Here's a clear, educational, and well-structured solution to the problem.

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1. Key Concepts and Formulas

• Matrix Power Calculation: For a square matrix $A$, $A^n$ is obtained by multiplying $A$ by itself $n$ times. Calculating the first few powers ($A^2, A^3, \dots$) is often crucial for identifying a pattern.
• Pattern Recognition for Matrix Powers: For matrices with specific structures (e.g., diagonal, triangular, or block matrices), their powers often exhibit predictable patterns. Identifying these patterns is essential for dealing with high powers like $A^{20}$ or $A^{19}$ without performing all multiplications.
• Matrix Equality: Two matrices are equal if and only if their corresponding elements are equal. This principle allows us to convert a matrix equation into a system of scalar linear equations, which can then be solved for unknown variables.

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2. Step-by-Step Solution

Step 1: Analyze the Matrix A and Calculate its First Few Powers The given matrix is A = \left[ {\matrix{ 1 & 0 & 0 \cr 0 & 2 & 0 \cr 3 & 0 & { - 1} \cr } } \right] We observe that $A$ is a lower triangular matrix. This structure often simplifies calculations of its powers. We will calculate $A^2$, $A^3$, and $A^4$ to identify a pattern.

1. Calculate $A^2 = A \cdot A$: A^2 = \left[ {\matrix{ 1 & 0 & 0 \cr 0 & 2 & 0 \cr 3 & 0 & { - 1} \cr } } \right] \left[ {\matrix{ 1 & 0 & 0 \cr 0 & 2 & 0 \cr 3 & 0 & { - 1} \cr } } \right] = \left[ {\matrix{ (1)(1)+(0)(0)+(0)(3) & (1)(0)+(0)(2)+(0)(0) & (1)(0)+(0)(0)+(0)(-1) \cr (0)(1)+(2)(0)+(0)(3) & (0)(0)+(2)(2)+(0)(0) & (0)(0)+(2)(0)+(0)(-1) \cr (3)(1)+(0)(0)+(-1)(3) & (3)(0)+(0)(2)+(-1)(0) & (3)(0)+(0)(0)+(-1)(-1) \cr } } \right] A^2 = \left[ {\matrix{ 1 & 0 & 0 \cr 0 & 4 & 0 \cr 0 & 0 & 1 \cr } } \right]

2. Calculate $A^3 = A^2 \cdot A$: A^3 = \left[ {\matrix{ 1 & 0 & 0 \cr 0 & 4 & 0 \cr 0 & 0 & 1 \cr } } \right] \left[ {\matrix{ 1 & 0 & 0 \cr 0 & 2 & 0 \cr 3 & 0 & { - 1} \cr } } \right] = \left[ {\matrix{ (1)(1) & (1)(0) & (1)(0) \cr (0)(1) & (4)(2) & (4)(0) \cr (0)(1)+(0)(0)+(1)(3) & (0)(0) & (1)(-1) \cr } } \right] A^3 = \left[ {\matrix{ 1 & 0 & 0 \cr 0 & 8 & 0 \cr 3 & 0 & { - 1} \cr } } \right]

3. Calculate $A^4 = A^3 \cdot A$: A^4 = \left[ {\matrix{ 1 & 0 & 0 \cr 0 & 8 & 0 \cr 3 & 0 & { - 1} \cr } } \right] \left[ {\matrix{ 1 & 0 & 0 \cr 0 & 2 & 0 \cr 3 & 0 & { - 1} \cr } } \right] = \left[ {\matrix{ (1)(1) & 0 & 0 \cr 0 & (8)(2) & 0 \cr (3)(1)+(0)(0)+(-1)(3) & 0 & (-1)(-1) \cr } } \right] A^4 = \left[ {\matrix{ 1 & 0 & 0 \cr 0 & 16 & 0 \cr 0 & 0 & 1 \cr } } \right]

