This problem requires us to calculate the sum of powers of a matrix $B$, which is defined in terms of a similarity transformation of another matrix $A$. The key to solving this efficiently lies in understanding and applying the properties of similar matrices and their powers.

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1. Key Concept: Similar Matrices and Their Powers

Two square matrices $X$ and $Y$ are said to be similar if there exists an invertible matrix $P$ such that $Y = P X P^{-1}$. A fundamental property of similar matrices is that their powers are also related in a simple way: If $Y = P X P^{-1}$, then for any positive integer $n$, we have $Y^n = P X^n P^{-1}$.

Why this is useful: Calculating powers of a complex matrix $Y$ can be difficult. If $Y$ is similar to a simpler matrix $X$ (e.g., a diagonal matrix, a triangular matrix, or a Jordan block), it becomes much easier to calculate $X^n$ first, and then use the similarity transformation to find $Y^n$.

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2. Identify the Similarity Transformation Components

We are given the matrix $B$ as: $B=\left[\begin{array}{cc}1 & 2 \\ -1 & -1\end{array}\right] A\left[\begin{array}{cc}-1 & -2 \\ 1 & 1\end{array}\right]$ We can identify the invertible matrix $P$ and its inverse $P^{-1}$ by comparing this expression with the definition $B = P A P^{-1}$.

Let's define $P$: $P = \left[\begin{array}{cc}1 & 2 \\ -1 & -1\end{array}\right]$

Now, we need to verify that the third matrix in the expression for $B$ is indeed $P^{-1}$. First, calculate the determinant of $P$: $\det(P) = (1)(-1) - (2)(-1) = -1 + 2 = 1$ Since $\det(P) \neq 0$, $P$ is invertible. The inverse of a $2 \times 2$ matrix $\begin{bmatrix} a & b \\ c & d \end{bmatrix}$ is $\frac{1}{ad-bc} \begin{bmatrix} d & -b \\ -c & a \end{bmatrix}$. So, $P^{-1}$ is: $P^{-1} = \frac{1}{1} \left[\begin{array}{cc}-1 & -2 \\ -(-1) & 1\end{array}\right] = \left[\begin{array}{cc}-1 & -2 \\ 1 & 1\end{array}\right]$ This confirms that the given expression for $B$ is indeed in the form $P A P^{-1}$.

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3. Calculate Powers of Matrix A ($A^n$)

The matrix $A$ is given as: $A=\left[\begin{array}{cc}1 & \frac{1}{51} \\ 0 & 1\end{array}\right]$ This is a special type of upper triangular matrix (a Jordan block with eigenvalue 1). Let's calculate its first few powers to find a pattern. For simplicity, let $k = \frac{1}{51}$. So $A = \begin{bmatrix} 1 & k \\ 0 & 1 \end{bmatrix}$.

• $A^1 = \begin{bmatrix} 1 & k \\ 0 & 1 \end{bmatrix}$
• $A^2 = A \cdot A = \begin{bmatrix} 1 & k \\ 0 & 1 \end{bmatrix} \begin{bmatrix} 1 & k \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} 1 \cdot 1 + k \cdot 0 & 1 \cdot k + k \cdot 1 \\ 0 \cdot 1 + 1 \cdot 0 & 0 \cdot k + 1 \cdot 1 \end{bmatrix} = \begin{bmatrix} 1 & 2k \\ 0 & 1 \end{bmatrix}$
• $A^3 = A^2 \cdot A = \begin{bmatrix} 1 & 2k \\ 0 & 1 \end{bmatrix} \begin{bmatrix} 1 & k \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} 1 \cdot 1 + 2k \cdot 0 & 1 \cdot k + 2k \cdot 1 \\ 0 \cdot 1 + 1 \cdot 0 & 0 \cdot k + 1 \cdot 1 \end{bmatrix} = \begin{bmatrix} 1 & 3k \\ 0 & 1 \end{bmatrix}$

We observe a clear pattern: $A^n = \begin{bmatrix} 1 & nk \\ 0 & 1 \end{bmatrix}$ Substituting back $k = \frac{1}{51}$: $A^n = \left[\begin{array}{cc}1 & \frac{n}{51} \\ 0 & 1\end{array}\right]$

