Key Concept: Conditions for Infinitely Many Solutions of a System of Linear Equations

For a system of linear equations $Ax=B$ (where $A$ is the coefficient matrix, $x$ is the variable vector, and $B$ is the constant vector), Cramer's Rule provides conditions for the nature of its solutions:

1. Unique Solution: If $\det(A) \neq 0$, the system has a unique solution.
2. No Solution or Infinitely Many Solutions: If $\det(A) = 0$, then:
   • If at least one of the determinants $\Delta_x, \Delta_y, \Delta_z$ (formed by replacing a column of $A$ with $B$) is non-zero, the system has no solution.
   • If $\det(A) = 0$ AND all determinants $\Delta_x, \Delta_y, \Delta_z$ are also zero, the system has infinitely many solutions.

In this problem, we are given that the system has infinitely many solutions. Therefore, we must satisfy the condition $\det(A) = 0$ and $\Delta_x = \Delta_y = \Delta_z = 0$.

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Step 1: Formulate the System of Equations and Identify the Coefficient Matrix

The given system of linear equations is:

1. $x+y+z=1$
2. $x+2y+4z=m$
3. $x+4y+10z=m^2$

We can write this system in matrix form $AX=B$, where: $A = \begin{pmatrix} 1 & 1 & 1 \\ 1 & 2 & 4 \\ 1 & 4 & 10 \end{pmatrix}$, $X = \begin{pmatrix} x \\ y \\ z \end{pmatrix}$, and $B = \begin{pmatrix} 1 \\ m \\ m^2 \end{pmatrix}$.

Step 2: Calculate the Determinant of the Coefficient Matrix ($\Delta$)

For infinitely many solutions, the determinant of the coefficient matrix, denoted as $\Delta$ or $\det(A)$, must be zero. Let's calculate $\Delta$:

$\Delta = \left|\begin{array}{ccc} 1 & 1 & 1 \\ 1 & 2 & 4 \\ 1 & 4 & 10 \end{array}\right|$

We expand the determinant along the first row: $\Delta = 1 \cdot \left|\begin{array}{cc} 2 & 4 \\ 4 & 10 \end{array}\right| - 1 \cdot \left|\begin{array}{cc} 1 & 4 \\ 1 & 10 \end{array}\right| + 1 \cdot \left|\begin{array}{cc} 1 & 2 \\ 1 & 4 \end{array}\right|$ $\Delta = 1(2 \cdot 10 - 4 \cdot 4) - 1(1 \cdot 10 - 4 \cdot 1) + 1(1 \cdot 4 - 2 \cdot 1)$ $\Delta = 1(20 - 16) - 1(10 - 4) + 1(4 - 2)$ $\Delta = 1(4) - 1(6) + 1(2)$ $\Delta = 4 - 6 + 2$ $\Delta = 0$

Since $\Delta = 0$, the system either has no solutions or infinitely many solutions. This confirms the first condition for infinitely many solutions.

Step 3: Apply the Condition $\Delta_x = 0$ to find values of $m$

For the system to have infinitely many solutions, in addition to $\Delta=0$, all determinants $\Delta_x, \Delta_y, \Delta_z$ must also be zero. We will calculate $\Delta_x$ and set it to zero to find the possible values of $m$.

$\Delta_x$ is the determinant of the matrix formed by replacing the first column of $A$ with the constant terms vector $B$: $\Delta_x = \left|\begin{array}{ccc} 1 & 1 & 1 \\ m & 2 & 4 \\ m^2 & 4 & 10 \end{array}\right|$

Expand $\Delta_x$ along the first column: $\Delta_x = 1 \cdot \left|\begin{array}{cc} 2 & 4 \\ 4 & 10 \end{array}\right| - m \cdot \left|\begin{array}{cc} 1 & 1 \\ 4 & 10 \end{array}\right| + m^2 \cdot \left|\begin{array}{cc} 1 & 1 \\ 2 & 4 \end{array}\right|$ $\Delta_x = 1(2 \cdot 10 - 4 \cdot 4) - m(1 \cdot 10 - 1 \cdot 4) + m^2(1 \cdot 4 - 1 \cdot 2)$ $\Delta_x = 1(20 - 16) - m(10 - 4) + m^2(4 - 2)$ $\Delta_x = 1(4) - m(6) + m^2(2)$ $\Delta_x = 4 - 6m + 2m^2$

Now, we set $\Delta_x = 0$ as required for infinitely many solutions: $2m^2 - 6m + 4 = 0$ Divide by 2: $m^2 - 3m + 2 = 0$ This is a quadratic equation in $m$. We can factor it: $(m-1)(m-2) = 0$ This gives us two values for $m$: $m=1$ or $m=2$.

Tip: While we only calculated $\Delta_x=0$, it is implicitly understood that for these values of $m$, $\Delta_y$ and $\Delta_z$ would also be zero. For problems involving parameters, finding the values that make one of these determinants zero is usually sufficient, as these are the only values of the parameter for which infinitely many solutions could exist.

Step 4: Identify $\alpha$ and $\beta$

The problem states that $\alpha, \beta$ are the values of $m$ for which the system has infinitely many solutions, and $\alpha \neq \beta$. From our calculation, the values of $m$ are $1$ and $2$. So, we can assign $\alpha = 1$ and $\beta = 2$ (or vice versa, it won't affect the final sum).

Step 5: Calculate the Required Summation

We need to find the value of $\sum\limits_{n=1}^{10} (n^{\alpha}+n^{\beta})$. Substitute $\alpha=1$ and $\beta=2$: $\sum\limits_{n=1}^{10} (n^{1}+n^{2})$

Using the property of summation $\sum (a_n+b_n) = \sum a_n + \sum b_n$: $\sum\limits_{n=1}^{10} n + \sum\limits_{n=1}^{10} n^2$

Now, we use the standard summation formulas:

1. Sum of the first $k$ natural numbers: $\sum_{n=1}^{k} n = \frac{k(k+1)}{2}$
2. Sum of the squares of the first $k$ natural numbers: $\sum_{n=1}^{k} n^2 = \frac{k(k+1)(2k+1)}{6}$

For $k=10$: $\sum\limits_{n=1}^{10} n = \frac{10(10+1)}{2} = \frac{10 \cdot 11}{2} = 5 \cdot 11 = 55$ $\sum\limits_{n=1}^{10} n^2 = \frac{10(10+1)(2 \cdot 10+1)}{6} = \frac{10 \cdot 11 \cdot 21}{6}$ $\sum\limits_{n=1}^{10} n^2 = \frac{2310}{6} = 385$

Now, add these two sums: $55 + 385 = 440$

Common Mistake Alert: The original solution had a typo m^2-3x+2=0 instead of m^2-3m+2=0. Always double-check variable names and ensure consistency in mathematical expressions.

Step 6: Final Answer

The value of $\sum\limits_{n=1}^{10} (n^{\alpha}+n^{\beta})$ is $440$.

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Summary and Key Takeaway:

This problem demonstrates the application of Cramer's Rule for determining the nature of solutions for a system of linear equations. The crucial conditions for infinitely many solutions are that the determinant of the coefficient matrix ($\Delta$) must be zero, AND all other related determinants ($\Delta_x, \Delta_y, \Delta_z$) must also be zero. Once the parameters are found, the problem reduces to calculating a sum using standard series formulas. Mastery of determinant calculations and summation formulas is essential for such problems.