Step 2: Identify the Pattern for $A^n$ Let's analyze the elements of $A^n$ based on the calculated powers:

• The elements $A^n_{11}$, $A^n_{12}$, $A^n_{13}$, $A^n_{21}$, $A^n_{23}$, $A^n_{32}$ follow simple patterns:
  • $A^n_{11} = 1^n = 1$
  • $A^n_{22} = 2^n$
  • $A^n_{33} = (-1)^n$
  • All other off-diagonal elements (except $A^n_{31}$) are 0.
• The element $A^n_{31}$ follows a specific pattern:
  • $A^1_{31} = 3$
  • $A^2_{31} = 0$
  • $A^3_{31} = 3$
  • $A^4_{31} = 0$ This indicates that $A^n_{31} = 3$ if $n$ is odd, and $A^n_{31} = 0$ if $n$ is even. This can be concisely expressed as $3 \cdot \frac{1 - (-1)^n}{2}$.

Combining these observations, the general form for $A^n$ is: A^n = \left[ {\matrix{ 1 & 0 & 0 \cr 0 & 2^n & 0 \cr 3 \cdot \frac{1 - (-1)^n}{2} & 0 & (-1)^n \cr } } \right]

Step 3: Substitute $A^{20}$, $A^{19}$, and $A$ into the Given Equation The given matrix equation is {A^{20}} + \alpha {A^{19}} + \beta A = \left[ {\matrix{ 1 & 0 & 0 \cr 0 & 4 & 0 \cr 0 & 0 & 1 \cr } } \right] Notice that the Right Hand Side (RHS) matrix is exactly $A^2$. So the equation is $A^{20} + \alpha A^{19} + \beta A = A^2$.

Let's use the general form for $A^n$ to find the specific matrices for $n=20$, $n=19$, and $n=1$:

• For $A^{20}$ (where $n=20$ is an even number): A^{20} = \left[ {\matrix{ 1 & 0 & 0 \cr 0 & 2^{20} & 0 \cr 3 \cdot \frac{1 - (-1)^{20}}{2} & 0 & (-1)^{20} \cr } } \right] = \left[ {\matrix{ 1 & 0 & 0 \cr 0 & 2^{20} & 0 \cr 3 \cdot \frac{1 - 1}{2} & 0 & 1 \cr } } \right] = \left[ {\matrix{ 1 & 0 & 0 \cr 0 & 2^{20} & 0 \cr 0 & 0 & 1 \cr } } \right]
• For $A^{19}$ (where $n=19$ is an odd number): A^{19} = \left[ {\matrix{ 1 & 0 & 0 \cr 0 & 2^{19} & 0 \cr 3 \cdot \frac{1 - (-1)^{19}}{2} & 0 & (-1)^{19} \cr } } \right] = \left[ {\matrix{ 1 & 0 & 0 \cr 0 & 2^{19} & 0 \cr 3 \cdot \frac{1 - (-1)}{2} & 0 & -1 \cr } } \right] = \left[ {\matrix{ 1 & 0 & 0 \cr 0 & 2^{19} & 0 \cr 3 & 0 & -1 \cr } } \right]
• For $A$ (where $n=1$ is an odd number): A = \left[ {\matrix{ 1 & 0 & 0 \cr 0 & 2 & 0 \cr 3 & 0 & { - 1} \cr } } \right]
• The RHS matrix is $A^2$: A^2 = \left[ {\matrix{ 1 & 0 & 0 \cr 0 & 4 & 0 \cr 0 & 0 & 1 \cr } } \right]

Now substitute these matrices into the given equation: \left[ {\matrix{ 1 & 0 & 0 \cr 0 & 2^{20} & 0 \cr 0 & 0 & 1 \cr } } \right] + \alpha \left[ {\matrix{ 1 & 0 & 0 \cr 0 & 2^{19} & 0 \cr 3 & 0 & -1 \cr } } \right] + \beta \left[ {\matrix{ 1 & 0 & 0 \cr 0 & 2 & 0 \cr 3 & 0 & { - 1} \cr } } \right] = \left[ {\matrix{ 1 & 0 & 0 \cr 0 & 4 & 0 \cr 0 & 0 & 1 \cr } } \right]

Step 4: Form and Solve the System of Equations for $\alpha$ and $\beta$ By equating the corresponding elements of the matrices on both sides of the equation, we can form a system of linear equations for $\alpha$ and $\beta$.