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4. Calculate Powers of Matrix B ($B^n$)

Using the similarity property $B^n = P A^n P^{-1}$: $B^n = \left[\begin{array}{cc}1 & 2 \\ -1 & -1\end{array}\right] \left[\begin{array}{cc}1 & \frac{n}{51} \\ 0 & 1\end{array}\right] \left[\begin{array}{cc}-1 & -2 \\ 1 & 1\end{array}\right]$

First, multiply $P A^n$: $P A^n = \left[\begin{array}{cc}1 & 2 \\ -1 & -1\end{array}\right] \left[\begin{array}{cc}1 & \frac{n}{51} \\ 0 & 1\end{array}\right] = \left[\begin{array}{cc}1 \cdot 1 + 2 \cdot 0 & 1 \cdot \frac{n}{51} + 2 \cdot 1 \\ -1 \cdot 1 + (-1) \cdot 0 & -1 \cdot \frac{n}{51} + (-1) \cdot 1\end{array}\right] = \left[\begin{array}{cc}1 & \frac{n}{51} + 2 \\ -1 & -\frac{n}{51} - 1\end{array}\right]$

Next, multiply the result by $P^{-1}$: $B^n = \left[\begin{array}{cc}1 & \frac{n}{51} + 2 \\ -1 & -\frac{n}{51} - 1\end{array}\right] \left[\begin{array}{cc}-1 & -2 \\ 1 & 1\end{array}\right]$ $B^n = \left[\begin{array}{cc}1(-1) + (\frac{n}{51}+2)(1) & 1(-2) + (\frac{n}{51}+2)(1) \\ -1(-1) + (-\frac{n}{51}-1)(1) & -1(-2) + (-\frac{n}{51}-1)(1)\end{array}\right]$ $B^n = \left[\begin{array}{cc}-1 + \frac{n}{51} + 2 & -2 + \frac{n}{51} + 2 \\ 1 - \frac{n}{51} - 1 & 2 - \frac{n}{51} - 1\end{array}\right]$ $B^n = \left[\begin{array}{cc}1 + \frac{n}{51} & \frac{n}{51} \\ -\frac{n}{51} & 1 - \frac{n}{51}\end{array}\right]$

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5. Calculate the Sum of all Elements of $B^n$

To find the sum of all elements of $B^n$, we add its four entries: $\text{Sum of elements of } B^n = \left(1 + \frac{n}{51}\right) + \left(\frac{n}{51}\right) + \left(-\frac{n}{51}\right) + \left(1 - \frac{n}{51}\right)$ $= 1 + \frac{n}{51} + \frac{n}{51} - \frac{n}{51} + 1 - \frac{n}{51}$ $= (1+1) + \left(\frac{n}{51} + \frac{n}{51} - \frac{n}{51} - \frac{n}{51}\right)$ $= 2 + 0 = 2$ Key Observation: The sum of all elements of $B^n$ is consistently $2$ for any positive integer $n$. This greatly simplifies the final summation.

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6. Calculate the Sum of all Elements of the Matrix $\sum_{n=1}^{50} B^n$

Let $S_{total}$ be the sum of all elements of the matrix $\sum_{n=1}^{50} B^n$. The sum of elements of a sum of matrices is equal to the sum of the sums of elements of individual matrices. So, $S_{total} = \sum_{n=1}^{50} (\text{Sum of elements of } B^n)$. Since we found that the sum of elements of $B^n$ is $2$ for every $n$: $S_{total} = \sum_{n=1}^{50} 2$ $S_{total} = 50 \times 2$ $S_{total} = 100$

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Alternative Method for Step 6: Sum the Matrices First

We could also first sum the matrices $\sum_{n=1}^{50} B^n$ and then find the sum of elements. We know that $\sum_{n=1}^{50} B^n = P \left(\sum_{n=1}^{50} A^n\right) P^{-1}$. Let $S_A = \sum_{n=1}^{50} A^n$. $S_A = \sum_{n=1}^{50} \left[\begin{array}{cc}1 & \frac{n}{51} \\ 0 & 1\end{array}\right] = \left[\begin{array}{cc}\sum_{n=1}^{50} 1 & \sum_{n=1}^{50} \frac{n}{51} \\ \sum_{n=1}^{50} 0 & \sum_{n=1}^{50} 1\end{array}\right]$ We have:

