristic equation. This is profound. It means that the matrix $A$ is not just a collection of scalars; it is a root of its own characteristic polynomial.

We start by rearranging our given equation into the standard form: $A^3 - 4A^2 + A + 21I = O$. By replacing the matrix $A$ with the scalar variable $\lambda$, we instantly derive the characteristic equation: 
$$\lambda^3 - 4\lambda^2 + \lambda + 21 = 0$$

We have successfully transformed a daunting matrix problem into a simple polynomial equation. This is the power of mathematical abstraction—we strip away the complexity to reveal the core structure.

### The Master Key

Now, we must recall the standard form of the characteristic equation for any $3 	imes 3$ matrix. It is given by:

$$\lambda^3 - 	ext{tr}(A)\lambda^2 + (M_{11} + M_{22} + M_{33})\lambda - \det(A) = 0$$

Here, $	ext{tr}(A)$ is the trace (the sum of the diagonal elements) and $\det(A)$ is the determinant. This equation is our master key. By comparing our derived equation, $\lambda^3 - 4\lambda^2 + \lambda + 21 = 0$, with this standard form, we can extract the vital statistics of matrix $A$ without ever performing a single matrix multiplication.

Comparing the coefficients of $\lambda^2$, we see that $-	ext{tr}(A) = -4$, which implies $	ext{tr}(A) = 4$. Comparing the constant terms, we see that $-\det(A) = 21$, which implies $\det(A) = -21$. Just like that, we have unlocked the two most important properties of our matrix.

### The Detective Work

Now, let us put on our detective hats. We know the trace is the sum of the diagonal elements. Looking at our matrix $A$, the diagonal elements are $2$, $3$, and $b$. Therefore:

$$	ext{tr}(A) = 2 + 3 + b = 5 + b$$

Since we already established that $	ext{tr}(A) = 4$, we have a simple linear equation: $5 + b = 4$. Solving this, we find **$b = -1$**. Half of our mystery is solved!

Next, we tackle the determinant. Expanding the determinant of $A$ along the first row is the most efficient path:

$$\det(A) = 2(3b - 5) - a(1 \cdot b - 0 \cdot 5) + 0$$

Simplifying this, we get $\det(A) = 6b - 10 - ab$. We know that $\det(A) = -21$. Substituting our known value of $b = -1$ into this expression:

$$\det(A) = 6(-1) - 10 - a(-1) = -6 - 10 + a = a - 16$$

Equating this to our known determinant value of $-21$, we get $a - 16 = -21$. Solving for $a$, we find **$a = -5$**.

### The Final Victory

We have arrived at the finish line. We have determined that $a = -5$ and $b = -1$. The problem asks us to calculate the value of $2a + 3b$. Substituting our values:

$$2(-5) + 3(-1) = -10 - 3 = -13$$

There it is. The **final answer is $-13$**.

Think about what we just achieved. We didn't get lost in the weeds of matrix multiplication. We didn't waste time calculating $A^2$ or $A^3$. Instead, we used the structural properties of matrices to dismantle the problem piece by piece. This is the essence of JEE Advanced preparation: it is not about how much you can calculate, but how well you can see the underlying patterns.

Animated Solution for Mathematics - Matrices and Determinants: Let $A = 210 a 35 01 b$ . If $A^{3} = 4 A^{2} - A - 21 I$ , where $I$ is the identity matrix of order $3 \times 3$ , then $2 a + 3 b$ is equal to:

Select Answer:

Visualized Solution

Given Matrix and Equation

Cayley-Hamilton Theorem

Characteristic Equation

Standard Form of Characteristic Equation

Comparing Coefficients

Solving for $b$ using Trace

Determinant Expansion

Substituting $b$ into Determinant

Solving for $a$

Target Expression Setup

Final Calculation

The Sigma Insight: Algebraic Operations on Matrices

Analyzing the Setup

The Bridge Between Worlds

The Master Key

The Detective Work

The Final Victory

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