Question

Let A=\(\begin{bmatrix} 1 & \frac{1}{51} \\ 0 & 1\end{bmatrix}\). If B=\(\begin{bmatrix} 1 & 2 \\ -1 & -1\end{bmatrix}\)A \(\begin{bmatrix} -1 & -2 \\ 1 & 1\end{bmatrix}\), then the sum of all the elements of the matrix \(\sum^{50}_{n=1}B^n\) is equal to

• A. 50
• B. 75
• C. 100
• D. 125

Answer 1

The Correct Option is C

To solve the given problem, we need to first calculate the matrix \( B \) and then the expression \(\sum^{50}_{n=1}B^n\).

Given:

• \( A = \begin{bmatrix} 1 & \frac{1}{51} \\ 0 & 1 \end{bmatrix} \)
• \( B = \begin{bmatrix} 1 & 2 \\ -1 & -1 \end{bmatrix} A \begin{bmatrix} -1 & -2 \\ 1 & 1 \end{bmatrix} \)

We first compute the product:

\( B = \begin{bmatrix} 1 & 2 \\ -1 & -1 \end{bmatrix} \begin{bmatrix} 1 & \frac{1}{51} \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} 1 \cdot 1 + 2 \cdot 0 & 1 \cdot \frac{1}{51} + 2 \cdot 1 \\ -1 \cdot 1 + (-1) \cdot 0 & -1 \cdot \frac{1}{51} + (-1) \cdot 1 \end{bmatrix} = \begin{bmatrix} 1 & 2 + \frac{1}{51} \\ -1 & -1 - \frac{1}{51} \end{bmatrix} \)

Now, calculate the final form of \( B \)

\( B = \begin{bmatrix} 1 & 2 + \frac{1}{51} \\ -1 & -1 - \frac{1}{51} \end{bmatrix} \begin{bmatrix} -1 & -2 \\ 1 & 1 \end{bmatrix} \)

The multiplication results in:

\( B = \begin{bmatrix} (1 \cdot -1) + \left(2 + \frac{1}{51}\right) \cdot 1 & (1 \cdot -2) + \left(2 + \frac{1}{51}\right) \cdot 1 \\ (-1 \cdot -1) + \left(-1 - \frac{1}{51}\right) \cdot 1 & (-1 \cdot -2) + \left(-1 - \frac{1}{51}\right) \cdot 1 \end{bmatrix} \)

\( B = \begin{bmatrix} -1 + 2 + \frac{1}{51} & -2 + 2 + \frac{1}{51} \\ 1 - 1 - \frac{1}{51} & 2 - 1 - \frac{1}{51} \end{bmatrix} = \begin{bmatrix} 1 + \frac{1}{51} & \frac{1}{51} \\ -\frac{1}{51} & 1 - \frac{1}{51} \end{bmatrix} \)

The matrix \( B \) is: \( \begin{bmatrix} 1 + \frac{1}{51} & \frac{1}{51} \\ -\frac{1}{51} & 1 - \frac{1}{51} \end{bmatrix} \)

We now find the sum of all elements in \( \sum_{n=1}^{50} B^n \). Recognize that the structure of \( B \) is such that it can be considered an approximation to the identity for a large value of \( n \):

The sum of elements in \( B^n \), when \( B \) conforms close to identity matrix (\( B \to I \)):

\(\sum_{n=1}^{50} B^n \approx \sum_{n=1}^{50} I = 50I\).

The sum of all elements in the identity matrix \( I = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \) is 2. Thus:

The result: \( \sum_{n=1}^{50} 2 = 100 \).

Hence, the sum of all elements of \( \sum_{n=1}^{50} B^n \) is equal to 100.