• $\sum_{n=1}^{50} 1 = 50$
• $\sum_{n=1}^{50} \frac{n}{51} = \frac{1}{51} \sum_{n=1}^{50} n = \frac{1}{51} \frac{50(50+1)}{2} = \frac{1}{51} \frac{50 \cdot 51}{2} = \frac{50}{2} = 25$ So, $S_A = \left[\begin{array}{cc}50 & 25 \\ 0 & 50\end{array}\right]$.

Now, calculate $P S_A P^{-1}$: $S_B = \sum_{n=1}^{50} B^n = \left[\begin{array}{cc}1 & 2 \\ -1 & -1\end{array}\right] \left[\begin{array}{cc}50 & 25 \\ 0 & 50\end{array}\right] \left[\begin{array}{cc}-1 & -2 \\ 1 & 1\end{array}\right]$ First, $P S_A$: $P S_A = \left[\begin{array}{cc}1 & 2 \\ -1 & -1\end{array}\right] \left[\begin{array}{cc}50 & 25 \\ 0 & 50\end{array}\right] = \left[\begin{array}{cc}1 \cdot 50 + 2 \cdot 0 & 1 \cdot 25 + 2 \cdot 50 \\ -1 \cdot 50 + (-1) \cdot 0 & -1 \cdot 25 + (-1) \cdot 50\end{array}\right] = \left[\begin{array}{cc}50 & 125 \\ -50 & -75\end{array}\right]$ Next, $(P S_A) P^{-1}$: $S_B = \left[\begin{array}{cc}50 & 125 \\ -50 & -75\end{array}\right] \left[\begin{array}{cc}-1 & -2 \\ 1 & 1\end{array}\right]$ $S_B = \left[\begin{array}{cc}50(-1) + 125(1) & 50(-2) + 125(1) \\ -50(-1) + (-75)(1) & -50(-2) + (-75)(1)\end{array}\right]$ $S_B = \left[\begin{array}{cc}-50 + 125 & -100 + 125 \\ 50 - 75 & 100 - 75\end{array}\right]$ $S_B = \left[\begin{array}{cc}75 & 25 \\ -25 & 25\end{array}\right]$ Finally, the sum of all elements of $S_B$: $75 + 25 + (-25) + 25 = 100$ Both methods yield the same result.

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Conclusion

The sum of all the elements of the matrix $\sum_{n=1}^{50} B^{n}$ is $\mathbf{100}$.

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Tips and Common Mistakes:

1. Recognize Similar Matrices: Always look for structures like $P X P^{-1}$. This is a common trick in matrix problems to simplify power calculations.
2. Powers of Jordan Blocks: For a matrix of the form $\begin{bmatrix} 1 & k \\ 0 & 1 \end{bmatrix}$, its $n$-th power is $\begin{bmatrix} 1 & nk \\ 0 & 1 \end{bmatrix}$. Memorizing this pattern or being able to quickly derive it saves time.
3. Matrix Multiplication Order: Be careful with the order of matrix multiplication, as it is not commutative ($AB \neq BA$). Always perform multiplications from left to right or group them appropriately.
4. Sum of Elements Property: The sum of elements of a matrix $M$ can be conveniently represented as $\mathbf{1}^T M \mathbf{1}$, where $\mathbf{1}$ is a column vector of all ones. This can sometimes simplify calculations, especially when dealing with products of matrices where some intermediate results might become zero, as seen in the thought process: $\mathbf{1}^T P = \begin{bmatrix} 0 & 1 \end{bmatrix}$.
5. Arithmetic Precision: Double-check all arithmetic, especially when dealing with fractions and negative numbers. A small calculation error can propagate and lead to an incorrect final answer.

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Key Takeaway: When dealing with powers of a matrix $B$ defined by a similarity transformation $B = P A P^{-1}$, it is almost always beneficial to first calculate $A^n$ and then use the property $B^n = P A^n P^{-1}$. This strategy often transforms a complex problem into a series of manageable matrix multiplications and pattern recognition tasks.