If the matrix A = [[1, 0, 0], [0, 2, 0], [3, 0, -1]] satisfies the equation A²⁰ + αA¹⁹ + βA = [[1, 0

Q: Can I use the Cayley-Hamilton theorem here to find the powers?

You could, but it is overkill. The Cayley-Hamilton theorem would give you a characteristic equation, but finding the pattern is significantly faster and less prone to calculation errors in an exam setting. Always look for the simplest path first!

Q: How do I know the pattern holds for all $n$?

Great question! You can prove it by mathematical induction. Once you see the pattern for $A^2, A^3, A^4$, you can formally prove it holds for all $n$. In the exam, once you see the pattern repeat, you can trust it.

Q: Why did the bottom-left element change back to 3?

It is the nature of the matrix multiplication. When you multiply by $A$, the specific structure of the matrix 'recycles' the values. It is a cyclic property of this specific matrix $A$. It is not random; it is the matrix's 'DNA'.

Question

If the matrix $A = \begin{bmatrix} 1 & 0 & 0 \ 0 & 2 & 0 \ 3 & 0 & -1 \end{bmatrix}$ satisfies the equation $A^{20} + \alpha A^{19} + \beta A = \begin{bmatrix} 1 & 0 & 0 \ 0 & 4 & 0 \ 0 & 0 & 1 \end{bmatrix}$ for some real numbers $\alpha$ and $\beta$, then $\beta - \alpha$ is equal to

Sigma Percentile · Accepted Answer

### The Mystery of the High Power

Imagine you are sitting in the examination hall. You see $A^{20}$. Your heart skips a beat. Is this a trap? Do I really have to multiply this matrix twenty times?

No, my friend. In the world of JEE Advanced, whenever you see a high power of a matrix, you are not looking at a calculation problem; you are looking at a pattern recognition puzzle. Let us embark on this journey together.

### Phase 1: The Detective Work

We start with our matrix:
$$A = \begin{bmatrix} 1 & 0 & 0 \ 0 & 2 & 0 \ 3 & 0 & -1 \end{bmatrix}$$

The first instinct is to find $A^2$. Let us perform the multiplication $A^2 = A \cdot A$:
$$A^2 = \begin{bmatrix} 1 & 0 & 0 \ 0 & 4 & 0 \ 0 & 0 & 1 \end{bmatrix}$$

Look closely! The middle element is $2^2 = 4$. The rest of the matrix looks suspiciously like the identity matrix. This is our first clue.

Now, let us test $A^3 = A^2 \cdot A$. Multiplying our result by the original matrix, we get:
$$A^3 = \begin{bmatrix} 1 & 0 & 0 \ 0 & 8 & 0 \ 3 & 0 & -1 \end{bmatrix}$$

The middle element is $2^3 = 8$. But look at the bottom-left and bottom-right elements. They have returned to $3$ and $-1$. The matrix has 'reset' its structure!

### Phase 2: The Generalization

This is the beauty of linear algebra. We have discovered a cycle.

For any even power $n$, the matrix takes the form:
$$A^n = \begin{bmatrix} 1 & 0 & 0 \ 0 & 2^n & 0 \ 0 & 0 & 1 \end{bmatrix}$$

For any odd power $n$, the matrix reverts to:
$$A^n = \begin{bmatrix} 1 & 0 & 0 \ 0 & 2^n & 0 \ 3 & 0 & -1 \end{bmatrix}$$

This distinction is the key to unlocking the entire problem.

### Phase 3: The Algebraic Siege

Now, we substitute these forms into our equation:
$$A^{20} + \alpha A^{19} + \beta A = \begin{bmatrix} 1 & 0 & 0 \ 0 & 4 & 0 \ 0 & 0 & 1 \end{bmatrix}$$

Since $20$ is even and $19$ is odd, we substitute the respective forms. By equating the elements, we find a system of equations.

The $(1,1)$ element gives us $1 + \alpha + \beta = 1$, which simplifies to $\alpha + \beta = 0$.

The $(2,2)$ element gives us:
$$2^{20} + \alpha 2^{19} + 2\beta = 4$$

With $\beta = -\alpha$, we solve for $\alpha$ and find $\alpha = -2$. Consequently, $\beta = 2$.

The final answer, **$\beta - \alpha$**, is $2 - (-2) = **4**$. You see? The 'impossible' problem was just a beautiful dance of patterns.

Animated Solution for Mathematics - Matrices and Determinants: If the matrix $A = 103020 00 - 1$ satisfies the equation $A^{20} + α A^{19} + β A = 100040001$ for some real numbers $α$ and $β$ , then $β - α$ is equal to

Enter Numerical Value:

Visualized Solution

Analyze Matrix $A$

Calculate $A^{2}$

Calculate $A^{3}$

Generalize $A^{n}$

Substitute into the Equation

Simplify the Matrix Sum

Compare Elements: $α + β$

Compare Middle Element $(2, 2)$

Solve for $α$

Find $β$ and Final Answer

The Sigma Insight: Algebraic Operations on Matrices

The Mystery of the High Power

Phase 1

Phase 2

Phase 3

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Enter Numerical Value:

Visualized Solution

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

Calculate .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } A2

Calculate .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } A3

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

Substitute into the Equation

Simplify the Matrix Sum

Compare Elements: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } α+β

Compare Middle Element .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } (2,2)

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

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