1. Equating the $(1,1)$ elements: $1 + \alpha(1) + \beta(1) = 1$ $1 + \alpha + \beta = 1$ $\alpha + \beta = 0 \quad \text{(Equation 1)}$

2. Equating the $(2,2)$ elements: $2^{20} + \alpha(2^{19}) + \beta(2) = 4 \quad \text{(Equation 2)}$

3. Equating the $(3,1)$ elements: $0 + \alpha(3) + \beta(3) = 0$ $3\alpha + 3\beta = 0$ $\alpha + \beta = 0$ (This equation is consistent with Equation 1, providing a check.)

4. Equating the $(3,3)$ elements: $1 + \alpha(-1) + \beta(-1) = 1$ $1 - \alpha - \beta = 1$ $-\alpha - \beta = 0 \implies \alpha + \beta = 0$ (This is also consistent with Equation 1, another check.)

From Equation 1, we have $\beta = -\alpha$. Substitute $\beta = -\alpha$ into Equation 2: $2^{20} + \alpha \cdot 2^{19} + (-\alpha) \cdot 2 = 4$ $2^{20} + \alpha \cdot 2^{19} - 2\alpha = 4$ Factor out $\alpha$: $\alpha (2^{19} - 2) = 4 - 2^{20}$ To solve for $\alpha$, we can factor out common powers of 2: $\alpha (2(2^{18} - 1)) = 4(1 - 2^{18})$ $\alpha = \frac{4(1 - 2^{18})}{2(2^{18} - 1)}$ $\alpha = \frac{2(1 - 2^{18})}{-(2^{18} - 1)}$ $\alpha = \frac{2(1 - 2^{18})}{-(1 - 2^{18})}$ Since $1 - 2^{18} \neq 0$, we can cancel the term $(1 - 2^{18})$ from the numerator and denominator: $\alpha = -2$

Now, substitute $\alpha = -2$ back into $\beta = -\alpha$: $\beta = -(-2)$ $\beta = 2$

Step 5: Calculate $\beta - \alpha$ We found $\alpha = -2$ and $\beta = 2$. Therefore, $\beta - \alpha = 2 - (-2) = 2 + 2 = 4$.

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3. Common Mistakes & Tips

• Matrix Multiplication Errors: Be meticulous with matrix multiplication. A single incorrect element can propagate errors throughout the solution.
• Pattern Generalization: Ensure the pattern you identify for $A^n$ is robust. Test it with several powers ($A^1, A^2, A^3, A^4$) to confirm its validity, especially for alternating elements.
• System of Equations: When forming equations from matrix equality, choose elements carefully. Using multiple elements that lead to the same relationship (like $\alpha + \beta = 0$ from $(1,1)$, $(3,1)$, and $(3,3)$ elements) serves as a valuable self-check.

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4. Summary

This problem required finding a general pattern for the powers of a given matrix $A$. By calculating the first few powers ($A^2, A^3, A^4$), we observed that $A^n$ maintains a specific lower triangular structure with predictable diagonal elements and an alternating pattern for the $(3,1)$ element. Substituting the derived forms of $A^{20}$, $A^{19}$, and $A$ into the given matrix equation allowed us to equate corresponding elements, yielding a system of linear equations for $\alpha$ and $\beta$. Solving this system gave $\alpha = -2$ and $\beta = 2$, leading to $\beta - \alpha = 4$.

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5. Final Answer

The final answer is $\boxed{4}$.